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\date{}
\author{V. Manuilov and K. Thomsen}
\title{E-theory is stabilized KK-theory}


\begin{document}

\maketitle

 \begin{abstract} Let $A$ and $B$ be $C^*$-algebras, $A$ separable, and $B$ $\sigma$-unital and stable. It is shown that there are natural isomorphisms
$$
E(A,B) = KK(SA,Q(B)) = [SA,Q(B) \otimes \mathbb K],
$$
where $Q(B) = M(B)/B$ is the generalized Calkin algebra and $\mathbb K$ denotes the $C^*$-algebra of compact operators of an infinite dimensional separable Hilbert space.

\end{abstract}

\section{Introduction}\label{s1}


Connes and Higson have introduced a variant of Kasparov's KK-theory, \cite{CH}, \cite{K}, called $E$-theory, originally with the purpose of realizing a theory developed by Higson, \cite{H2}, which is half-exact with respect to arbitrary extensions of (separable) $C^*$-algebras. E-theory relates directly to Kasparov's theory via a natural map
$$
KK(A,B) \to E(A,B),
$$
which is an isomorphism when $A$ is nuclear, but which is not always injective; reflecting the fact that KK-theory is not half-exact in general, \cite{S}. By specializing, this map gives us also a natural map
$$
KK(SA,Q(B \otimes \mathbb K)) \to E(SA, Q(B \otimes \mathbb K)),
$$
where $SA = C_0(0,1) \otimes A$ is the suspension of $A$, and $Q(B \otimes \mathbb K) = M(B\otimes \mathbb K)/B\otimes \mathbb K$ is the generalized Calkin algebra, or 'corona' algebra, of the stabilized $C^*$-algebra $B\otimes \mathbb K$. Thanks to the unrestricted excision properties of E-theory, combined with Bott-periodicity and stability, there is a natural isomorphism $E(SA,Q(B \otimes \mathbb K)) = E(A,B)$, and we have therefore a natural map
\begin{equation}\label{1!}
KK(SA,Q(B \otimes \mathbb K)) \to E(A, B) .
\end{equation}
The main result of the present paper is that this is an isomorphism, when $A$ is separable and $B$ $\sigma$-unital. The method of proof is easily explained. In \cite{MT3} we introduced a method which produces an genuine $*$-homomorphism $\tilde{\varphi}: A \to Q(B \otimes \mathbb K)$ out of an asymptotic homomorphism $\varphi = \left( \varphi_t\right)_{t \in [1,\infty)} : A \to Q(B)$. We will show that this construction gives rise to a group homomorphism 
$$
[[SA, Q(B) \otimes \mathbb K]] \to [SA, Q(B \otimes \mathbb K) \otimes \mathbb K] .
$$
Composed with the obvious map $[SA, Q(B \otimes \mathbb K) \otimes \mathbb K] \to KK(SA, Q(B \otimes \mathbb K))$, we obtain a natural map $E(A,B) = [[SA, Q(B) \otimes \mathbb K]] \to KK(SA, Q(B \otimes \mathbb K))$, which we show is an inverse for the map (\ref{1!}). As a bi-product of the proof we also obtain the description of E-theory as homotopy classes of $*$-homomorphisms spelled out in the abstract.

Despite the simplicity in the idea of proof, the actual realization of the approach is quite technical, and it will occupy the remaining part of the paper.  

It must be pointed out that an alternative translation between KK-theory and E-theory was obtained by the second-named author in \cite{T}; specifically, it was shown that
$$
E(A,B) = KK(A, C_b([1,\infty), B \otimes \mathbb K)/C_0([1,\infty), B \otimes \mathbb K)),
$$
under the same assumptions on $A$ and $B$ as here, i.e. separability of $A$ and $\sigma$-unitality of $B$. And in \cite{MT3} we came very close to the conclusion that $E(A,B) = [SA, Q(B) \otimes \mathbb K]$, in that we proved the equality $E(A,B) = \varinjlim_n [SA, Q(B \otimes \mathbb K) \otimes M_n]$. By combining this with the result in the present paper we see that $\varinjlim_n [SA, Q(B \otimes \mathbb K) \otimes M_n] = [SA, Q(B \otimes \mathbb K) \otimes \mathbb K]$. This equality seems so plausible that one is tempted to write down a direct proof. However, we are unaware of any proof which avoids the use of asymptotic homomorphisms. 

\section{Fundamental notation and terminology}\label{s2}

Let $A$ and $B$ be $C^*$-algebras, $M(B)$ the multiplier algebra of $B$ and $q_B : M(B) \to M(B)/B = Q(B)$ the quotient map. An asymptotic homomorphism $\varphi = (\varphi_t)_{t\in[1,\infty)}:A\to Q(B)$ will also be called \emph{an asymptotic extension} of $A$ by $B$. Two asymptotic extensions $\varphi, \psi :A\to Q(B)$ are \emph{unitarily equivalent} when there is a normcontinuous path $u_t,t \in [1, \infty)$, of unitaries in $M(B)$ such that $\lim_{t \to \infty} \Ad q_B(u_t) \circ \psi_t(a) - \varphi_t(a) = 0$ for all $a \in A$. An asymptotic extension $\varphi: A \to Q(B)$ is
said to be {\it split} when there exists an
asymptotic homomorphism $(\lambda_t)_{t\in[1,\infty)}:A\to M(B)$ such that
$\lim_{t \to \infty} q_B\circ \lambda_t(a) -\varphi_t(a) = 0$ for all $a \in A$. (We will show along the way that this is equivalent to the apparently stronger condition that there exists an asymptotic homomorphism $(\lambda'_t)_{t\in[1,\infty)}:A\to M(B)$ such that $q_B\circ \lambda'_t(a) = \varphi_t(a)$ for all $a,t$, provided only that $\varphi$ is equi-continuous, cf. Lemma \ref{crux2}.) An asymptotic homomorphism $\varphi =  (\varphi_t)_{t\in[1,\infty)}: A \to B$ is \emph{equi-continuous} when the family of maps $\varphi_t : A \to B, t \in [1,\infty)$, is an equi-continuous family, i.e. has the property that for all $a \in A$ and all $\epsilon > 0$ there is a $\delta > 0$ such that $\|\varphi_t(a) - \varphi_t(b)\| \leq \epsilon$ for all $t \in [1,\infty)$ and all $b \in A$ with $\|a-b\| \leq \delta$. It is well-known that any asymptotic homomorphism $\varphi : A \to B$ is asymptotically identical with an equi-continuous asymptotic homomorphism $\psi$, in the sense that $\lim_{t \to \infty} \varphi_t(a) - \psi_t(a) = 0$ for all $a \in A$. Thanks to this it is almost always possible to restrict the attention to asymptotic homomorphisms that are equi-continuous. This is very useful, also for asymptotic extensions, and to make it even more useful we make the following observation.


\begin{lem}\label{obs} Let $\varphi : A \to Q(B)$ be an equi-continuous asymptotic extension. It follows that there is an equi-continuous family of maps $\hat{\varphi}_t : A \to M(B), t \in [1,\infty)$, such that $q_B \circ \hat{\varphi}_t = \varphi_t$ for all $t \in [1,\infty)$ and $\sup_{t \in [1,\infty)} \|\hat{\varphi}_t(a)\| < \infty$ for all $a \in A$.
\end{lem}
\begin{proof} Since $\varphi$ is an asymptotic homomorphism, $\limsup_{t \to \infty} \|\varphi_t(a)\| \leq \|a\|$ for all $a \in A$. Since $\varphi$ is equi-continuous, we can define a continuous map $\Phi : A \to C_b([1,\infty), Q(B))$ such that $\Phi(a)(t) = \varphi_t(a)$. There is a surjective $*$-homomorphism $Q : C_b([1,\infty), M(B)) \to C_b([1,\infty), Q(B))$ such that $Q(f)(t) = q_B(f(t))$. By the Bartle-Graves selection theorem there is a continuous section $S : C_b([1,\infty), Q(B)) \to  C_b([1,\infty), M(B))$ for $Q$, and we set $\hat{\varphi}_t(a) = \left(S \circ \Phi(a)\right)(t)$. 
\qed
\end{proof}

We will refer to $\hat{\varphi} = \left( \hat{\varphi}_t\right)_{t \in [1,\infty)}$ as \emph{an equi-continuous lift} of $\varphi$. Recall, cf. \cite{MT1}, that a \emph{discretization}, $\left( \varphi_{t_n}\right)_{n =1}^{\infty}$, of an asymptotic homomorphism $\varphi = \left( \varphi_t\right)_{t \in [1,\infty)} : A \to B$ is given by a sequence $t_1 < t_2 < t_3 < \dots$ in $[1,\infty)$ such that
\begin{enumerate}
\item[d1)] $\lim_{n \to \infty} t_n = \infty$, and
\item[d2)] $\lim_{n \to \infty} \sup_{t \in [t_n, t_{n+1}]} \|\varphi_t(a) - \varphi_{t_n}(a)\| = 0$ for all $a \in A$.
\end{enumerate}
When $A$ is separable discretizations always exist, cf. \cite{MT1}.





For any $C^*$-algebra $A$ we denote by $SA$ the suspension of $A$, i.e. $SA = C_0(0,1) \otimes A$, and we denote by $IA$ the $C^*$-algebra $C[0,1] \otimes A$. For any $s \in [0,1]$, $\ev_s : IA \to A$ is then the $*$-homomorphism given by evaluation at $s$, i.e. $\ev_s(f) = f(s)$. 


Concerning extensions we will adopt the terminology from \cite{MT3}; in particular, we will call an extension $\varphi : A \to Q(B)$ \emph{asymptotically split} when there is an asymptotic homomorphism $\pi : A \to M(B)$ such that $\varphi(a) = q_B \circ \pi_t(a)$ for all $a,t$. Concerning asymptotic homomorphism we shall use the standard notation and terminology; in particular $[[A,B]]$ will denote the homotopy classes of asymptotic homomorphisms from $A$ to $B$, and $[[A,B]]_{cp}$ will denote the homotopy classes of completely positive asymptotic homomorphsims from $A$ to $B$, where we call an asymptotic homomorphism $\varphi = \left( \varphi_t\right)_{t \in [1,\infty)} : A \to B$ \emph{completely positive} when the individual maps $\varphi_t$ are all completely positive contractions. It is fundamental to our approach that KK-theory can be realized by using completely positive asymptotic homomorphisms, specifically that $KK(A,B) = [[SA, SB \otimes \mathbb K]]_{cp}$, cf. \cite{H-LT}. 



%, and if also $\varphi$ is equi-continuous and $\left(\hat{\varphi}_t\right)_{t \in [1,\infty)}$ is an equi-continuous lift of $\varphi$, we can arrange that
%\begin{enumerate}
%\item[d3)] $\lim_{n \to \infty} \sup_{t \in [t_n, t_{n+1}]} \|\hat{\varphi}_t(a) - \hat{\varphi}_{t_n}(a)\| = 0$ for all $a \in A$.
%\end{enumerate}
 %An asymptotic extension $\varphi : A \to Q(B)$ is {\it semi-invertible} when there exists another asymptotic extension
%$\psi = (\psi_t)_{t\in[1,\infty)}:A\to Q(B)$ such that the  



%$\varphi \oplus\psi$ is split. Two semi-invertible asymptotic extensions
%$\tau,\sigma D \to Q(B)$ are said to be \emph{stably equivalent} when there exists a split asymptotic
%extension $\lambda : D \to Q(B)$ such that
%$\tau\oplus\lambda$ and $\sigma\oplus\lambda$ are unitarily
%equivalent. To see that this is an equivalence relation we need a lemma. 




\section{The basic construction}\label{s3} 






Let $D$ and $E$ be $C^*$-algebras, $D$ separable, $E$ $\sigma$-unital. Let $(\varphi_t)_{t\in[1,\infty)}:D\to Q(E)$ be an equi-continuous asymptotic extension. We will construct an extension $\varphi^f$ of $D$ by $E$ out 
of $\varphi$. The construction was introduced and used in \cite{MT3}, and uses Voiculescu's tri-diagonal
projection trick from \cite{V}. 
Let $b$ be a strictly positive element in $E$ of norm $\leq 1$. A {\it unit sequence} (cf.
\cite{MT3})
in $E$ is a sequence $\{u_n\}_{n=0}^{\infty} \subseteq
E$ such that
\begin{enumerate}
\item[u1)] there is a continuous function $f_n : [0,1] \to [0,1]$ which
is zero in a neighbourhood of $0$ and $u_n = f_n(b)$, and
\item[u2)] $u_{n+1}u_n = u_n $ for all $n$, and
\item[u3)] $\lim_{n \to \infty} u_n x = x$ for any $x \in E$.
\end{enumerate}
Unit sequences exist by elementary spectral theory. Given a unit sequence $\{u_n\}$ we set $\Delta_0 = \sqrt{u_0}$ and $\Delta_j = \sqrt{u_j - u_{j-1}}, j \geq 1$. Note that u2) implies that
\begin{equation}\label{BB}
\Delta_i\Delta_j = 0, \ |i-j|\geq 2.
\end{equation} 


\begin{lem}\label{E1} For any norm-bounded sequence $\{m_j\} \subseteq M(E)$, and any $k \in \mathbb N$, the sum $\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k}$ converges in the strict topology to an element of $M(E)$, 
$$
\|\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k} \| \leq \sup_j \|m_j\| 
$$
and
$$
\|q_E\left(\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k}\right)\| \leq \limsup_n \|m_n\|.
$$
In particular, $\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k} \in E$ when $\lim_{j \to \infty} \|m_j\| = 0$. 
\end{lem}
\begin{proof} For any $e \in E$ and any $N < M$ in $\mathbb N$ we have the estimates
$$
\|\sum_{j=N}^M \Delta_jm_j\Delta_{j+k}e\|^2 \leq \|\sum_{j=N}^M \Delta_jm_jm_j^*\Delta_{j}\|\|\sum_{j=N}^M e^*\Delta_{j+k}^2e\| \leq \|\sum_{j=N}^M e^*\Delta_{j+k}^2e\| \max_{N \leq i\leq M}\|m_i\|^2, 
$$
and
$$
\|\sum_{j=N}^M \Delta_{j+k}m_j^*\Delta_{j}e\|^2 \leq \|\sum_{j=N}^M \Delta_{j+k}m_j^*m_j\Delta_{j+k}\|\|\sum_{j=N}^M e^*\Delta_{j}^2e\| \leq \|\sum_{j=N}^M e^*\Delta_{j}^2e\| \max_{N \leq i\leq M}\|m_i\|^2 .
$$
It follows that $\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k}$ converges in the strict topology, as asserted. We can then take $N = 0$ and let $M$ tend to infinity in the first estimate to conclude that
$$
\|\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k} \| \leq \sup_j \|m_j\| . 
$$
If we only let $M$ go to infinity we see that
$$
\|q_E\left(\sum_{j=0}^{\infty} \Delta_j m_j \Delta_{j+k}\right)\| \leq  \|\sum_{j=N}^{\infty} \Delta_j m_j \Delta_{j+k}\| \leq \sup_{i \geq N} \|m_i\|
$$
for all $N$. 


\qed
\end{proof}



Let $\hat{\varphi}_t : D \to M(E)$ be an equi-continuous lift of $\varphi$. We will say that the pair 
$$\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}\right)
$$ 
is a \emph{compatible pair} for $\varphi$ when
\begin{enumerate}
\item[p1)]  $\lim_{n \to \infty} \sup_{t \in [1, n+3]} \|u_n\hat{\varphi}_t(a) - \hat{\varphi}_{t}(a)u_n\| = 0$
\end{enumerate}
for all $a \in D$.


\begin{lem}\label{ex} For any unit sequence $\{v_n\}_{n = 0}^{\infty}$ in $E$ there is a unit sequence $\{u_n\}_{n =0}^{\infty}$ such that $u_n$ is in the convex hull $\co \{v_i : i \geq n\}$ for all $n$, and  
$$\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_{n}\}_{n = 0}^{\infty}\right)
$$ 
is a compatible pair.
\end{lem}
\begin{proof} Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \dots$ be a sequence of finite sets with dense union in $D$. Since $\hat{\varphi}_t, t\in [1,\infty)$, is an equi-continuous family of maps it suffices to construct a sequence $\{u_n\}_{n =0}^{\infty}$ such that
\begin{enumerate}
\item[1)] $\|u_{n} \hat{\varphi}_{t}(a) - \hat{\varphi}_{t}(a)u_{n}\| \leq \frac{1}{n}$ for all $ a \in F_n$ and all $t \in [1,n+1]$,
\item[2)] $u_{n}u_{n -1} = u_{n-1}, n \geq 1$,
\item[3)] $u_n \in \co\{v_i : i \geq n\}$,
\end{enumerate}
for all $n$. Since the argument for the induction start is contained in the argument for the induction step, we will only give the latter. So assume that $u_{n-1}$ has been found. Since $u_{n-1} \in \co \{v_i\}$, there is an $N \geq n$ such that $v_i u_{n-1} = u_{n-1}$ for all $i \geq N$. Since $\co \{v_i : i \geq N\}$ is a convex approximate unit in $E$, the existence of $u_{n}$ follows from \cite{A} or \cite{P}.
\qed
\end{proof}   



\begin{lem}\label{ex0} Let $\left(\hat{\varphi}_t\right)_{t \in [1,\infty)}$ be an equi-continuous lift of $\varphi$. There exists a sequence $t_1 < t_2 < t_3 < \dots$ in $[1,\infty)$ such that $\{\varphi_{t_n}\}_{n = 1}^{\infty}$ is a discretization of $\varphi$ and 
\begin{enumerate}
\item[t1)] $\lim_{n \to \infty} \sup_{t \in [t_n,t_{n+1}]} \|\hat{\varphi}_t(a) - \hat{\varphi}_{t_n}(a)\| = 0$ for all $a \in D$, and
\item[t2)] $t_n \leq n$ for all $n \in \mathbb N$.
\end{enumerate}
\end{lem}
\begin{proof} Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \dots$ be a sequence of finite sets with dense union in $D$. By uniform continuity there are $N_n$'s in $\mathbb N$, $N_n \geq 1$, such that $s,t \in [n,n+1], |s -t| \leq \frac{1}{N_n}  \Rightarrow \|\hat{\varphi}_t(a)- \hat{\varphi}_s(a)\| \leq \frac{1}{n}$ for all $a \in F_n$. Set $N_0 = 0$ and $t_j = k + \frac{j -  N_0 + N_1 + \dots + N_k}{N_{k+1}}$ when $N_0 + N_1 + \dots + N_k \leq j \leq  N_0 + N_1 + \dots + N_k + N_{k+1}$. Then $\lim_{n \to \infty} \sup_{t \in [t_n, t_{n+1}]} \|\hat{\varphi}_t(a) - \hat{\varphi}_{t_n}(a)\| = 0$ for all $a \in \bigcup_n F_n$. By equi-continuity of $\{\hat{\varphi}_t\}_{t \in [1,\infty)}$, the same must hold for all $a \in D$. Since $\lim_{n \to \infty} t_n = \infty$, $\{\varphi_{t_n}\}_{n =1}^{\infty}$ is a discretization of $\varphi$. Note that $t_n \leq n$ for all $n$ by construction.
\qed
\end{proof}   




\begin{lem}\label{E5}  Let $\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}\right)$ be a compatible pair for $\varphi$. There is then a sequence $n_0 < n_1 < n_2 < \dots $ in $\mathbb N$ such that
\begin{enumerate}
\item[t3)] $n_i - n_{i-1} > i + 1$ for all $i \geq 1$,
\item[t4)] $\lim_{i \to \infty} \sup_{j \geq n_i} \sup_{t \in [1,i+3]} \|\left(1-u_j\right)\left( \hat{\varphi}_t(a)\hat{\varphi}_t(b) -  \hat{\varphi}_t(ab)\right)\| -  \|\varphi_t(a)\varphi_t(b) - \varphi_t(ab)\| = 0$,
\item[t5)] $ \lim_{i \to \infty} \sup_{j \geq n_i} \sup_{t \in [1, i+3]} \|\left(1 -u_j\right) \left( \hat{\varphi}_t(a) + \lambda \hat{\varphi}_t(b) - \hat{\varphi}_t(a + \lambda b)\right)\| -  \|\varphi_t(a) + \lambda \varphi_t(b) - \varphi_t(a + \lambda b)\| = 0$, 
\item[t6)]  $ \lim_{i \to \infty} \sup_{j \geq n_i} \sup_{t \in [1, i+3]} \|\left(1-u_j\right)\left(\hat{\varphi}_t(a^*) - \hat{\varphi}_t(a)^*\right)\| -  \|\varphi_t(a^*) - \varphi_t(a)^*\| = 0$ 
\end{enumerate}
for all $a,b \in D$ and all $\lambda \in \mathbb C$. 
\end{lem}
\begin{proof} Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \dots$ be a sequence of finite sets with dense union in $D$, and let $K_i$ be the compact subset of $M(E)$ which is the union of $\{ \hat{\varphi}_t(a)\hat{\varphi}_t(b) -  \hat{\varphi}_t(ab) : t \in [1,i+3], a,b \in F_i\}$, $\{ \hat{\varphi}_t(a) + \lambda \hat{\varphi}_t(b) - \hat{\varphi}_t(a + \lambda b) : t \in [1,i+3], a,b \in F_i, \lambda \in \mathbb C, |\lambda| \leq i\}$, and $\{\hat{\varphi}_t(a^*) - \hat{\varphi}_t(a)^* : t \in [1,i+3], a \in F_i\}$. Since $\{u_n\}$ is an approximate unit for $E$ there is an $l_i \in \mathbb N$ so large that 
$$
\|\left(1 - u_j\right) m\| \leq \|q_E(m)\| + \frac{1}{i+1}
$$
for all $m \in K_i$ and all $j \geq l_i$. Set $n_0 = l_0$ and $n_i = \max \{l_i, n_{i-1} + i +2\}, i \geq 1$. Then t3) holds, and t4)-t6) hold for all $a,b \in \bigcup_n F_n$ and all $\lambda \in \mathbb C$. We can therefore conclude by appealing to the equi-continuity of $\hat{\varphi}_t, t \in [1,\infty)$, and $\varphi_t, t \in [1,\infty)$. 



\qed
\end{proof}














Let $\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}\right)$ be a compatible pair for $\varphi$. Let $\{\varphi_{t_n}\}_{n = 1}^{\infty}$ be a discretization of $\{\varphi_t\}_{t \in [1,\infty)}$ such that t1) and t2) hold. Let $\{n_i\}_{i=0}^{\infty}$ be any sequence in $\mathbb N$ such that t3)-t6) hold. In this setting, set 
$$
\Delta_0 = \sqrt{u_{n_0}} 
$$ 
and
$$
\Delta_j = \sqrt{u_{n_j} - u_{n_{j-1}}}, \ \ \ \ \ \ \ j \geq 1.
$$ 




\begin{lem} There is then a $*$-homomorphism $\varphi^f : D \to Q(E)$ such that
$$
\varphi^f(d) = q_E\left(\sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(d) \Delta_j\right)
$$
for all $d \in D$.
\end{lem}
\begin{proof} Since $\{\|\hat{\varphi}_t(d)\| : t \in [1, \infty)\}$ is bounded, we can apply Lemma \ref{E1} to see that $\sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(d) \Delta_j$ converges in the strict topology. To check that $\varphi^f$ is a $*$-homomorphism, we calculate modulo $E$, with the aid of Lemma \ref{E1}: 
\begin{equation*}\label{BB3}
\begin{split}
&  \left(\sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a)  \Delta_j \right) \left(\sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(b) \Delta_j\right) \\
&\\
& = \sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a) \Delta_j^2 \hat{\varphi}_{t_{j+1}}(b) \Delta_j +  \sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a) \Delta_j\Delta_{j+1} \hat{\varphi}_{t_{j+2}}(b) \Delta_{j+1} \\
& \ \ \ \ \ \  \ \  \ \  \ \ \  \ \ \  \    \ \  \ \ \  \ \ \ +  \sum_{j=0}^{\infty} \Delta_{j+1} \hat{\varphi}_{t_{j+2}}(a) \Delta_{j+1}\Delta_j \hat{\varphi}_{t_{j+1}}(b) \Delta_j \ \ \ \ \ \ \ \ \ \ \ \text{(by (\ref{BB}))} \\ 
& \\
& =  \sum_{j=1}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a) \hat{\varphi}_{t_{j+1}}(b) \Delta_j^2  \Delta_j  +  \sum_{j=1}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a) \hat{\varphi}_{t_{j+2}}(b) \Delta_{j+1}\Delta_{j}\Delta_{j+1} \\
& \ \ \ \ \ \  \ \  \ \  \ \ \  \ \ \  \  \ \  \ \  \ \  \ \ \  \ \ \   \ \  \ \ \  \ \ \ +  \sum_{j=1}^{\infty} \Delta_{j} \hat{\varphi}_{t_{j+1}}(a) \hat{\varphi}_{t_{j}}(b) \Delta_{j-1} \Delta_j \Delta_{j-1} \ \ \ \ \ \ \ \ \ \text{(using t2) and p1))} \\
& \\
& =  \sum_{j=1}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a) \hat{\varphi}_{t_{j+1}}(b) \Delta_j^2  \Delta_j  +  \sum_{j=1}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(a) \hat{\varphi}_{t_{j+1}}(b) \Delta_{j+1}\Delta_{j} \Delta_{j+1}\\
& \ \ \ \ \ \  \ \  \ \  \ \ \  \ \ \  \  \ \  \ \  \ \  \ \ \  \ \ \   \ \  \ \ \  \ \ \ +  \sum_{j=1}^{\infty} \Delta_{j} \hat{\varphi}_{t_{j+1}}(a) \hat{\varphi}_{t_{j+1}}(b) \Delta_{j-1} \Delta_j  \Delta_{j-1}\ \ \ \ \ \text{(using t1))} 
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& =  \sum_{j=1}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(ab) \Delta_j^2  \Delta_j  +  \sum_{j=1}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(ab) \Delta_{j+1}\Delta_{j} \Delta_{j+1} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \  \ \  \ \  \ \ \  \ \ \  \  \ \  \ \  \ \  \ \ \  \ \ \   \ \  \ \ \  \ \ \ +  \sum_{j=1}^{\infty} \Delta_{j} \hat{\varphi}_{t_{j+1}}(ab) \Delta_{j-1} \Delta_j  \Delta_{j-1} \ \ \ \ \ \ \ \ \ \text{(using t4))} \\
& \\
& =  \sum_{j=0}^{\infty} \Delta_j \hat{\varphi}_{t_{j+1}}(ab)  \Delta_j
 \ \ \  \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \text{(since $(\Delta_j^2 + \Delta_{j+1}^2 + \Delta_{j-1}^2)\Delta_j = \Delta_j,  j \geq 2$)}.  
\end{split}
\end{equation*}
Hence $\varphi^f(ab) = \varphi^f(a)\varphi^f(b)$ for all $a,b \in D$. The other conditions for $\varphi^f$ to be a $*$-homomorphism (i.e. linearity and self-adjointness) are established in the same way.  





\qed
\end{proof}


We will refer to $\varphi^f$ as \emph{a folding} of $\varphi$. The quadruple 
$$\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}, \{n_i\}_{i=0}^{\infty}, \{t_i\}_{i=1}^{\infty}\right)
$$ 
which goes into the construction of $\varphi^f$ will be called \emph{the folding data}.








 At first sight it is not clear how much a folding depends on the folding data chosen for its construction. Furthermore, it is not difficult to vary the construction in different ways, for example by omitting condition t3). A major part of the proof consists of showing that in the appropriate setting and modulo the appropriate equivalence relations the construction is in fact independent of all choices made. Specifically, we will show that it gives rise to a group homomorphism
$$
f : [[SD,Q(E) \otimes \mathbb K]]  \to [SD, Q(E \otimes \mathbb K)\otimes \mathbb K].
$$
For this purpose, condition t3) will come in handy.




\section{Preparing the ground}\label{s4}


Let $A$ and $B$ be $C^*$-algebras, $A$ separable, $B$ $\sigma$-unital. We say that an asymptotic homomorphism $\varphi = (\varphi_t)_{t \in [1, \infty)} : A \to B$ is \emph{uniformly continuous} when the function $[1, \infty) \ni t \mapsto \varphi_t(a)$ is uniformly continuous for all $a \in A$, i.e. when the following holds :
$$
\forall a \in A \ \forall \epsilon > 0 \ \exists \delta > 0 \ : \ |t-s| \leq \delta \ \Rightarrow \ \|\varphi_t(a) - \varphi_s(a)\| \leq \epsilon .
$$ 

\begin{lem}\label{000} Let $\varphi, \psi : A \to B$ be asymptotic homomorphisms such that $\lim_{t \to \infty} \varphi_t(a) - \psi_t(a) = 0$ for all $a \in A$. If $\psi$ is uniformly continuous then so is $\varphi$.
\end{lem}
\begin{proof} Left to the reader.
\qed
\end{proof}


\begin{lem}\label{repar1} Let $\varphi : A \to B$ be an asymptotic homomorphism. There exists a continuous increasing function $r : [1, \infty) \to [1, \infty)$ such that $\lim_{t \to \infty}r(t) = \infty$ and $\{\varphi_{r(t)}\}_{t \in [1, \infty)} : A \to B$ is uniformly continuous.
\end{lem}  


\begin{proof} By Lemma \ref{000} we may assume that $\varphi$ is equi-continuous. Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \cdots $ be a sequence of finite sets with dense union in $A$. There is a $\delta_n > 0$ so small that $\|\varphi_t(a) - \varphi_s(a)\| \leq \frac{1}{n}$ for all $a \in F_n$ and all $t,s \in [1, n]$ with $|t-s| \leq \delta_n$. For each $n$ we choose $k_n \in \Bbb N, k_n \geq 1$, so large that $\delta_{n+4} \geq \frac{1}{k_n}$. Set $N_j = 1 + \sum_{n=1}^j k_n , j = 1,2, \cdots $, and $N_0 = 1$. Define $r : [1, \infty) \to [1, \infty)$ such that
$$
r(t) = j + \frac{t- N_{j-1}}{k_j} \ , \ t \in [N_{j-1}, N_j] .
$$ 
To check that $\{\varphi_{r(t)}\}_{t \in [1, \infty)}$ is uniformly continuous, let $a \in A$ and $\epsilon >0$. Since $\varphi$ is equi-continuous there is an $n \in \mathbb N$ and an element $b \in F_n$ such that $\frac{1}{n} \leq \frac{\epsilon}{3}$ and $\sup_{t \in [1, \infty)} \|\varphi_t(a) - \varphi_t(b)\| \leq \frac{\epsilon}{3}$. For $x \geq N_n$ and $|x-y| \leq 1$, we have that $|r(x)-r(y)| \leq \frac{1}{k_j} \leq \delta_{j+4}$ when $x \in [N_j,N_{j+1}]$. Since $r(x),r(y) \in [1,j+4]$ in this case, we see that $\|\varphi_{r(x)}(b) - \varphi_{r(y)}(b)\| \leq \frac{1}{n+4} \leq \frac{\epsilon}{3}$ for all $x,y$ with $x \geq N_n$ and $|x-y| \leq 1$. Choose $\delta \in ]0,1]$ so small that $\|\varphi_{r(x)}(b) - \varphi_{r(y)}(b)\| \leq \frac{\epsilon}{3}$ for all $x,y \in [1,N_n + 1]$ with $|x-y| \leq \delta$. Then $\|\varphi_{r(x)}(b) - \varphi_{r(y)}(b)\| \leq \frac{\epsilon}{3}$ for all $x,y \in [1,\infty)$ with $|x-y| \leq \delta$. It follows that $\|\varphi_{r(x)}(a) - \varphi_{r(y)}(a)\| \leq \epsilon$ for all $x,y \in [1,\infty)$ with $|x-y| \leq \delta$. 



\qed

\end{proof}

In the following we will write $\sim$ between two expressions that depend on $t$ when their difference tends to $0$ in norm as $t$ tends to infinity.
   

\begin{lem}\label{crux2} Let $\lambda : A \to Q(B)$ be an equi-continuous asymptotic extension. Assume that $\lambda$ is homotopic to a split asymptotic extension, i.e. that there is a split asymptotic extension defining the same class in $[[A,Q(B)]]$ as $\lambda$. It follows that there is an equi-continuous asymptotic homomorphism $\delta : A \to M(B)$ such that $\lambda_t = q_B \circ \delta_t$ for all $t$.
\end{lem}
\begin{proof} The proof is based on an idea of Voiculescu, cf. \cite{V}, and is a refinement of the proof of Lemma 2.1 in \cite{MT3}. By assumption there is an asymptotic homomorphism $\psi' : A \to M(B)$ such that $\lambda$ is homotopic to the asymptotic extension $q_B \circ \psi'_t,t \in [1,\infty)$. Let $\psi : A \to M(B)$ be an equi-continuous asymptotic homomorphism such that $\lim_{t \to \infty} \psi'_t(a) - \psi_t(a) = 0$ for all $a \in A$. There is then an equi-continuous asymptotic homomorphism $\Phi : A \to IQ(B)$ which is a homotopy connecting $\lambda$ and $q_B \circ \psi$, i.e. $\ev_0 \circ \Phi_t = \lambda_t$ and $\ev_1 \circ \Phi_t = q_B \circ \psi_t$ for all $t$. Define $\Lambda : A \to C_b([1,\infty), IQ(B))$ by $\Lambda(a)(t) = \Phi_t(a)$. $\Lambda$ is continuous since $\Phi$ is equi-continuous. Let $Q : C_b([1,\infty), IM(B)) \to C_b([1,\infty),IQ(B))$ be the surjective $*$-homomorphism induced by $\id_{C[0,1]} \otimes q_B : IM(B) \to IQ(B)$, i.e. $Q(f)(t)  = \left(\id_{C[0,1]} \otimes q_B\right)(f(t)), t \in [1,\infty)$. By the Bartle-Graves selection theorem there is a continuous section $S$ for $Q$. We set
$$
\mu^s_t(a) = \ev_s\left(S \circ \Lambda(a)(t)\right)
$$
for all $a \in A$, all $t \in [1,\infty)$ and all $s \in [0,1]$. Then
$$
q_B\left( \mu^0_t(a)\right) = \ev_0\left(Q \circ S \circ \Lambda(a)(t)\right) = \lambda_t(a)
$$
for all $a,t$. Hence $\mu^s_t, s \in [0,1], t \in [1,\infty)$, is an equi-continuous family of maps such that 
\begin{enumerate}
\item[1)] $\{\mu_t^1\}_{t \in [1,\infty)}$ is an asymptotic homomorphism, and 
\item[2)] $q_B \circ \mu_t^0(a) = \lambda_t(a)$ for all $a \in A$ and all $t$. 
\end{enumerate}
Furthermore,
\begin{enumerate}
\item[3)] $\lim_{t \to \infty} \sup_{s \in [0,1]} \|q_B \circ \mu^s_t(a^*) - q_B \circ \mu^s_t(a)^*\| = 0$,
\item[4)] $\lim_{t \to \infty} \sup_{s \in [0,1]} \| q_B \circ \mu^s_t(a)  q_B \circ \mu^s_t(b) -  q_B \circ \mu^s_t(ab)\| = 0$, and
\item[5)] $\lim_{t \to \infty} \sup_{s \in [0,1]} \| q_B \circ \mu^s_t(a)  + \lambda q_B \circ \mu^s_t(b) -  q_B \circ \mu^s_t(a +\lambda b)\| = 0$,
\end{enumerate}
for all $a,b \in A$ and all $\lambda \in \mathbb C$.
Choose continuous functions $f_i : [1, \infty) \to [0,1], i = 0,1,2,\cdots$, such that
\begin{enumerate}
\item[6)] $f_0(t) =0$ for all $t \in [1,\infty)$,
\item[7)] $f_n \leq f_{n+1}$ for all $n$,
\item[8)] for each $ n \in \Bbb N$, there is an $m_n \in \Bbb N$ such that $f_i(t) = 1$ for all $i \geq m_n$, and all $t \in [1,n+1]$, 
\item[9)] $\lim_{t \to \infty} \max_i |\mu_t^{f_i(t)}(a) - \mu_t^{f_{i+1}(t)}(a)| = 0$ for all $a \in A$.
\end{enumerate}
The last requirement, 9), can be fulfilled thanks to the separability of $A$ and the equi-continuity of the family $\mu^s_t, s \in [0,1], t \in [1, \infty)$.
Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \cdots $ be an increasing sequence of finite subsets with dense union in $A$. For each $n$, choose $m_n \in \Bbb N$ as in 8). We may assume that $m_{n+1} > m_n$. By general facts on quasi-central approximate units (\cite{A}, \cite{P}) we can choose elements 
$$
X^n_{0}   \geq X^n_{1} \geq X^n_{2} \geq  \cdots    
$$
in $B$ such that $0 \leq X^n_i \leq 1$ for all $i$ and $X^n_i = 0$ for $i \geq  m_n$, and
\begin{enumerate}
\item[i)] $X^n_{i}X^n_{i+1} = X^n_{i+1}$ for all $i$,
\item[ii)] $\|X^n_i x - x\| \leq \frac{1}{n} + \|q_B(x)\|$ for all $i = 0,1,2, \cdots , m_n - 1$, and all $x \in S_n$,
\item[iii)] $\|X^n_i y - y X^n_i\| \leq \frac{1}{n}$ for all $ i$ and all $y \in L_n$,
\end{enumerate}
where $L_n$ and $S_n$ are the compact sets $L_n = \{ \mu_t^s(a) : \ s \in [0,1], \ a \in F_n, \ t \in [1,n +1] \}$ and
\begin{equation*}
\begin{split}
&S_n = \{\mu_t^s(a) + \lambda \mu_t^s(b) - \mu_t^s(a + \lambda b) : \ a,b \in F_n, \ s \in [0,1], \ t \in [1,n+1], \ \lambda \in \mathbb C, |\lambda| \leq n \} \\
& \cup \{\mu_t^s(ab) - \mu_t^s(a)\mu_t^s(b) : \ a,b \in F_n, \ s \in [0,1], \ t \in [1,n+1] \}  \\
& \cup \{\mu_t^s(a^*) - {\mu_t^s(a)}^* : \ a \in F_n, \ s \in [0,1], \ t \in [1, n+1] \}. 
\end{split}
\end{equation*}
Since we choose the $X^n_i$'s recursively we can arrange that $X^{n+1}_i X^n_k = X^n_k$ for all $k$ and all $i$. By connecting first $X^n_0$ to $X^{n+1}_0$ via the straight line between them, then $X^n_{1}$ to $X^{n+1}_{1}$ via a straight line, then $X^n_{2}$ to $X^{n+1}_{2}$ etc., we obtain norm-continuous pathes, $X(t,i), t \in [n,n+1], i = 0,1,2,3, \dots$, in $B$ such that $X(n,i) = X^n_i, \ X(n+1,i) = X^{n+1}_i$ for all $i$ and 
\begin{enumerate}
\item[10)] $X(t,{i})X(t,i+1) = X(t,i+1), \  t \in [n,n+1]$, for all $i$,
\item[11)] $\|X(t,i) x - x\| \leq \frac{1}{n} + \|q_B(x)\|$ for all $i = 0,1,2, \dots , m_n - 1, \ t \in [n,n+1]$ and all $x \in S_n$,
\item[12)] $\|X(t,i) y - y X(t,i)\| \leq \frac{1}{n}$ for all $ i $, all $t \in [n,n+1]$ and all $y \in L_n$.
\end{enumerate}
In addition, $X(t,i) = 0, i \geq m_{n+1}, t \in [n,n+1]$. Set $\Lambda^t_0 = \sqrt{1 - X(t,0)}$, and $\Lambda^t_j = \sqrt{X(t,j-1)-X(t,j)}, j \geq 1$, and define $\delta_t : A \to M(B)$ by
$$
\delta_t(a) = \sum_{j = 0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a) \Lambda^t_j .
$$  
Note that the sum is finite for $t$ in a compact set, and that $t \mapsto \delta_t(a)$ is norm-continuous. Observe also that
\begin{enumerate}
\item[13)] $\Lambda^t_j\Lambda^t_i = 0, |i-j| \geq 2$, for all $t$, and 
\item[14)] $\sum_{j=0}^{\infty} \left(\Lambda^t_j\right)^2 = 1$ for all $t$.
\end{enumerate}
It follows from 6) that $q_B \circ \delta_t(a) = q_B \circ \mu^0_t(a)$ for all $a,t$. Thanks to 2) it now only remains to show that $\delta = \left( \delta_t\right)_{t \in [1, \infty)}$ is an asymptotic homomorphism. We check that it is multiplicative. Because $\{\delta_t\}$ is an equi-continuous family, it suffices to consider $a,b \in F_n$, and show that $\lim_{t \to \infty} \delta_t(a)\delta_t(b) - \delta_t(ab) = 0$. For this purpose observe that for any sequence of functions $g_i : [1,\infty) \to M(B), i = 0,1,2, \dots$, we have the estimates
\begin{equation}\label{ebbe}
\| \sum_{j=0}^{\infty} \Lambda^t_j g_i(t)\Lambda^t_j\| \leq \sup_i \|g_i(t)\|, \ \ \| \sum_{j=0}^{\infty} \Lambda^t_j g_i(t)\Lambda^t_{j+1}\| \leq \sup_i \|g_i(t)\|
\end{equation}
for all $t$. This follows as in the proof of Lemma \ref{E1}. (\ref{ebbe}) will be used repeatedly in the following calculation.
\begin{equation*}
\begin{split}
& \left( \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\Lambda^t_j \right) \left( \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(b)\Lambda^t_j \right) \\
&\\
&=  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\left(\Lambda^t_j\right)^2 \mu_t^{f_j(t)}(b)\Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a){\Lambda^t_j}{\Lambda^t_{j+1}} \mu_t^{f_{j+1}(t)}(b)\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \ +  \sum_{j=0}^{\infty} \Lambda^t_{j+1} \mu_t^{f_{j+1}(t)}(a){\Lambda^t_{j+1}}{\Lambda^t_{j}} \mu_t^{f_{j}(t)}(b)\Lambda^t_{j} \ \ \  \ \ \ \ \ \text{(using 13))} \\
& \\
& \sim  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a) \mu_t^{f_j(t)}(b)\left(\Lambda^t_j\right)^2 \Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\mu_t^{f_{j+1}(t)}(b){\Lambda^t_j}{\Lambda^t_{j+1}}\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \  \ \ \  \ \ \  \ \  \ +  \sum_{j=1}^{\infty} \Lambda^t_{j} \mu_t^{f_{j}(t)}(a)\mu_t^{f_{j-1}(t)}(b){\Lambda^t_{j}}{\Lambda^t_{j-1}}\Lambda^t_{j-1} \ \ \  \ \ \ \ \ \text{(using 12))} \\
& \\
& \sim  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a) \mu_t^{f_j(t)}(b)\left(\Lambda^t_j\right)^2 \Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\mu_t^{f_{j}(t)}(b){\Lambda^t_j}{\Lambda^t_{j+1}}\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \  \ \ \  \ \ \  \ \  \ +  \sum_{j=1}^{\infty} \Lambda^t_{j} \mu_t^{f_{j}(t)}(a)\mu_t^{f_{j}(t)}(b){\Lambda^t_{j}}{\Lambda^t_{j-1}}\Lambda^t_{j-1} \ \ \  \ \ \ \ \ \text{(using 9))} \\
& \\
& \sim  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(ab)\left(\Lambda^t_j\right)^2 \Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(ab){\Lambda^t_j}{\Lambda^t_{j+1}}\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \  \ \ \  \ \ \  \ \  \ +  \sum_{j=1}^{\infty} \Lambda^t_{j} \mu_t^{f_{j}(t)}(ab){\Lambda^t_{j}}{\Lambda^t_{j-1}}\Lambda^t_{j-1} \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \text{(using 11),4) and 1))} \\
\end{split}
\end{equation*}
\begin{equation*}
 =  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(ab)\Lambda^t_j \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(using 13) and 14)).} 
\end{equation*}
Similar considerations show that $\delta$ is also asymptotically linear and asymptotically self-adjoint.



\qed

\end{proof}



\begin{lem}\label{XX} Let $\varphi, \psi : A \to Q(B)$ be equi-continuous asymptotic extensions. Assume that the asymptotic extension 
$$
\left( \begin{smallmatrix} \varphi & 0 \\ 0 & \psi \end{smallmatrix} \right) : A \to Q(M_2(B))
$$
is split. It follows that for any foldings, $\varphi^f$ and $\psi^f$, of $\varphi$ and $\psi$, respectively, the extension
$$
\left( \begin{smallmatrix} \varphi^f & 0 \\ 0 & \psi^f \end{smallmatrix} \right) : A \to Q(M_2(B))
$$   
is asymptotically split.
\end{lem} 
\begin{proof} Let $\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}, \{n_i\}_{i =0}^{\infty}, \{t_i\}_{i=1}^{\infty}, \right)$ be the folding data used to define $\varphi^f$ and $\left( (\hat{\psi}_t)_{t \in [1,\infty)}, \{u'_n\}_{n = 0}^{\infty}, \{n'_i\}_{i=0}^{\infty}, \{t'_i\}_{i=1}^{\infty}\right)$ the folding data used to define $\psi^f$.  To simplify notation, set $w_i = u_{n_i}$ and $w'_i = u'_{n'_i}$ for all $i$. By assumption there is a an asymptotic homomorphism $\pi : A \to M_2(M(B))$ such that
\begin{equation}\label{EF4}
\lim_{t \to \infty} q_{M_2(B)} \circ \pi_t(a) - \left( \begin{smallmatrix} \varphi_t(a) & \\ & \psi_t(a) \end{smallmatrix} \right) = 0    
\end{equation}
for all $a \in A$. We may assume that $\pi$ is an equi-continuous asymptotic homomorphism. Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \dots$ be a sequence of finite sets with dense union in $A$. Since $\left( \left(\begin{smallmatrix} u_n & \\  & u'_n \end{smallmatrix} \right) \right)_{n \in \mathbb N}$ is an approximate unit in $M_2(B)$ we can choose a sequence $r_1 < r_2 < r_3 < \dots $ in  $\Bbb N$ such that
\begin{equation}\label{A}
\|\left ( \begin{smallmatrix} 1- u_j & {} \\ {} & 1- u'_j \end{smallmatrix} \right )\left[ \left ( \begin{smallmatrix} \hat{\varphi}_{t}(d) & {} \\ {} & \hat{\psi}_{t}(d) \end{smallmatrix} \right ) - \pi_{t}(d)\right]\| \leq \frac{1}{n} + \| \left ( \begin{smallmatrix} \varphi_{t}(d) & {} \\ {} & \psi_{t}(d) \end{smallmatrix} \right ) - q_{M_2(B)} \circ \pi_{t}(d)\|
\end{equation}
for all $t \in [1,n+2], j \geq r_{n-2}, d \in F_n$. It follows from conditions p1) and t4)-t6) on the folding data that we can arrange that
\begin{enumerate}
\item[a)] $\|u_j \hat{\varphi}_t(a) - \hat{\varphi}_t(a)u_j\| \leq \frac{1}{i+1}, \ t \in [1,i+2], \ j \geq r_i$,
\item[b)] $\|\left(1-u_j\right) \left( \hat{\varphi}_t(a) \hat{\varphi}_t(b) -  \hat{\varphi}_t(ab)\right)\| \leq \|\varphi_t(a)\varphi_t(b) - \varphi_t(ab)\| + \frac{1}{i+1}, \ t \in [1,i+2], j \geq r_i$,
\item[c)] $\|\left(1 - u_j\right) \left( \hat{\varphi}_t(a) + \lambda \hat{\varphi}_t(b) -  \hat{\varphi}_t(a + \lambda b)\right)\| \leq \|\varphi_t(a) + \lambda \varphi_t(b) - \varphi_t(a + \lambda b)\| + \frac{1}{i+1}, \ t \in [1,i+2], j \geq r_i$,
\item[d)] $\|\left( 1-u_j\right) \left( \hat{\varphi}_t(a^*) - \hat{\varphi}_t(a)^*\right)\| \leq \|\varphi_t(a^*) - \varphi_t(a)^*\| + \frac{1}{i+1}, \ t \in [1, i+2], \ j \geq r_i$
\end{enumerate}  
and
\begin{enumerate}
\item[a')] $\|u'_j \hat{\psi}_t(a) - \hat{\psi}_t(a)u'_j\| \leq \frac{1}{i+1}, \ t \in [1,i+2], \ j \geq r_i$,
\item[b')] $\|\left(1-u'_j\right) \left( \hat{\psi}_t(a) \hat{\psi}_t(b) -  \hat{\psi}_t(ab)\right)\| \leq \|\psi_t(a)\psi_t(b) - \psi_t(ab)\| + \frac{1}{i+1}, \ t \in [1,i+2], j \geq r_i$,
\item[c')] $\|\left(1 - u'_j\right) \left( \hat{\psi}_t(a) + \lambda \hat{\psi}_t(b) -  \hat{\psi}_t(a + \lambda b)\right)\| \leq \|\psi_t(a) + \lambda \psi_t(b) - \psi_t(a + \lambda b)\| + \frac{1}{i+1}, \ t \in [1,i+2], j \geq r_i$,
\item[d')] $\|\left( 1-u'_j\right) \left( \hat{\psi}_t(a^*) - \hat{\psi}_t(a)^*\right)\| \leq \|\psi_t(a^*) - \psi_t(a)^*\| + \frac{1}{i+1}, \ t \in [1, i+2], \ j \geq r_i$,
\end{enumerate}
for all $a,b \in F_i$ and all $\lambda \in \mathbb C, |\lambda| \leq i$.










Set $r_0 =0$. We will construct continuous paths $w_i(t), w'_i(t), t \in [0,\infty), i = 0,1,2, \dots $, in $B$ such that $\{w_i(t)\}_{i =0}^{\infty}$ and $\{w'_i(t)\}_{i =0}^{\infty}$ are unit sequences for all $t$, $w_i(0) = w_i, w'_i(0) = w'_i$ for all $i$, $w_i(t) = w_i, w'_i(t) = w'_i, i \geq r_{n+1}$, for all $t \in [n,n+1]$, and $w_0(n) \geq w_{r_{n-1}}, \ w'_0(n) \geq w'_{r_{n-1}}$. We will arrange that
\begin{equation}\label{bb2}
w_i(t) \in \co \{ u_j : j \geq n_i\}, \ w'_i(t) \in \co \{ u'_j : j \geq n'_i\}
\end{equation}
for all $i,t$, and
\begin{equation}\label{bb1}
w_i(t) \in \co \{ u_j : j \geq r_{n-1}\}, \ w'(t) \in \co \{u'_j : j \geq r_{n-1}\}
\end{equation}
for all $i$ when $t \in [n,n+1]$. To see how to do that, assume that we have constructed $\{w_i(t\}_{i=0}^{\infty}$ and $\{w'_i(t)\}_{i=0}^{\infty}$ for all $t \in [0,n]$. Then $w_{r_{n-1}} \leq w_0(n) \leq w_1(n) \leq \dots \leq w_{r_{n} -1}(n) \leq w_{r_{n}}(n) =  w_{r_{n}}$, and $w_i(n) = w_i, i \geq r_{n}$. Thanks to condition t3) on the folding data there are elements $v_0 < v_1 < \dots < v_{r_{n}}$ from $\{u_i : r_{n}+1 > i > r_{n}\}$ such that
$$
w_0(n) \leq w_1(n) \leq \dots \leq w_{r_{n}-1}(n) \leq w_{r_n} \leq v_0 \leq v_1 \leq \dots \leq v_{r_{n}} \leq w_{r_{n} +1} \leq w_{r_{n}+2}  \leq \dots 
$$  
is a unit sequence. We define $w_i(t), t \in [n,n+1]$, such that $w_i(t) = w_i, i \geq r_{n}+1$. To define $w_i(t)$ for $i \leq r_n$ and $t \in [n,n+1]$, set $I_j = [n + \frac{j}{r_n+1}, n + \frac{j+1}{r_n+1}], j = 0,1, \dots, r_n$. On the interval $I_j$, $w_{r_n -j}$ is the line from $w_{r_n-j}(n)$ to $v_{r_n-j}$, while $w_k$ is constant on $I_j$ for $k \neq r_n-j$. This completes the construction of $\{w_i(t)\}_{i =0}^{\infty}, t \in [n,n+1]$. Proceed inductively to construct $\{w_i(t)\}_{i =0}^{\infty}, t \in [1,\infty)$.  $\{w'_i(t)\}_{i =0}^{\infty}, t \in [1,\infty)$, is constructed in the same way.



Note that a)-d) and a')-d') in combination with (\ref{bb1}) imply that
\begin{enumerate}
\item[A)] $ \sup_j \|w_j(s) \hat{\varphi}_t(a) - \hat{\varphi}_t(a)w_j(s)\| \leq \frac{1}{i}, \ s,t \in [i,i+1]$,
\item[B)] $\sup_j \|\left(1-w_j(s)\right) \left( \hat{\varphi}_t(a) \hat{\varphi}_t(b) -  \hat{\varphi}_t(ab)\right)\| \leq \|\varphi_t(a)\varphi_t(b) - \varphi_t(ab)\| + \frac{1}{i}, \ s,t \in [i,i+1]$,
\item[C)] $\sup_j \|\left(1 - w_j(s)\right) \left( \hat{\varphi}_t(a) + \lambda \hat{\varphi}_t(b) -  \hat{\varphi}_t(a + \lambda b)\right)\| \leq \|\varphi_t(a) + \lambda \varphi_t(b) - \varphi_t(a + \lambda b)\| + \frac{1}{i}, \ s,t \in [i,i+1], j \geq r_i$,
\item[D)] $\sup_j\|\left( 1-w_j(s)\right) \left( \hat{\varphi}_t(a^*) - \hat{\varphi}_t(a)^*\right)\| \leq \|\varphi_t(a^*) - \varphi_t(a)^*\| + \frac{1}{i}, \ s,t \in [i, i+1], \ j \geq r_i$
\end{enumerate}  
and
\begin{enumerate}
\item[A')] $\sup_j\|w'_j(s) \hat{\psi}_t(a) - \hat{\psi}_t(a)w'_j(s)\| \leq \frac{1}{i}, \ s,t \in [i,i+1]$,
\item[B')] $\sup_j\|\left(1-w'_j(s)\right) \left( \hat{\psi}_t(a) \hat{\psi}_t(b) -  \hat{\psi}_t(ab)\right)\| \leq \|\psi_t(a)\psi_t(b) - \psi_t(ab)\| + \frac{1}{i}, \ s,t \in [i,i+1]$,
\item[C')] $\sup_j\|\left(1 - w'_j(s)\right) \left( \hat{\psi}_t(a) + \lambda \hat{\psi}_t(b) -  \hat{\psi}_t(a + \lambda b)\right)\| \leq \|\psi_t(a) + \lambda \psi_t(b) - \psi_t(a + \lambda b)\| + \frac{1}{i}, \ s,t \in [i,i+1]$,
\item[D')] $\sup_j\|\left( 1-w'_j(s)\right) \left( \hat{\psi}_t(a^*) - \hat{\psi}_t(a)^*\right)\| \leq \|\psi_t(a^*) - \psi_t(a)^*\| + \frac{1}{i}, \ s,t \in [i, i+1]$,
\end{enumerate}
for all $a,b \in F_i$ and all $\lambda \in \mathbb C, |\lambda| \leq i$.










Set $\Delta_0(t) = \sqrt{w_0(t)}, \Delta'_0(t) = \sqrt{w'_0(t)}, \Delta_i(t) = \sqrt{w_i(t) - w_{i-1}(t)},  \Delta'_i(t) = \sqrt{w'_i(t) - w'_{i-1}(t)}, i \geq 1$, and define $f^i,g^i : [1,\infty) \to [1,\infty)$ by $f^i(t) = \max\{t, t_{i+1}\}$ and $g^i(t) = \max \{t, t'_{i+1}\}, i = 1,2, \dots $. Let
$$
\pi_t = \left ( \begin{matrix} \pi_t^{11} &  \pi_t^{12} \\ \pi_t^{21} & \pi_t^{22} \end{matrix} \right )
$$
be the matrix decomposition of $\pi_t$. We define $\Psi_t : A \to M_2(M(B))$ by
\begin{equation*}
\begin{split}
&\rho_t(a) = \left( \begin{matrix} \sum_{i=1}^{\infty} \Delta_i(t) \hat{\varphi}_{f^i(t)}(a) \Delta_i(t)  & 0  \\  0  & \sum_{i=1}^{\infty} \Delta'_i(t) \hat{\varphi}_{g^i(t)}(a) \Delta'_i(t)  \end{matrix} \right) \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +  \left( \begin{matrix} \Delta_0(t) \pi_t^{11}(a)\Delta_0(t) &  \Delta_0(t) \pi_t^{12}(a)\Delta'_0(t) \\   \Delta'_0(t) \pi_t^{21}(a)\Delta_0(t)   &  \Delta'_0(t) \pi_t^{22}(a)\Delta'_0(t)  \end{matrix} \right) .
\end{split}
\end{equation*}
Note first of all that 
\begin{equation}\label{Uii1}
q_{M_2(B)} \circ \rho_t = \left( \begin{smallmatrix} \varphi^f & 0 \\ 0 & \psi^f \end{smallmatrix} \right)
\end{equation}
for all $t$. It suffices therefore to show that $\rho$ is an asymptotic homomorphism. 
 To simplify the following calculations set $\Phi_t(a) =  \sum_{i=1}^{\infty} \Delta_i(t) \hat{\varphi}_{f^i(t)}(a) \Delta_i(t), \Psi_t(a) =  \sum_{i=1}^{\infty} \Delta'_i(t) \hat{\psi}_{g^i(t)}(a) \Delta'_i(t)$ so that
$$
\rho_t(a) = \left( \begin{smallmatrix} \Phi_t(a) & 0 \\ 0 & \Psi_t(a) \end{smallmatrix} \right) +  \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right)\pi_t(a) \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right) .
$$
Note that $\Phi_t, t \in [1,\infty)$, is an equicontinuous family by Lemma \ref{E1} since $\hat{\varphi}_t, t\in [1,\infty)$, is. As $\pi$ is also equi-continuous, we find that $\rho_t, t \in [1,\infty)$, is an equi-continuous family. To show that $\rho_t(a) \rho_t(b) \sim \rho_t(ab)$ for all $a,b$ it suffices therefore to check for $a,b \in F_n$. 
We find that
\begin{equation*}
\begin{split}
&\Phi_t(a) \Phi_t(b)  \\
& = \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a) \Delta_i(t)^2 \hat{\varphi}_{f^i(t)}(b)\Delta_i(t) +  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a) \Delta_i(t)\Delta_{i+1}(t) \hat{\varphi}_{f^{i+1}(t)}(b)\Delta_{i+1}(t)\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum_{i=2}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a) \Delta_i(t)\Delta_{i-1}(t) \hat{\varphi}_{f^{i-1}(t)}(b)\Delta_{i-1}(t)\\
& \\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& \sim  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a)\hat{\varphi}_{f^i(t)}(b)\Delta_i(t)^2\Delta_i(t) +  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a) \hat{\varphi}_{f^{i+1}(t)}(b) \Delta_i(t)\Delta_{i+1}(t)^2 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum_{i=2}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a) \hat{\varphi}_{f^{i-1}(t)}(b) \Delta_i(t)\Delta_{i-1}(t)^2 \ \ \ \ \ \ \ \text{(by A), (\ref{bb2}) and p1))}\\
& \\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& \sim  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a)\hat{\varphi}_{f^i(t)}(b)\Delta_i(t)^2\Delta_i(t) +  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(a) \hat{\varphi}_{f^{i}(t)}(b) \Delta_i(t)\Delta_{i+1}(t)^2 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum_{i=2}^{\infty} \Delta_i(t)\hat{\varphi}_{f^{i}(t)}(a) \hat{\varphi}_{f^{i}(t)}(b) \Delta_i(t)\Delta_{i-1}(t)^2 \ \ \ \ \ \ \ \text{(by t1))}\\
& \\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& \sim  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(ab)\Delta_i(t)^2\Delta_i(t) +  \sum_{i=1}^{\infty} \Delta_i(t)\hat{\varphi}_{f^i(t)}(ab) \Delta_i(t)\Delta_{i+1}(t)^2 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum_{i=2}^{\infty} \Delta_i(t)\hat{\varphi}_{f^{i}(t)}(ab)\Delta_i(t)\Delta_{i-1}(t)^2 \ \ \ \ \ \ \ \text{(by D), (\ref{bb2}) and t4))}\\
& \\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split} 
& = \Phi_t(ab) - \Delta_1(t)^2 \Delta_0(t)^2\hat{\varphi}_{f^1(t)}(ab) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(since $\sum_{j=0}^{\infty} \Delta_j(t)^2 =1$).}
\end{split}
\end{equation*}
In the same way we see that
$$
\Psi_t(a) \Psi_t(b) \sim \Psi_t(ab) -  \Delta'_1(t)^2 \Delta'_0(t)^2\hat{\psi}_{g^1(t)}(ab).
$$ 
It follows from (\ref{A}) and (\ref{bb1}) that
\begin{equation}\label{poul}
\left( \begin{smallmatrix} \Delta_1(t)^2 & \\  & \Delta'_1(t)^2 \end{smallmatrix} \right)  \left( \begin{smallmatrix} \hat{\varphi}_{f^1(t)}(ab) & \\  & \hat{\psi}_{g^1(t)}(ab) \end{smallmatrix} \right) \sim \left( \begin{smallmatrix} \Delta_1(t)^2 & \\  & \Delta'_1(t)^2 \end{smallmatrix} \right) \pi_t(ab).
\end{equation}
Thus 
\begin{equation}\label{Ruth}
\left( \begin{smallmatrix} \Phi_t(a) & \\  & \Psi_t(a) \end{smallmatrix} \right) \left( \begin{smallmatrix} \Phi_t(b) & \\  & \Psi_t(b) \end{smallmatrix} \right) \sim \left( \begin{smallmatrix} \Phi_t(ab) & \\  & \Psi_t(ab) \end{smallmatrix} \right) - \left( \begin{smallmatrix} \Delta_1(t)^2\Delta'_0(t)^2 & \\  & \Delta'_1(t)^2\Delta'_0(t)^2 \end{smallmatrix} \right) \pi_t(ab).
\end{equation}
It follows from A), A') and (\ref{bb1}) that $ \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right)$ asymptotically commutes $\pi_t(a)$ for all $a \in  \bigcup_n F_n$. Hence
\begin{equation}\label{niver}
 \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right)\pi_t(a) \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right) \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right)\pi_t(b) \left( \begin{smallmatrix} \Delta_0(t) & 0 \\ 0 & \Delta'_0(t) \end{smallmatrix} \right) \sim  \left( \begin{smallmatrix} \Delta_0(t)^2 & 0 \\ 0 & \Delta'_0(t)^2 \end{smallmatrix} \right)\pi_t(ab) .
\end{equation}
Finally, it follows easily from (\ref{poul}) that
\begin{equation}\label{kliver}
\left( \begin{smallmatrix} \Phi_t(a) & \\  & \Psi_t(a) \end{smallmatrix} \right)  \left( \begin{smallmatrix} \Delta_0(t)^2 & 0 \\ 0 & \Delta'_0(t)^2 \end{smallmatrix} \right) \pi_t(b) \sim  \left( \begin{smallmatrix} \Delta_0(t)^2 \Delta_1(t)^2 & 0 \\ 0 & \Delta'_0(t)^2\Delta'_1(t)^2 \end{smallmatrix} \right)\pi_t(ab)
\end{equation}
and
\begin{equation}\label{hans}
  \left( \begin{smallmatrix} \Delta_0(t)^2 & 0 \\ 0 & \Delta'_0(t)^2 \end{smallmatrix} \right) \pi_t(a) \left( \begin{smallmatrix} \Phi_t(b) & \\  & \Psi_t(b) \end{smallmatrix} \right)  \sim  \left( \begin{smallmatrix} \Delta_0(t)^2 \Delta_1(t)^2 & 0 \\ 0 & \Delta'_0(t)^2\Delta'_1(t)^2 \end{smallmatrix} \right)\pi_t(ab). 
\end{equation}
Now the desired conclusion, that $\rho_t(a)\rho_t(b) \sim \rho_t(ab)$, follows by combining (\ref{Ruth}),(\ref{niver}),(\ref{kliver}) and (\ref{hans}). That $\rho$ is also asymptotically linear and self-adjoint follows fairly straightforwardly from C),C'),D),D'),t5),t6), (\ref{bb1}) and the fact that $\pi$ is asymptotically linear and self-adjoint.
\qed


\end{proof}




\begin{lem}\label{unitary} Let $\varphi, \psi : A \to Q(B)$ be asymptotic extension that are unitarily equivalent. It follows that there are foldings, $\varphi^f$ and $\psi^f$, of $\varphi$ and $\psi$, respectively, such that $\left( \begin{smallmatrix} \varphi^f & 0 \\ 0 & 0 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} \psi^f & 0 \\ 0 & 0 \end{smallmatrix} \right)$ are unitarily equivalent.
\end{lem}
\begin{proof} By assumption there is a norm-continuous path $v_t, t \in [1,\infty)$, of unitaries in $M(B)$ such that $\lim_{t \to \infty} \Ad q_B(v_t) \circ \varphi_t(a) - \psi_t(a) = 0$ for all $a \in A$. It is easy to see that we can folding data $\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}, \{n_i\}_{i =0}^{\infty}, \{t_i\}_{i=1}^{\infty}, \right)$ for $\varphi$ and $\left( \left(\hat{\psi}_t\right)_{t \in [1,\infty)}, \{u'_n\}_{n = 0}^{\infty}, \{n'_i\}_{i=0}^{\infty}, \{t'_i\}_{i=1}^{\infty}\right)$ for $\psi$ such that $u_n = u'_n, t_n= t'_n$ for all $n$ and $n_i = n'_i$ for all $i$, and such that, furthermore, $\lim_{i \to \infty} v_{t_{i+1}} - v_{t_i} = 0$ and $\lim_{i \to \infty} u_{n_i}v_{t_i} - v_{t_i}u_i = 0$. Let $\varphi^f$ and $\psi^f$ be the foldings obtained with these choices. Set $\Delta_0 = \sqrt{u_{n_0}}$ and $\Delta_j = \sqrt{u_{n_j} - u_{n_{j-1}}}, \ j \geq 1$. Then $\sum_{j=0}^{\infty} \Delta_j v_{t_{j+1}}\Delta_j$ converges in the strict topology to an element $v \in M(B)$. Using Lemma \ref{E1} we see that $q_B(v)$ is a unitary in $Q(B)$ such that $\Ad q_B(v) \circ \varphi^f = \psi^f$. Let $w \in M(M_2(B))$ be a unitary lift of $\left( \begin{smallmatrix} q_B(v) & 0 \\ 0 & q_B(v)^* \end{smallmatrix} \right)$. Then 
$$
Ad q_{M_2(B)}(w) \circ \left( \begin{smallmatrix} \varphi^f & 0 \\ 0 & 0 \end{smallmatrix} \right) = \left( \begin{smallmatrix} \psi^f & 0 \\ 0 & 0 \end{smallmatrix} \right).
$$
\qed
\end{proof}




We assume now that $B$ is stable. This gives space to add extensions and asymptotic extensions: Let $W_1,W_2$ be isometries in $M(B)$ such that $W_1W_1^* + W_2W_2^* = 1$. When $\psi, \varphi : A \to Q(B)$ are asymptotic extensions we set $(\psi \oplus \varphi)_t(a) = q_B(W_1) \psi_t(a)q_B(W_1^*) + q_B(W_2)\varphi_t(a)q_B(W_2^*)$. Up to unitary equivalence this addition is independent of the choice of isometries, $W_1,W_2$. 


\begin{lem}\label{cp} Let $\psi : A \to Q(B)$ be a completely positive asymptotic extension. Assume that $\psi$ is homotopic to a split extension. It follows that there are
\begin{enumerate}
\item[1)] a completely positive asymptotic extension $\mu : A \to Q(B)$ which is split,
\item[2)] a continuous increasing function $r : [1, \infty) \to [1,\infty)$ with $\lim_{t \to \infty} r(t) = \infty$ , and
\item[3)] a norm-continuous path $U_t, t \in [1, \infty)$, of unitaries in $M_2(M(B))$ 
\end{enumerate}
such that
$$
\lim_{t \to \infty} \Ad q_{M_2(B)}\left(U_t\right) \circ \left( \begin{smallmatrix} \psi_{r(t)}(a) \\  & \mu_t(a) \end{smallmatrix} \right) -  \left( \begin{smallmatrix} 0              \\  & \mu_t(a) \end{smallmatrix} \right) = 0
$$
for all $a \in A$.
\end{lem}
\begin{proof} Since $\psi$ is equi-continuous Lemma \ref{crux2} tells us that there is an equi-continuous asymptotic homomorphism $\varphi : A \to M(B)$ such that $q_B \circ \varphi_t = \psi_t$ for all $t \in [1,\infty)$. By Lemma \ref{repar1} there is a continuous increasing function $r_0 : [1, \infty) \to [1,\infty)$ such that $\lim_{t \to \infty} r_0(t) = \infty$ and such that $\varphi_{r_0(t)}, t \in [1, \infty)$, is a uniformly continuous asymptotic homomorphism. Since $q_B \circ \varphi_{r_0(t)} = \psi_{r_0(t)}$, we may as well assume that $\varphi$ is uniformly continuous. Set $t_n = \sum_{k=1}^n \frac{1}{k}$. Let $\{V_i\}_{i=1}^{\infty}$ be a sequence of isometries in $M(B)$ such that $\sum_{i=1}^{\infty} V_iV_i^* = 1$, in the strict topology. Set
$$
\tilde{\nu}_t(a) = \sum_{i=2}^{\infty} V_i\varphi_{t_i + t}(a)V_i^* .
$$
It is clear that $\tilde{\nu}: A \to M(B)$ is an asymptotic homomorphism since $\varphi$ is. We claim that $q_B \circ \tilde{\nu}_t$ is a completely positive contraction for all $t \in [1,\infty)$. We prove first that $\|1-q_B \circ \tilde{\nu}_t(a)\| \leq 1$ when $0 \leq a \leq 1$ in $A$. To this end, let $\{b_n\}_{n =1}^{\infty}$ be an approximate unit in $B$. Then 
\begin{equation}\label{EF2}
\begin{split}
&\|1 - q_B \circ \tilde{\nu}_t(a)\| \\
& \leq  \|\left (1 -\sum_{k=1}^n V_kb_mV_k^*\right) \left( 1  - \sum_{i=2}^{\infty} V_i\varphi_{t_i + t}(a)V_i^*\right)\left( 1 -\sum_{k=1}^n V_kb_mV_k^*\right)\| \\
& = \|\sum_{j > n} V_j (1 - \varphi_{t_j + t}(a))V_j^* + V_1 (1 - b_m)V_1^* + \sum_{j =2}^n V_j(1- b_m)\left( 1  - \varphi_{t_j + t}(a) \right)(1- b_m)V_j^* \|
\end{split}
\end{equation} 
for all $n,m$.
For a given $\epsilon > 0$ there is a $K \in \mathbb N$ such that $\|1 - \varphi_{t_j + s}(a))\| \leq 1 + \epsilon$ for all $j \geq K$ and all $s \in [1, \infty)$. This is because $\varphi$ is an asymptotic homomorphism. In particular,
$$
 \|\sum_{j > K} V_j (1 - \varphi_{t_j + t}(a))V_j^*\| \leq 1 + \epsilon.
$$
We see therefore from (\ref{EF2}) that
\begin{equation}\label{EF3}
\|1 - q_B \circ \tilde{\nu}_t(a)\| \leq \max \{ 1 + \epsilon, \  \|\sum_{j =2}^K V_j(1- b_m)\left( 1  - \varphi_{t_j + t}(a) \right)(1- b_m)V_j^*\| \}
\end{equation}
for all $m \in \mathbb N$. Since $q_B \circ \varphi_{t_j + t}$ is a completely positive contraction for all $j = 2,3, \dots, K$, there is an $m \in \mathbb N$ such that
$$
\|(1- b_m)\left( 1  - \varphi_{t_j + t}(a) \right)(1- b_m)\| \leq 1 + \epsilon
$$
for all $j \in \{2,3,\dots, K\}$. Since $\epsilon > 0$ is arbitrary, it follows therefore from (\ref{EF3}) that $\|1 - q_B \circ \tilde{\nu}_t(a)\| \leq 1$. Similar arguments show that $q_B \circ \tilde{\nu}_t$ is a linear self-adjoint contraction, and combined with what we just established this implies that $q_B \circ \tilde{\nu}_t$ a positive linear contraction. By applying the same argument to the maps $M_n(A) \to M_n(Q(B))$ induced by $q_B \circ \tilde{\nu}_t$, we see that $q_B \circ \tilde{\nu}_t$ is a completely positive contraction for all $t$. Let $r : [1, \infty) \to [1,\infty)$ be a continuous function such that 
\begin{equation}\label{EF7}
t_n + t \leq r(t) \leq t_{n+1} + t
\end{equation}
for $t \in [n,n+1]$. Set
\begin{equation*}
\begin{split}
S^n_t = 1 - V_nV_n^* - V_{n+1}V_{n+1}^*& +  \cos \left(\frac{\pi}{2}(t-n)\right) V_nV_n^* +  \cos \left(\frac{\pi}{2}(t-n)\right) V_{n+1}V_{n+1}^* \\
&+  \sin \left(\frac{\pi}{2}(t-n)\right) V_{n+1}V_n^* -  \sin \left(\frac{\pi}{2}(t-n)\right) V_{n}V_{n+1}^*
\end{split}
\end{equation*}
for $t \in [n,n+1]$. Then $S_t = S^n_tS^{n-1}_n\dots S^1_2, t \in [n, n+1]$, is a norm-continuous path of unitaries in $M(B)$, constructed such that (\ref{EF7}) and the uniform continuity of $\varphi$ ensure that 
$$ 
\lim_{t \to \infty} S_t \left( V_1 \varphi_{r(t)}(a)V_1^* + \sum_{i=2}^{\infty} V_i \varphi_{t_{i} + t}(a)V_i^*\right)S_t^* - \sum_{i = 1}^{\infty} V_i \varphi_{t_{i+1} + t}(a)V_i^* = 0
$$ 
for all $a \in A$. Since $V = \sum_{i=2}^{\infty} V_iV_{i-1}^*$ is a an isometry in $M(B)$ such that $V \left(\sum_{i = 1}^{\infty} V_i \varphi_{t_{i+1} + t}(a)V_i^*\right)V^* = \tilde{\nu}_t(a)$ for all $a,t$, we can use $\mu = \nu \oplus 0$.   

\qed
\end{proof}





\begin{lem}\label{basic} Let $\psi : A \to Q(B)$ be an asymptotically split extension. It follows that there is a asymptotically split extension $\nu = (\nu_t)_{t \in [1,\infty)} : A \to Q(B)$ and a unitary $U \in M_2(M(B))$ such that 
$$
\Ad q_{M_2(B)}(U) \circ \left( \begin{smallmatrix} \psi & 0 \\ 0 & \nu \end{smallmatrix} \right) =  \left( \begin{smallmatrix} 0 & 0 \\ 0 & \nu \end{smallmatrix} \right).
$$
\end{lem}
\begin{proof}  By assumption there is an asymptotic homomorphism $\tilde{\varphi} =(\tilde{\varphi}_t)_{t \in [1,\infty)}: A \to M(B)$ such that $q_B \circ \tilde{\varphi}_t = \psi$ for all $t \in [1,\infty)$. Set
$$
\mathcal B = \{ f \in C_b([1,\infty), M(B)) : q_B(f(t)) = q_B(f(1)), t \in [1,\infty) \}.
$$
Then $\tilde{\varphi}$ defines a $*$-homomorphism $\widehat{\varphi} : A \to \mathcal B/C_0([1,\infty), B)$. By the Bartle-Graves selection theorem there is a continuous section $\chi :  \mathcal B/C_0([1,\infty), B) \to \mathcal B$ for the quotient map $\mathcal B \to \mathcal B/C_0([1,\infty), B)$. Set $\varphi_t(a) = \chi\left(\widehat{\varphi}(a)\right) (t)$. Then $q_B \circ \varphi_t = \psi$. Compared to $\tilde{\varphi}$, $\varphi$ has the property of being equi-continuous. This will be helpful in the construction of $\nu$. Let $d_i, i \in \mathbb N$, be a dense sequence in $A$. Let $\varphi_{t_i}, i \in \mathbb N$, be a discretization of $\varphi$, cf. \cite{MT1}, chosen such that $\|\varphi_{t_i}(d_k) - \varphi_{t_{i-1}}(d_k)\| \leq \frac{1}{i}, k \leq i$, for all $i$. For each $i \in \mathbb N$, choose $\alpha_i \geq 1$ such that 
\begin{equation}\label{A2}
\|\varphi_{t_i + \frac{t}{\alpha_i}}(d_k) - \varphi_{t_i}(d_k)\| \leq \frac{1}{i}
\end{equation}
for all $t \in [1,i]$ and all $k \leq i$. Let $V_i, i \in \mathbb N$, be a sequence of isometries in $M(B)$ such that $\sum_{i =1}^{\infty} V_iV_i^* = 1$, in the strict topology. Set
$$
\tilde{\nu}_t(d) = \sum_{i=2}^{\infty} V_{i} \varphi_{t_i + \frac{t}{\alpha_i}}(d)V_{i}^* .
$$
To check that $\tilde{\nu}$ is an asymptotic homomorphism we must check that
\begin{equation}\label{A1}
\lim_{t \to \infty} \sup_{i \in \mathbb N} \|\varphi_{t_i + \frac{t}{\alpha_i}}(a) \varphi_{t_i + \frac{t}{\alpha_i}}(b) - \varphi_{t_i + \frac{t}{\alpha_i}}(ab)\| = 0
\end{equation}
for any pair $a,b \in A$. Let $\epsilon > 0$. Since $\varphi$ is an asymptotic homomorphism, there is an $N \in \mathbb N$ so large that
$$
\|\varphi_{t_i + \frac{t}{\alpha_i}}(a) \varphi_{t_i + \frac{t}{\alpha_i}}(b) - \varphi_{t_i + \frac{t}{\alpha_i}}(ab)\| \leq \epsilon
$$
for all $t \in [1,\infty)$, when $i \geq N$. Choose $T \in [1,\infty)$ so large that 
$$
\sup_{i \leq N} \|\varphi_{t_i + \frac{t}{\alpha_i}}(a) \varphi_{t_i + \frac{t}{\alpha_i}}(b) - \varphi_{t_i + \frac{t}{\alpha_i}}(ab)\| \leq \epsilon
$$
when $t > T$. Then 
$$
\sup_{i \in \mathbb N} \|\varphi_{t_i + \frac{t}{\alpha_i}}(a) \varphi_{t_i + \frac{t}{\alpha_i}}(b) - \varphi_{t_i + \frac{t}{\alpha_i}}(ab)\| \leq \epsilon 
$$
when $t > T$, proving (\ref{A1}). The other asymptotic algebraic identities follow in the same way. Since $\varphi_t(a) - \varphi_s(a) \in B$ for all $a,s,t$, (\ref{A2}) ensures that $\tilde{\nu}_t(d_k) - \tilde{\nu}_1(d_k) \in B$ for all $k$ and $t$. Thanks to the equicontinuity of $\varphi$, $q_B \circ \tilde{\nu}_t$ is a continuous map for each $t$, so the density of $\{d_i\}$ in $A$ ensures that $\nu = q_B \circ \tilde{\nu}_t$ is independent of $t$, and hence defines an extension which is asymptotically split by construction. Note that $\psi \oplus \nu$ is unitarily equivalent to $\mu = q_B \circ \tilde{\mu}$, where
$$
\tilde{\mu}_t(a) = V_1 varphi_{t_1 + \frac{t}{\alpha_1}}(a)V_1^* + \sum_{i=2}^{\infty} V_{i} \varphi_{t_i + \frac{t}{\alpha_i}}(a)V_{i}^* .
$$     
Set $S = \sum_{i = 2}^{\infty}V_{i-1}V_i^*$. Then $S$ is an isometry in $M(B)$ which by (\ref{A2}) has the property that $\Ad S \circ \tilde{\nu}_t(d_k) - \tilde{\mu}_t(d_k) \in B$ for all $t,k$. It follows that $\Ad q_B(S) \circ \nu = \mu$. Hence $\psi \oplus \nu \oplus 0$ is unitarily equivalent to $\nu \oplus 0$. Since $\nu$ is unitarily equivalent to $0 \oplus \nu$ it follows that $\psi \oplus \nu$ is unitarily equivalent to $0 \oplus \nu$, which is the statement of the lemma.  

\qed

\end{proof} 
 We will say that an asymptotic extension $\varphi : A \to Q(B)$ is \emph{semi-invertible} when there is another asymptotic extension $\psi : A \to Q(B)$ such that $\varphi \oplus \psi$ is split. 

\begin{lem}\label{gomot}
Let $\varphi,\psi:A\to Q(B)$ be two semi-invertible asymptotic extensions that are homotopic. Then there exists a split asymptotic extension $\mu$
such that $\varphi\oplus\mu$ and $\psi\oplus\mu$ are unitarily
equivalent. In fact, there is a unitary $U \in M_2(M(B))$ such that
$$
\lim_{t \to \infty} \Ad q_{M_2(B)}\left(U \right) \circ \left( \begin{smallmatrix} \varphi_t(a) & 0 \\ 0 & \mu_t(a) \end{smallmatrix} \right) -  \left( \begin{smallmatrix} \psi_t(a) & 0 \\ 0 & \mu_t(a) \end{smallmatrix} \right) = 0
$$
for all $a \in A$.

\end{lem}

\begin{proof} We may assume that $\varphi$ and $\psi$ are equi-continuous. Choose equi-continuous asymptotic extensions $\varphi_1,\psi_1 : A \to Q(B)$ such that $\varphi \oplus \varphi_1$ and $\psi \oplus \psi_1$ are split. Then $\varphi \oplus \psi_1$ is homotopic to the split asymptotic extension $\psi \oplus \psi_1$, and hence $\mu_1 = \varphi \oplus \psi_1$ is split by Lemma \ref{crux2}. Set $\mu_2 = \psi \oplus \psi_1$, and note that $\psi \oplus \mu_1$ is unitarily equivalent to $\varphi \oplus \mu_2$. It follows from Lemma \ref{basic} that there are split asymptotic extensions $\nu_i$ such that $\mu_i \oplus \nu_i$ is unitarily equivalent to $\nu_i, i = 1,2$. Set $\mu = \nu_1 \oplus \nu_2$.

\qed
\end{proof}






























\section{Proof of the main results}




For any $C^*$-algebra $D$ we denote by $s_D$ the stabilizing $*$-homomorphism $s_D : D \to D \otimes \mathbb K$ given in standard notation by $s_D(d) = d \otimes e_{11}$, and we let $\beta_B : Q(B) \otimes \mathbb K \to Q(B \otimes \mathbb K)$ be the canonical embedding.











\begin{lem}\label{w1} There is a group homomorphism 
$$f : [[SA,Q(B) \otimes \mathbb K]] \to [SA, Q(B \otimes \mathbb K) \otimes \mathbb K]
$$ 
such that $f[\varphi] = [s_{Q(B\otimes \mathbb K)} \circ (\beta_B \circ \varphi)^f]$, when $\varphi$ is an equi-continuous asymptotic homomorphism $\varphi : SA \to Q(B) \otimes \mathbb K$.  
\end{lem}
\begin{proof} We prove first that the class of $s_{Q(B\otimes \mathbb K)} \circ (\beta_B \circ \varphi)^f$ in $[SA,  Q(B \otimes \mathbb K) \otimes \mathbb K]$ is independent of the choices made in the construction of $(\beta_B \circ \varphi)^f$. The philosophy underlying the approach is due to Higson, cf. Theorem 3.4.3 in \cite{H1}. Since $[[SA,Q(B) \otimes \mathbb K]]$ is a group there is an equi-continuous asymptotic homomorphism $\psi : SA \to Q(B) \otimes \mathbb K$ such that 
$$
\left ( \begin{smallmatrix} \varphi & 0 \\ 0 & \psi \end{smallmatrix} \right) : SA \to Q(B) \otimes \mathbb K
$$
is homotopic to $0$. It follows that 
$$
\left ( \begin{smallmatrix} \beta_B \circ \varphi & 0 \\ 0 & \beta_B \circ \psi \end{smallmatrix} \right)
$$
is homotopic to $0$, and hence split by Lemma \ref{crux2}. Then Lemma \ref{XX} tells us that 
$$
\left ( \begin{smallmatrix} \left(\beta_B \circ \varphi\right)^f & 0 \\ 0 & \left(\beta_B \circ \psi\right)^f \end{smallmatrix} \right)
$$
is asymptotically split. It follows then from Lemma \ref{basic} that $[s_{Q(B \otimes \mathbb K)} \circ (\beta_B \circ \varphi)^f] = [s_{Q(B \otimes \mathbb K)} \circ (\beta_B \circ \psi)^f]$ in $ [SA, Q(B \otimes \mathbb K) \otimes \mathbb K]$. Thus $[s_{Q(B \otimes \mathbb K)} \circ (\beta_B \circ \varphi)^f] = - [s_{Q(B \otimes \mathbb K)} \circ (\beta_B \circ \psi)^f]$, regardless of the folding data used in the construction of $(\beta_B \circ \varphi)^f$. Assume next that $\varphi_1, \psi_1 : SA \to Q(B) \otimes \mathbb K$ are homotopic as asymptotic homomorphisms. By Lemma \ref{gomot} there is a (split) asymptotic extension $\mu : SA \to Q(B \otimes \mathbb K)$ such that $(\beta_B \circ \varphi_1) \oplus \mu$ and $(\beta_B \circ \psi_1) \oplus \mu$ are unitarily equivalent. Since we are now free to choose the foldings at will we conclude from Lemma \ref{unitary} that $[s_{Q(B \otimes \mathbb K)} \circ \left( \beta_B \circ \varphi_1\right)^f] = [s_{Q(B \otimes \mathbb K)} \circ \left( \beta_B \circ \psi_1\right)^f]$. Hence $f : [[SA,Q(B) \otimes \mathbb K]] \to [SA, Q(B \otimes \mathbb K) \otimes \mathbb K]$ is well-defined, and by using the freedom of choice of foldings again it follows that $f$ is a homomorphism. 




\qed
\end{proof}


Let $ i : [[SA, Q(B) \otimes \mathbb K]]_{cp} \to  [[SA, Q(B) \otimes \mathbb K]]$ and $I : [SA, Q(B \otimes \mathbb K) \otimes \mathbb K] \to [[SA, Q(B \otimes \mathbb K) \otimes \mathbb K]]_{cp}$ be the obvious (forgetful) maps.

\begin{lem}\label{w2} The composition

\begin{equation*}
\begin{xymatrix}{
[[SA, Q(B) \otimes \mathbb K]]_{cp} \ar[r]^-{i} & [[SA, Q(B) \otimes \mathbb K]]  \ar[r]^-{I \circ f} & [[SA, Q(B \otimes \mathbb K) \otimes \mathbb K]]_{cp} }
\end{xymatrix}
\end{equation*}
agrees with $(\widehat{s_B} \otimes \id_{\mathbb K})_*$.
\end{lem}
\begin{proof} Let $\varphi', \psi' : SA \to Q(B) \otimes \mathbb K$ be completely positive asymptotic homomorphisms such that $\psi'$ represents the inverse of $[\varphi']$ in $[[SA, Q(B) \otimes \mathbb K]]_{cp}$. Set $\varphi = \beta_B \circ \varphi'$ and $\psi = \beta_B \circ \psi'$. It follows from Lemma \ref{crux2} that there is an equi-continuous asymptotic homomorphism $\pi : SA \to M_2(M(B))$ such that
\begin{equation}\label{anne}
\left( \begin{smallmatrix} \varphi_t & 0 \\ 0 & \psi_t \end{smallmatrix} \right) = q_{M_2(B \otimes \mathbb K)} \circ \pi_t
\end{equation}
for all $t$. Let $\hat{\psi}_t, \hat{\varphi}_t : SA \to M(B), t \in [1,\infty)$, be equi-continuous lifts of $\psi$ and $\varphi$, respectively. 









Let $\left( \left(\hat{\varphi}_t\right)_{t \in [1,\infty)}, \{u_n\}_{n = 0}^{\infty}, \{n_i\}_{i =0}^{\infty}, \{t_i\}_{i=1}^{\infty}, \right)$ be the folding data used to define $\varphi^f$, so that 
$$
\varphi^f(a) = q_{B \otimes \mathbb K}\left( \sum_{j=0}^{\infty} \Delta_j\hat{\varphi}_{t_{j+1}}(a) \Delta_j \right),
$$ 
where $\Delta_0 = \sqrt{u_{n_0}}$ and $\Delta_j = \sqrt{u_{n_j} - u_{n_{j-1}}}, j \geq 1$.  Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \dots$ be a sequence of finite sets with dense union in $SA$. Since we can choose the folding data at will we can apply Lemma \ref{ex} to arrange that
\begin{equation}\label{AAA} 
\|\Delta_j \hat{\psi}_t(a) - \hat{\psi}_t(a) \Delta_j \| \leq 2^{-j}
\end{equation}
for all $a \in F_j, t \in [1,j+3]$, and all $j$. It follows then that
$$
\psi_t(a) = q_{B\otimes \mathbb K} \circ \hat{\psi}_t(a) = q_{B \otimes \mathbb K} \left( \sum_{j=0}^{\infty} \Delta_j \hat{\psi}_t(a) \Delta_j \right) ,
$$
for all $t \in [1,\infty)$, first for all $a \in \bigcup_n F_n$, and then by continuity for all $a \in SA$. We now proceed much as in the proof of Lemma \ref{XX}: Using (\ref{anne}) we choose a sequence $r_1 < r_2 < r_3 < \dots $ in  $\Bbb N$ such that
\begin{equation}\label{AXA}
\|\left ( \begin{smallmatrix} 1- u_j & {} \\ {} & 1- u_j \end{smallmatrix} \right )\left[ \left ( \begin{smallmatrix} \hat{\varphi}_{t}(d) & {} \\ {} &  \hat{\psi}_{t}(d)) \end{smallmatrix} \right ) - \pi_{t}(d)\right]\| \leq \frac{1}{n}
\end{equation}
for all $t \in [1,n+3], j \geq r_{n-2}, d \in F_n$, and such that a)-d) and a')-d') from the proof of Lemma \ref{XX} hold, with $u'_j = u_j$. We then construct the same path $\{w_i(t)\}_{i=0}^{\infty}, t \in [1,\infty)$ of unit sequences as in the proof of Lemma \ref{XX}, but this time we set
\begin{equation*}
\begin{split}
&\rho_t(a) = \left( \begin{matrix}  \sum_{i=1}^{\infty} \Delta_i(t) \hat{\varphi}_{f^i(t)}(a) \Delta_i(t)   & 0  \\  0  &  \sum_{i=1}^{\infty} \Delta_i(t) \hat{\psi}_{t}(a) \Delta_i(t) \end{matrix} \right) \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +  \left( \begin{matrix} \Delta_0(t) &  0 \\  0   &  \Delta_0(t)  \end{matrix} \right)\pi_t(a)\left( \begin{matrix} \Delta_0(t) &  0 \\  0   &  \Delta_0(t)  \end{matrix} \right) .
\end{split}
\end{equation*} 
As in the proof of Lemma \ref{XX} we see that $\rho$ is an asymptotic homomorphism. Since 
$$
q_{B \otimes \mathbb K} \circ \rho_t =  \left( \begin{smallmatrix} \varphi^f & 0 \\ 0 & \psi_t \end{smallmatrix} \right)
$$
for all $t$, it follows from Lemma \ref{cp} that $[s_{Q(B \otimes \mathbb K)} \circ \varphi^f] = - [s_{Q(B \otimes \mathbb K)} \circ  \psi]$ in $[[ SA, Q(B  \otimes \mathbb K) \otimes \mathbb K]]_{cp}$. Since $ - [s_{Q(B \otimes \mathbb K)} \circ  \psi] = [s_{Q(B \otimes \mathbb K)} \circ  \varphi]$, we conclude that $I \circ f \circ i =  \left( s_{Q(B \otimes \mathbb K)} \circ \beta_B\right)_*$. Note that $ s_{Q(B \otimes \mathbb K)} \circ \beta_B$ and $\widehat{s_B} \otimes \id_{\mathbb K}$ both maps $Q(B) \otimes \mathbb K$ into $Q(B) \otimes \mathbb K \otimes \mathbb K$, and that there is a unitary $U \in M(Q(B) \otimes \mathbb K \otimes \mathbb K)$ such that $ \Ad U \circ s_{Q(B \otimes \mathbb K)} \circ \beta_B =\widehat{s_B} \otimes \id_{\mathbb K}$. Since the unitary group of $M(Q(B) \otimes \mathbb K \otimes \mathbb K)$ is connected in the strict topology, it follows that $ s_{Q(B \otimes \mathbb K)} \circ \beta_B$ and $\widehat{s_B} \otimes \id_{\mathbb K}$ are homotopic. Hence $I \circ f \circ i = (\widehat{s_B} \otimes \id_{\mathbb K})_*$.

\qed
\end{proof}







Note that since $B$ is stable there is an isometry $V \in Q(B \otimes \mathbb K)$ such that $Q(B) \ni x \mapsto V^*\widehat{s_B}(x)V$ is an isomorphism, which we will denote by $\gamma$, and such that $\Ad V \circ \gamma = \widehat{s_B}$. We can therefore improve Lemma \ref{w2} as follows.



\begin{lem}\label{w3} The composition
\begin{equation*}
\begin{xymatrix}{
[[SA, Q(B) \otimes \mathbb K]]_{cp} \ar[d]_-{i} & [[SA, Q(B \otimes \mathbb K) \otimes \mathbb K]]_{cp} \ar[d]^-{\left(\gamma^{-1} \otimes \id_{\mathbb K}\right)_* } & \\
[[SA, Q(B) \otimes \mathbb K]]  \ar[ur]_-{I \circ f} &  [[SA, Q(B) \otimes \mathbb K]]_{cp} }
\end{xymatrix}
\end{equation*}
is the identity.
\end{lem}     



In a similar way we can prove


\begin{lem}\label{w4} The composition
\begin{equation*}
\begin{xymatrix}{
[[SA, Q(B) \otimes \mathbb K]] \ar[d]_-{I \circ f} & [[SA, Q(B \otimes \mathbb K) \otimes \mathbb K]] \ar[d]^-{(\gamma^{-1} \otimes \id_{\mathbb K})_* } & \\
[[SA, Q(B\otimes \mathbb K) \otimes \mathbb K]]_{cp}  \ar[ur]_-{i} &  [[SA, Q(B) \otimes \mathbb K]]_{cp} }
\end{xymatrix}
\end{equation*}
is the identity.
\end{lem}     

In fact, the proof is somewhat simpler because we don't have to worry about complete positivity of the asymptotic homomorphisms.


It follows from Lemma \ref{w3} that $i : [[SA,Q(B) \otimes \mathbb K]]_{cp} \to [[SA,Q(B) \otimes \mathbb K]]$ is injective. Since the $\gamma^{-1}_* \circ i = i \circ \gamma^{-1}_*$ we conclude from Lemma \ref{w4} that $i$ is also surjective. This gives us the following result.




\begin{thm}\label{THM} Let $A$ be a separable $C^*$-algebra, and $B$ a stable $\sigma$-unital $C^*$-algebra. There is then a natural isomorphism
$$
E(A,B) = KK(SA, Q(B)).
$$
\end{thm}
\begin{proof} Note that $E(A,B) = E(SA, Q(B)) = [[SA, Q(B) \otimes \mathbb K]]$ by \cite{CH} and \cite{DL} and that $KK(SA,Q(B)) = [[SA, Q(B) \otimes \mathbb K]]_{cp}$ by \cite{H-LT} and \cite{DL}. Combine Lemma \ref{w2} and Lemma \ref{w3}.
\qed
\end{proof}



\begin{cor}\label{cor}  Let $A$ be a separable $C^*$-algebra, and $B$ a stable $\sigma$-unital $C^*$-algebra. Then $E(A,B) = [SA, Q(B) \otimes \mathbb K]$.
\end{cor}
\begin{proof} It follows from Lemma \ref{w3} that $I : [SA, Q(B \otimes \mathbb K) \otimes \mathbb K] \to  [[SA, Q(B \otimes \mathbb K) \otimes \mathbb K]]_{cp}$ is surjective. To show that it is also injective, observe that Lemma 4.3 of \cite{MT3} tells us that the composition 
\begin{equation*}
\begin{xymatrix}{
[SA, Q(B) \otimes \mathbb K] \ar[d]_-{i \circ I} & [SA, Q(B \otimes \mathbb K) \otimes \mathbb K] \ar[d]^-{\left(\gamma^{-1} \otimes \id_{\mathbb K}\right)_* } & \\
[[SA, Q(B) \otimes \mathbb K]]  \ar[ur]_-{f} &  [SA, Q(B) \otimes \mathbb K] }
\end{xymatrix}
\end{equation*}
is the identity.






 
\qed
\end{proof}







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\end{document}


\begin{lem}\label{We} In the setting of Section \ref{s3}, assume that $\varphi : D \to Q(E)$ be a uniformly continuous completely asymptotic homomorphism, and that instead of 4)-7) we have that
\begin{enumerate}
\item[4')] $ \|u_n   \chi\left(\varphi_{2t}(a)\right) -  \chi\left(\varphi_{2t}(a)\right)u_n\| \leq \frac{1}{n}, \ a \in F_n, t \in [1,n +3]$,
\item[5')] $\|\left(1 - u_{n}\right)\left( \chi\left(\varphi_{2t}(ab)\right) - \chi\left(\varphi_{2t}(a)\right)\chi\left(\varphi_{2t}(b)\right)\right)\| \leq \|\varphi_{2t}(ab) - \varphi_{2t}(a)\varphi_{2t}(b)\| + \frac{1}{n}, \ t \in [1,n+3], \ a,b \in F_n$, 
\item[6')] $\| \left(1 - u_{n}\right)\left(\chi\left(\varphi_{2t}(a + b)\right) - \chi\left(\varphi_{2t}(a)\right) - \chi\left(\varphi_{2t}(b)\right)\right)\| \leq \|\varphi_{2t}(a + b) - \varphi_{2t}(a) - \varphi_{2t}(b)\| + \frac{1}{n},  \ t \in [1,n+3], a,b \in F_n $,
\item[7')] $\|\left(1-u_n\right)\left(\chi(\varphi_{2t}(a)) - \chi(\varphi_{2s}(b))\right)\| \leq \|\varphi_t(a) - \varphi_s(b)\| + \frac{1}{n}$ for all $s,t \in [1,n+3]$ and all $a,b \in F_n$.
\end{enumerate}
It follows that there is a subsequence $\{u_{k_n}\}$ of $\{u_n\}$ and a completely positive asymptotic homomorphism $\Pi_t : D \to IQ(E)$ such that
\begin{equation}\label{Uttt}
\Pi_t(a)(s) = q_B \left( \sum_{j=0}^{\infty} \Delta'_j \chi\left( \varphi_{st_j + t}(a) \right) \Delta'_j \right) 
\end{equation}
for all $s \in [0,1], a \in \bigcup_n F_n$, where $\Delta'_0 = \sqrt{u_{k_0}}$ and $\Delta'_j = \sqrt{u_{k_j} - u_{k_{j-1}}}$.

\end{lem}
\begin{proof} For each $a \in F_n$, we can write
$$
\chi\left( \psi_{st_n + t}(a)\right) = w(s,t,n,a) + k(s,t,n,a),
$$
for $s \in [0,1], t \in [1, n + 3]$, where $\|w(s,n,a)\| = \| \psi_{st_n + t}(a)\|, k(s,n,a) \in E$, and both $(s,t) \mapsto w(s,t,n,a)$ and $(s,t) \mapsto k(s,t,n,a)$ are continuous, and we can write  
$$
\chi\left( \psi_{st_n + t}(aa^*)\right) = w'(s,t,n,a) + k'(s,t,n,a),
$$
for $s \in [0,1], t \in [1, n + 3]$, where $ 0 \leq w'(s,n,aa^*) \leq \|a\|^2$, $k(s,n,a) \in E$, and both $(s,t) \mapsto w'(s,t,n,a)$ and $(s,t) \mapsto k'(s,t,n,a)$ are continuous. We choose the subsequence $\{u_{k_n}\}$ such that 
$$
\lim_{n \to \infty} \left(\sup_{s,t,a} \|(1- u_{k_n})k'(s,t,n,a)\|\right) =\lim_{n \to \infty} \left(\sup_{s,t,a} \|(1- u_{k_n})k(s,t,n,a)\|\right) = 0.
$$ 
The estimates that gave (\ref{Ut}) show also that
\begin{equation}\label{Utt}
\begin{split}
&\|q_B \left( \sum_{j=0}^{\infty} \Delta'_j \chi\left( \varphi_{st_j + t}(a) \right) \Delta'_j \right) - q_B \left( \sum_{j=0}^{\infty} \Delta'_j \chi\left( \varphi_{s't_j + t}(b) \right) \Delta'_j \right)\| \\
& \leq \sqrt{3} \limsup_j \|\psi_{st_j + t}(a) - \psi_{s't_j + t'}(b) \|
\end{split}
\end{equation}
for all $a,b \in \bigcup_n F_n$, all $s,s' \in [0,1]$ and all $t,t' \in [1,\infty)$. When we take $s' = s$, we see that there is a map $\Pi_t : A \to IQ(B)$ such that (\ref{Uttt}) holds, and also
$$
\|\Pi_t(a) - \Pi_{t'}(b)\| \leq \sqrt{3} \limsup_j \sup_{s \in [0,1]} \|\psi_{st_j + t}(a) - \psi_{st_j + t'}(b) \|
$$
for all $a,b \in A$ and all $t,t' \in [1,\infty)$. In particular, $\Pi_t(a)$ depends continuously on $t$ since $\psi$ is uniformly continuous. The calculations from the proof of Lemma \ref{E2} shows that $\Pi$ is an asymptotic homomorphism; for example, the estimates that gave multiplicitivity of $\varphi^f$ give now that
$$
\|\Pi_t(d)\Pi_t(d_1) - \Pi_t(dd_1)\| \leq \limsup_j \sup_{s \in [0,1]} \|\psi_{st_j + t}(d)\psi_{st_j + t}(d_1) -\psi_{st_j + t}(dd_1) \|,
$$
for all $d,d_1  \in \bigcup_n F_n$ and all $t$. Since 
$$\lim_{t \to \infty} \left(\limsup_j \sup_{s \in [0,1]} \|\psi_{st_j + t}(d)\psi_{st_j + t}(d_1) -\psi_{st_j + t}(dd_1) \|\right) = 0, 
$$
this shows that $\Pi$ is asymptotically multiplcative. Since each $\psi_t$ is linear, we find that $\Pi_t$ is linear. To see that $\Pi_t$ is a contraction, that by our carteful c
       
\qed
\end{proof}


\begin{lem}\label{crux3} Let $\lambda : A \to Q(B)$ be an equi-continuous extension which splits. It follows that there is equi-continuous asymptotic homomorphism $\delta : A \to M(B)$ such that $\lambda_t(a) = q_B \circ \delta_t(a)$ for all $a,t$.
\end{lem}
\begin{proof} Define $\Lambda : A \to C_b([1,\infty), Q(B))$ by
$$
\Lambda(a)(t = \lambda_t(a) .
$$
$\Lambda$ is continuous since $\lambda$ is equi-continuous. Let $Q_B : C_b([1,\infty), M(B)) \to C_b([1,\infty),Q(B))$ be the surjective $*$-homomorphism induced by $q_B : M(B) \to Q(B)$. By the Bartle-Graves selection theorem there is continuous section $S$ for $Q_B$. We set
$$
\rho_t(a) = S \circ \Lambda(a)(t)
$$
for all $a \in A$ and all $t$. Then
$$
q_B\left( \rho_t(a)\right) = Q_B\left( S \circ \Lambda(a)\right)(t) = \Lambda(a)(t) = \lambda_t(a)
$$
for all $a,t$, and $\rho_t, t \in [1,\infty)$ is an equi-continuous family of maps. By assumption there is an asymptotic homomorphism $\pi : A \to M(B)$ such that $\lim_{t \to \infty} \lambda_t(a) - q_B \circ \pi_t(a)$ for all $a \in A$. We may assume that $\pi$ is equi-continuous. Set  
$$
\mu_t^s (a) = (1-s) \left(\rho_t(a)\right) + s\pi_t(a).
$$
Then $\mu_t^s :A \to M(B), t \in [1,\infty), s \in [0,1]$, is an equi-continuous family of maps such that 
\begin{enumerate}
\item[1)] $\{\mu_t^1\}_{t \in [1,\infty)}$ is an asymptotic homomorphism. 
\end{enumerate}
Furthermore,
\begin{enumerate}
\item[3)] $\lim_{t \to \infty} \sup_{s \in [0,1]} \| q_B \circ \mu^s_t(a)  q_B \circ \mu^s_t(b) -  q_B \circ \mu^s_t(ab)\| = 0$ for all $a,b \in A$, and
\item[4)] $\lim_{t \to \infty} \sup_{s \in [0,1]} \| q_B \circ \mu^s_t(a)  + q_B \circ \mu^s_t(b) -  q_B \circ \mu^s_t(a +b)\| = 0$ for all $a,b \in A$.
\end{enumerate}
Choose continuous functions $f_i : [1, \infty) \to [0,1], i = 0,1,2,\cdots$, such that
\begin{enumerate}
\item[5)] $f_0(t) =0$ for all $t \in [1,\infty)$,
\item[6)] $f_n \leq f_{n+1}$ for all $n$,
\item[7)] for each $ n \in \Bbb N$, there is an $m_n \in \Bbb N$ such that $f_i(t) = 1$ for all $i \geq m_n$, and all $t \in [1,n+1]$, 
\item[8)] $\lim_{t \to \infty} \max_i |\mu_t^{f_i(t)}(a) - \mu_t^{f_{i+1}(t)}(a)| = 0$ for all $a \in A$.
\end{enumerate}
The last requirement, 8), can be fulfilled thanks to the separability of $A$ and the equi-continuity of the family $\mu^s_t, s \in [0,1], t \in [1, \infty)$.
Let $F_1 \subseteq F_2 \subseteq F_3 \subseteq \cdots $ be an increasing sequence of finite subsets with dense union in $A$. For each $n$, choose $m_n \in \Bbb N$ as in 3). We may assume that $m_{n+1} > m_n$. By general facts on quasi-central approximate units (\cite{A}, \cite{P}) we can choose elements 
$$
X^n_{0}   \geq X^n_{1} \geq X^n_{2} \geq  \cdots    
$$
in $B$ such that $0 \leq X^n_i \leq 1$ for all $i$ and $X^n_i = 0$ for $i \geq  m_n$, and
\begin{enumerate}
\item[i)] $X^n_{i}X^n_{i+1} = X^n_{i+1}$ for all $i$,
\item[ii)] $\|X^n_i x - x\| \leq \frac{1}{n} + \|q_B(x)\|$ for all $i = 0,1,2, \cdots , m_n - 1$, and all $b \in S_n$,
\item[iii)] $\|X^n_i y - y X^n_i\| \leq \frac{1}{n}$ for all $ i$ and all $y \in L_n$,
\end{enumerate}
where $L_n$ and $S_n$ are the compact sets $L_n = \{ \mu_t^s(a) : \ s \in [0,1], \ a \in F_n, \ t \in [1,n +1] \}$ and
\begin{equation*}
\begin{split}
&S_n = \{\mu_t^s(a) + \mu_t^s(b) - \mu_t^s(a + b) : \ a,b \in F_n, \ s \in [0,1], \ t \in [1,n+1] \} \\
& \cup \{\mu_t^s(ab) - \mu_t^s(a)\mu_t^s(b) : \ a,b \in F_n, \ s \in [0,1], \ t \in [1,n+1] \} \ .
\end{split}
\end{equation*}
Since we choose the $X$'s recursively we can arrange that $X^{n+1}_i X^n_k = X^n_k$ for all $k$ and all $i$. By connecting first $X^n_0$ to $X^{n+1}_0$ via the straight line between them, then $X^n_{1}$ to $X^{n+1}_{1}$ via a straight line, then $X^n_{2}$ to $X^{n+1}_{2}$ etc., we obtain norm-continuous pathes, $X(t,i), t \in [n,n+1], i = 0,1,2,3, \cdots$, in $B$ such that $X(n,i) = X^n_i, \ X(n+1,i) = X^{n+1}_i$ for all $i$ and 
\begin{enumerate}
\item[9)] $X(t,{i})X(t,i+1) = X(t,i+1), \  t \in [n,n+1]$, for all $i$,
\item[10)] $\|X(t,i) x - x\| \leq \frac{1}{n} + \|q_B(x)\|$ for all $i = 0,1,2, \cdots , m_n - 1, \ t \in [n,n+1]$ and all $x \in S_n$,
\item[11)] $\|X(t,i) y - y X(t,i)\| \leq \frac{1}{n}$ for all $ i $, all $t \in [n,n+1]$ and all $y \in L_n$.
\end{enumerate}
In addition, $X(t,i) = 0, i \geq m_{n+1}, t \in [n,n+1]$. Set $\Lambda^t_0 = \sqrt{1 - X(t,0)}$, and $\Lambda^t_j = \sqrt{X(t,j-1)-X(t,j)}, j \geq 1$, and define $\delta_t : A \to M(B)$ by
$$
\delta_t(a) = \sum_{j = 0}^{\infty} \Lambda^t_j \mu_j^{f_j(t)}(a) \Lambda^t_j .
$$  
Note that there are no convergence problems here since the sum is finite for $t$ in a compact set, and that $t \mapsto \delta_t(a)$ is norm-continuous. Observe also that
\begin{enumerate}
\item[12)] $\Delta^t_j\Delta^t_i = 0, |i-j| \geq 2$, for all $t$, and 
\item[13)] $\sum_{j=0}^{\infty} \left(\Delta^t_j\right)^2 = 1$.
\end{enumerate}
Since $q_B \circ \delta_t(a) = q_B \circ \mu^0_t(a) = \lambda_t(a)$ for all $a,t$, 2) tells us that it only remains to show that $\delta = \left( \delta_t\right)_{t \in [1, \infty)}$ is an asymptotic homomorphism. It is clearly asymptotically homogeneous and self-adjoint. We check that it is multiplicative. Because $\{\delta_t\}$ is an equi-continuous family, it suffices to consider $a,b \in F_n$, and show that $\lim_{t \to \infty} \delta_t(a)\delta_t(b) - \delta_t(ab) = 0$. For this purpose observe that for any function $g : [1,\infty) \to M(B)$ we have the estimates
$$ 
\| \sum_{j=0}^{\infty} \Lambda^t_j g(t)\Lambda^t_j\| \leq \|g(t)\|, \ \ \| \sum_{j=0}^{\infty} \Lambda^t_j g(t)\Lambda^t_{j+1}\| \leq \|g(t)\|
$$
for all $t$. This is used repeatedly in the following calculation, where we write $\sim$ between two expressions that depend on $t$ when their difference tends to $0$ in norm as $t$ tends to infinity.
\begin{equation*}
\begin{split}
& \left( \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\Lambda^t_j \right) \left( \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(b)\Lambda^t_j \right) \\
&\\
&=  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\left(\Lambda^t_j\right)^2 \mu_t^{f_j(t)}(b)\Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a){\Lambda^t_j}{\Lambda^t_{j+1}} \mu_t^{f_{j+1}(t)}(b)\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \ +  \sum_{j=0}^{\infty} \Lambda^t_{j+1} \mu_t^{f_{j+1}(t)}(a){\Lambda^t_{j+1}}{\Lambda^t_{j}} \mu_t^{f_{j}(t)}(b)\Lambda^t_{j} \ \ \  \ \ \ \ \ \text{(using 12))} \\
& \\
& \sim  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a) \mu_t^{f_j(t)}(b)\left(\Lambda^t_j\right)^2 \Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\mu_t^{f_{j+1}(t)}(b){\Lambda^t_j}{\Lambda^t_{j+1}}\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \  \ \ \  \ \ \  \ \  \ +  \sum_{j=1}^{\infty} \Lambda^t_{j} \mu_t^{f_{j}(t)}(a)\mu_t^{f_{j-1}(t)}(b){\Lambda^t_{j}}{\Lambda^t_{j-1}}\Lambda^t_{j-1} \ \ \  \ \ \ \ \ \text{(using 11))} \\
& \\
& \sim  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a) \mu_t^{f_j(t)}(b)\left(\Lambda^t_j\right)^2 \Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(a)\mu_t^{f_{j}(t)}(b){\Lambda^t_j}{\Lambda^t_{j+1}}\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \  \ \ \  \ \ \  \ \  \ +  \sum_{j=1}^{\infty} \Lambda^t_{j} \mu_t^{f_{j}(t)}(a)\mu_t^{f_{j}(t)}(b){\Lambda^t_{j}}{\Lambda^t_{j-1}}\Lambda^t_{j-1} \ \ \  \ \ \ \ \ \text{(using 8))} \\
& \\
& \sim  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(ab)\left(\Lambda^t_j\right)^2 \Lambda^t_j +  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(ab){\Lambda^t_j}{\Lambda^t_{j+1}}\Lambda^t_{j+1} \\
& \ \ \ \  \ \ \ \ \ \ \ \  \ \ \  \ \  \  \ \ \  \ \ \  \ \  \ +  \sum_{j=1}^{\infty} \Lambda^t_{j} \mu_t^{f_{j}(t)}(ab){\Lambda^t_{j}}{\Lambda^t_{j-1}}\Lambda^t_{j-1} \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \text{(using 10),3) and 1))} \\
\end{split}
\end{equation*}
\begin{equation*}
 =  \sum_{j=0}^{\infty} \Lambda^t_j \mu_t^{f_j(t)}(ab)\Lambda^t_j \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(using 12) and 13))} 
\end{equation*}
Similar considerations show that $\delta$ is also asymptotically linear.



\qed

\end{proof}






By passing to a subsequence of $\{u_n\}$ is necessary we may assume the unit sequence $\{u_n\}_{n =1}^{\infty}$ in $B \otimes \mathbb K$ satisfies conditions 4)-7) of Section \ref{s3} with $\chi \left(\varphi_t (\cdot)\right)$ replaced by $\hat{\psi}_{2t}$. Set
$$
\Psi_t(a)  = \sum_{j =0}^{\infty} \Delta_j \hat{\psi}_{t_j +t}(a) \Delta_j .
$$
Since $\hat{\psi}_t, t \in [1,\infty)$, is an equi-continuous family, it follows from Lemma \ref{E1} that so is $\Psi_t, t \in [1,\infty)$. Furthermore, each $q_{B \otimes \mathbb K} \circ \Psi_t$ is an extension; in fact, calculations as in the proof of Lemma \ref{E2} show that $q_{B \otimes \mathbb K} \circ \Psi_t$ is a $*$-homomorphism. Condition 7) of Section \ref{s3} ensures that 
$$   
\|q_{B \otimes \mathbb K} \circ \Psi_t(a) - q_{B \otimes \mathbb K} \circ \Psi_s(a)\| \leq \sqrt{3}\limsup_j\|\psi_{t_j + t}(a) -\psi_{t_j + s}(a)\|
$$
for all $a \in \bigcup_n F_n$ and all $s,t \in [1,\infty)$, cf. the first part of the proof of Lemma \ref{E2}. By equi-continuity this estimate extends to all of $SA$. Thus the uniform continuity of $\psi$ implies that $q_{B \otimes \mathbb K} \circ \Psi_t, t \in [1,\infty)$, is a continuous path of $*$-homomorphisms, in particular, a completely postive asymptotic homomorphism. 
Note that $q_{B \otimes \mathbb K} \circ \rho_t = \left( \begin{smallmatrix} q_{B \otimes \mathbb K} \circ \Psi_t & 0 \\ 0 & \varphi^f \end{smallmatrix} \right)$. The arguments from the proof of Lemma \ref{XX} shows that $\rho$ satisfies the identities of a $*$-homomorphism asymptotically when the elements come from $\bigcup_n F_n$. But $\rho_t, t \in [1,\infty)$, is an equi-continuous family, so we conclude that $\rho$ is an asymptotic homomorphism. We can now conclude from Lemma \ref{cp} that $[s_{Q(B \otimes \mathbb K)} \circ \varphi^f] = - [s_{Q(B \otimes \mathbb K)} \circ  q_{B \otimes \mathbb K} \circ \Psi]$ in $[[ SA, Q(B  \otimes \mathbb K) \otimes \mathbb K]]_{cp}$. 



If we arrange, as we can, that
$$
 \|(1- u_j)[\hat{\psi}_{t_j + t}(a) - \chi\left( \psi_{t_j + t}(a)\right) \| \leq \|q_{B \otimes \mathbb K} \circ \hat{\psi}_{t_j + t}(a) - \psi_{t_j + t}(a) \| + \frac{1}{j} 
$$
for all $a \in F_j$, then the estimates from the first part of Lemma \ref{E2} shows us that $q_{B \otimes \mathbb K} \circ \Psi_t(a) = q_{B \otimes \mathbb K} \left( \sum_{j =0}^{\infty} \Delta_j \chi\left( \psi_{t_j + t}(a)\right) \Delta_j \right)$ for all $t$ and all $a \in \bigcup_n F_n$. 



For each $a \in \bigcup_n F_n$, define $\Pi_t(a) \in IQ(B \otimes \mathbb K)$ by
$$
\Pi_t(a)(s) = q_{B \otimes \mathbb K}\left( \sum_{j =0}^{\infty} \Delta'_j \chi \left( \psi_{st_j + t}(a)\right) \Delta'_j \right), \ s \in [0,1],
$$ 
where $\Delta'_j$ arises from an appropriate subsequence of $\{u_n\}$. For each $a \in \bigcup_n F_n$ and each $j \in \mathbb N$, we can write
$$
\chi
$$
    





Since 
$$
\chi\left( \psi_t(a)\right) - \hat{\psi}_t(a) = \sum_{j=0}^{\infty} \Delta_j\left( \Delta_j \chi\left( \psi_t(a)\right) - \chi\left( \psi_t(a)\right) \Delta_j\right)
$$
for all $a,t$, and since (\ref{AAA}) ensures that
$$
\sum_{j=0}^{\infty}  \|\Delta_j\left( \Delta_j \chi\left( \psi_t(a)\right) - \chi\left( \psi_t(a)\right) \Delta_j\right)\| < \infty
$$
for all $a \in \bigcup_n F_n$ and all $t$, we see first that $\chi\left( \psi_t(a)\right) - \hat{\psi}_t(a) \in B \otimes \mathbb K$ for all $a \in \bigcup_n F_n$, and then by continuity for all $a \in SA$, and all $t$. Hence $\psi_t = q_{B \circ \mathbb K} \circ \hat{\psi}_t$ for all $t$. The virtue of this alternative description of $\psi$, as $q_{B \otimes \mathbb K} \circ \hat{\psi}$, is that we can use the argument from the proof of Lemma \ref{XX} to show that the completely positive asymptotic extension $\psi \oplus \varphi^f : SA \to Q(B \otimes \mathbb K)$ is strongly split. We adopt the notation from the proof of Lemma \ref{XX}, but this time we Note first of all that
$$
q_{M_2(B \otimes \mathbb K)} \circ \rho_t = \left( \begin{smallmatrix} \mu_t & 0 \\ 0 & \varphi^f \end{smallmatrix} \right)
$$
for all $t$, where $\mu_t(a) = q_{B \otimes \mathbb K}\left(\sum_{j=0}^{\infty} \Delta_j \hat{\psi}_{t_j+t}(a)\Delta_j\right)$. (An argument that $\mu_t$ is a completely positive contraction for all $t$ must be inserted.) We assert that $\{\rho_t\}$ is an asymptotic homomorphism. On elements from $a,b \in \bigcup_n F_n$, this follows as in the proof of Lemma \ref{XX}. The equi-continuity her now gives that $\rho$ is an asymptotic homomorphism.  Now, for each $i$ and $a \in F_n$  



To this end note that Lemma \ref{E1} ensures that
$$
\|\Phi_t(a) - \Phi_t(b)\| \leq \|\chi\left( \psi_t(a)\right) - \chi\left( \psi_t(b)\right) \|
$$
and

















It follows then from Lemma \ref{cp} that $[s_{Q(B \otimes \mathbb K)} \circ \varphi^f] = - [s_{Q(B \otimes \mathbb K)} \circ  \psi]$ in $[[ SA, Q(B  \otimes \mathbb K) \otimes \mathbb K]]_{cp}$. Since $  - [s_{Q(B \otimes \mathbb K)} \circ  \psi] = -\left( s_{Q(B \otimes \mathbb K)} \circ \beta_B\right)_*[\psi'] = \left( s_{Q(B \otimes \mathbb K)} \circ \beta_B\right)_*[\varphi']$, the proof is complete.   


It is staightforward to check that
$$
 \beta \circ s_{Q(B)} = \widehat{s_B}.
$$ 
It follows that
$$
s_{Q(B \otimes \mathbb K \otimes \mathbb K)} \circ  \widehat{s_{B \otimes \mathbb K}} \circ \beta \circ s_{Q(B)} = s_{Q(B \otimes \mathbb K \otimes \mathbb K)} \circ \widehat{s_{B \otimes \mathbb K} \circ s_B} .
$$ 
It is easy to check that $s_{Q(B \otimes \mathbb K \otimes \mathbb K)} \circ \widehat{s_{B \otimes \mathbb K} \circ s_B} = \left((\widehat{s_{B \otimes \mathbb K} \circ s_B}) \otimes \id_{\mathbb K}\right) \circ s_{Q(B)}$, so we conclude that $s_{Q(B \otimes \mathbb K \otimes \mathbb K)} \circ  \widehat{s_{B \otimes \mathbb K}} \circ \beta$ equals $(\widehat{s_{B \otimes \mathbb K} \circ s_B}) \otimes \id_{\mathbb K} $ in KK-theory.


Since $D$ is separable we can choose a discretization $\{\varphi_{t_n}\}_{n \in \Bbb N}$ of $\varphi$, i.e. $t_1 \leq t_2 \leq t_3 \leq \dots$ is a sequence in $[1, \infty)$ such that 
\begin{enumerate}
\item[a)] $\lim_{n \to \infty} \sup_{t \in [t_n,t_{n+1}]} \|\varphi_t(a) - \varphi_{t_n}(a)\| = 0$ for all $a \in D$, and
\item[b)] $\lim_{n \infty \infty} t_n = \infty$.
\end{enumerate}
In general the sequence $\{t_n\}$ will have to grow slowly to infinity, but in some cases one can take it to grow quickly. In order to adjust it to unit sequences, it is convenient to empose the following condition.
\begin{enumerate}
\item[c)] $t_n \leq n$ for all $n \in \Bbb N$.
\end{enumerate}

Since $\hat{\varphi}_t,t \in [1,\infty)$, is an equi-continuous family, the map $d \mapsto \sum_{j=0}^{\infty} \Delta_j \left(\hat{\varphi}_{t_{j}}(d)\right) \Delta_j$ is continuous by Lemma \ref{E1}. 


Note that 
\begin{equation}\label{BB1}
\begin{split}
&  \|q_E\left(\sum_{j=0}^{\infty} \Delta_j \varphi_{t_{j+1}}(d) \Delta_j\right) - q_E\left(\sum_{j=0}^{\infty} \Delta_j \varphi_{t_{j+1}}(d_1) \Delta_j\right)\|^2 \\
& = \|q_E\left( \sum_{j=0}^{\infty}  \Delta_j X_j \Delta_j\right)\|^2,  
\end{split}
\end{equation}
where $X_j =  \chi\left(\varphi_{t_{j+1}}(d)\right) - \chi\left(\varphi_{t_{j+1}}(d_1)\right)$. Using (\ref{BB}) and Lemma \ref{BB1} we find that  
\begin{equation}\label{BB2}
\begin{split}
& \|q_E\left( \sum_{j=0}^{\infty}  \Delta_j X_j \Delta_j\right)^* q_E\left( \sum_{j=0}^{\infty}  \Delta_j X_j \Delta_j\right)\|\\
& \leq \|q_E \left( \sum_{j=0}^{\infty} \Delta_j X_j^* \Delta_j^2 X_j \Delta_j\right)\| + \|q_E\left( \sum_{j=0}^{\infty}\Delta_j X_j^* \Delta_j \Delta_{j+1}X_{j+1}\Delta_{j+1}\right)\| \\
& \ \ \ \ \  \ \  \ \  \ \  \ \ \  \ \ \  \ \ \ \ \  \ \  \ \  \ \  \ \ \  \ \ \   \ \ \  \ \ \ + \|q_E\left(\sum_{j=0}^{\infty} \Delta_{j+1}X_{j+1}^*\Delta_{j+1}\Delta_j X_j \Delta_j\right)\| \\
& \leq \limsup_j \| X_j^* \Delta_j^2 X_j\| + \limsup_j \| X_j^* \Delta_j \Delta_{j+1}X_{j+1}\| + \limsup_j \|X_{j+1}^*\Delta_{j+1}\Delta_jX_j\|.
\end{split}
\end{equation}
It follows from condition 7) that each of the three last terms are dominated by $\limsup_j \| \varphi_{t_j}(d) - \varphi_{t_j}(d_1)\|^2$. Since $\varphi$ is an asymptotic homomorphism, $\limsup_j \| \varphi_{t_j}(d) - \varphi_{t_j}(d_1)\|^2 \leq \|d-d_1\|$, so we conclude that  
\begin{equation}\label{Ut}
\|q_E\left(\sum_{j=0}^{\infty} \Delta_j \varphi_{t_{j+1}}(d) \Delta_j\right) - q_E\left(\sum_{j=0}^{\infty} \Delta_j \varphi_{t_{j+1}}(d_1) \Delta_j\right)\| \leq \sqrt{3}\|d - d_1\|.
\end{equation}
We can therefore extend the map 
$$
\bigcup_{n=1}^{\infty} F_n \ni d \mapsto q_E\left(\sum_{j=0}^{\infty} \Delta_j \varphi_{t_{j+1}}(d) \Delta_j\right)
$$
by continuity to map $d(\varphi) : D \to Q(E)$. We must check that $\varphi^f$ is a $*$-homomorphism. Let $d,d_1 \in F_n$. Then we can use Lemma \ref{E1} a number of times to conclude that modulo $E$ we have that




a bit; for example as follows. The given asymptotic homomorphism $\varphi$ defines a $*$-homomorphism $\Phi : D \to C_b([1,\infty), Q(E))/C_0([1,\infty),Q(E))$ in a canonical way. Let $\Lambda :  C_b([1,\infty), Q(E))/C_0([1,\infty),Q(E)) \to  C_b([1,\infty), M(E))$ be a continuous section for the canonical surjective $*$-homomorphism $  C_b([1,\infty), M(E)) \to C_b([1,\infty), Q(E))/C_0([1,\infty),Q(E))$. As usual the existence follows from the Bartle-Graves selection theorem. Then $\hat{\varphi}_t(d) = \Lambda(d)(t), t \in [1,\infty)$, is an equi-continuous family of maps such that $\lim_{t \to \infty} \varphi_t(d) -q_E \circ \hat{\varphi}_t(d) = 0$ for all $d \in D$. Choose now a unit sequence such that 4)-6) hold with $\chi \circ \varphi_t$ replaced by $\hat{\varphi}_t$, and let $t_1 < t_2 < t_3 < \dots $ be a sequence in $[1,\infty)$ such that
\begin{enumerate}
\item[a)] $\lim_{n \to \infty} t_n = \infty$,
\item[b)] $t_n \leq n$ for all $n$, and
\item[c)] $\lim_{n \to \infty} \sup_{t \in [t_n,t_{n+1}]} \|\hat{\varphi}_t(a) - \hat{\varphi}_{t_n}(a)\| = 0$ for all $a \in D$.
\end{enumerate}
Then 
\begin{equation}\label{bub1}
\hat{\varphi}(a) =  \sum_{i=0}^{\infty}  \Delta_j\hat{\varphi}_{t_{j+1}}(a)\Delta_j
\end{equation}
is a continuous map because the family $\{\hat{\varphi}_t\}_{t \in [1,\infty)}$ is equi-continuous. The same considerations as for $bB\varphi)$ show that $q_{E} \circ \hat{\varphi}$ satisfies the algebraic relations defining a $*$-homomorphism when the elements involved come from $\bigcup_n F_n$. So $q_{E} \circ \hat{\varphi}$ is a $*$-homomorphism, i.e. an extension of $D$ by $E$, by continuity.



The latter construction has the advantage over the first that we do not need 7) and that the defining equation, (\ref{bub1}), is valid on $D$; not only on a dense subset of $D$, as with $\varphi^f$. If we choose the unit sequence $\{u_n\}$ so that it meets all the requirements needed to define $\varphi^f$, together with the requirements that ensure that $q_{E} \circ \hat{\varphi}$ is an extension, and on top of it also that
$$
\|(1-u_n) \left(\chi(\varphi_t(a)) - \hat{\varphi}_t(a)\right) \| \leq \frac{1}{n}, \ t \in [1, n+3], a \in F_n,
$$ 
then $\varphi^f = q_{E} \circ \hat{\varphi}$. So in a setting where the actual choice of unit sequence is irrelevant, the two approaches give the same result.
By using the appropiate version of conditions 7) in Section \ref{s3} in combination with Lemma \ref{E1} plus the fact that $\pi$ is an asymptotic homomorphism, it follows that
$$
\limsup_{t \to \infty} \|\Psi_t(a) - \Psi_t(b)\| \leq 2\| a-b\|
$$
for all $a, b \in   \bigcup_n F_n \cup F'_n$. We can therefore define a map
$$
\Psi : A \to \mathcal B/C_0([1,\infty), M_2(B)) 
$$
as the extension by continuity of the obvious map defined on $\bigcup_n F_n \cup F_n'$, where $\mathcal B = \{h \in C_b([1,\infty), M_2(M(B))) : q_{M_2(B)}(h(t)) =  q_{M_2(B)}(h(1)) \ \forall t \}$. We claim $\Psi$ is a $*$-homomorphism. To check that $\Psi$ is multiplicative we use the same notation as in the proof of Lemma \ref{crux2} and write $\sim$ between two $t$ dependent expressions when their difference tends to $0$ as $t$ tends to infinity.




\begin{lem}\label{basicx} Let $\psi =(\psi_t)_{t \in [1,\infty)}: A \to Q(B)$ be a split asymptotic extension. It follows that there is a split asymptotic extension $\nu = (\nu_t)_{t \in [1,\infty)} : A \to Q(B)$ and a unitary $U \in M_2(M(B))$ such that 
$$
\lim_{t \to \infty} \Ad q_{M_2(B)}(U) \circ \left( \begin{smallmatrix} \psi_t(a) & 0 \\ 0 & \nu_t(a) \end{smallmatrix} \right) -  \left( \begin{smallmatrix} 0 & 0 \\ 0 & \nu_t(a) \end{smallmatrix} \right) = 0
$$
for all $a \in A$.
\end{lem}
\begin{proof} By assumption there is an asymptotic homomorphism $\tilde{\psi} =(\tilde{\psi}_t)_{t \in [1,\infty)}: A \to M(B)$ such that $\lim_{t \to \infty} q_B \circ \tilde{\psi}_t(a) - \psi_t(a) = 0$ for all $a \in A$. Let $V_i, i \in \mathbb N$, be a sequence of isometries in $M(B)$ such that $\sum_{i =1}^{\infty} V_iV_i^* = 1$, in the strict topology. Set
$$
\tilde{\nu}_t(d) = \sum_{i=2}^{\infty} V_{i} \tilde{\psi}_{t}(d)V_{i}^* .
$$
Then $\tilde{\nu} : A \to M(B)$ is an asymptotic homomorphism and $\nu = q_B \circ \tilde{\nu} : A \to Q(B)$ is a split asymptotic extension. We leave the reader to check that $\nu$ has the stated property. 
\qed
\end{proof}
