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\title[ Quasidiagonal extensions]{ Quasidiagonal extensions and
sequentially trivial asymptotic homomorphism}
\author{V. M. Manuilov and K. Thomsen}

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        %\email{matkt@mi.aau.dk}
        %\address{Institut for matematiske fag, Ny Munkegade, 8000
Aarhus C, Denmark}
        %\dedicatory{Dedicated to S\o ren Have Hansen on the occasion of
his birth}
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        %\commby{Jens Peter S\ae r}

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\begin{document}

        %\email{matkt@mi.aau.dk}

 \address{Department of Mech. and Math., Moscow State University,
Moscow, 119899, Russia}
\address{Institut for matematiske fag, Ny Munkegade, 8000 Aarhus C, Denmark}

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%\end{abstract}

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%----------Selve teksten --------------------------------------------

\section{Introduction.}
N. Salinas defined an extension of separable $C^*$-algebras,
\begin{equation}\label{0.1}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E   \ar[r]  & A  \ar[r] & 0 \ , }
\end{xymatrix}
\end{equation}
to be quasidiagonal when $B$ contains an approximate unit of projections
which is quasi-central in $E$, cf. \cite{Sa}.\footnote{Strictly speaking
Salinas only considered essential and unital extensions, but the definition
has been extended to the general case by Brown and Dadarlat in \cite{BD}.}
He identified, under certain conditions, the subgroup of $KK^1(A,B)$ which
the quasidiagonal extensions correspond to under Kasparov's isomorphism
$\Ext^{-1}(A,B) \ \simeq \ KK^1(A,B)$. C. Schochet has removed some of
Salinas' conditions in \cite{S}, the result being that when $A$ is a
unital $C^*$-algebra in the bootstrap category for which the UCT holds,
and there exists a unital absorbing quasidiagonal extension of $A$ by
$B\otimes \mathcal K$, the subgroup of the $\Ext^{-1}(A,B)$ corresponding
to the quasidiagonal extensions can be identified with the group
$$
\Pext(K_*(A),K_*(B)) \ = \ \Pext(K_0(A),K_0(B)) \oplus
\Pext(K_1(A),K_1(B)) \ .
$$
This subgroup of $KK^1(A,B)$ plays an important role in the version of
the UCT theorem which was obtained by Dadarlat and Loring in \cite{DL2},
and it is very interesting to reverse the point of view and ask which
extensions this subgroup of $KK^1(A,B)$ corresponds to under Kasparov's
isomorphism. It is clear that despite the work of Salinas and Schochet,
the answer can in general not be 'the quasidiagonal extensions', because
there may not be any (e.g. when $B$ does not have an approximate unit
consisting of projections). To formulate the answer which we offer in
this paper, call an extension of the form (\ref{0.1}) \it weakly
quasidiagonal \rm when there is an increasing sequence of projections
$P_1 \leq P_2 \leq P_3 \leq \cdots $ in the multiplier $M(E)$ of $E$
such that
\begin{enumerate}
\item[1)] $P_n E \subseteq B \ ,  \ n \in \Bbb N $,
\item[2)] $\lim_{n \to \infty} P_ne - eP_n \ = 0 \ , \ e \in E$,
\item[3)] $\lim_{n \to \infty} P_nb \ = \ b \ , \ b \in B$.
\end{enumerate}
For unital extensions (i.e. $E$ when is unital) the two notions of
quasidiagonality coincide. But contrary to quasidiagonal extensions
weakly quasidiagonal extensions always exist, and our main result shows
that when $A$ is $KK$-equivalent to an abelian separable $C^*$-algebra
there is a natural group isomorphism between $\Pext(K_*(A),K_{*-1}(B))$
and the subgroup of $\Ext^{-1}(SA,B)$ represented by the weakly
quasidiagonal extensions of $SA$ by $B \otimes \mathcal K$. Furthermore,
we show without any restriction on $A$ that the subgroup of
$\Ext^{-1}(SA,B)$ represented by the weakly quasidiagonal extensions of
$SA$ by $B \otimes \mathcal K$ coincides with the subgroup represented by
the quasidiagonal extensions whenever there exists a quasidiagonal
extension of $SA$ by $B \otimes \mathcal K$, i.e. when
$B \otimes \mathcal K$ has an approximate unit consisting of projections.
It may be, in this case, that the identification of the extensions which
correspond to $\Pext(K_*(A),K_{*-1}(B))$ can be obtained by combining the
methods of Salinas and Schochet with the result of Voiculescu that the
suspension of any $C^*$-algebra is quasidiagonal, cf. Theorem 5 of \cite{V}.
However, to handle the general case it is necessary to introduce the
weakly quasidiagonal extensions.

\bigskip

The main tool we use to investigate extensions, and in this paper mainly
the weakly quasidiagonal extensions, is the Connes-Higson construction
which allows us to translate considerations about extensions to
considerations about asymptotic homomorphisms. This leads us to introduce
a subclass of asymptotic homomorphisms which are important for other
purposes also, namely asymptotic homomorphisms
$\varphi = (\varphi_t)_{t \in [0, \infty)} \ : \ A \to B$ with the
property that
$$
\lim_{n \to \infty} \varphi_n(a) \ = \ 0 \ \ , \ \ a \in A \ .
$$
We call these asymptotic homomorphisms \it sequentially trivial. \rm
To get anywhere with this approach we need to be able to go back from
(sequentially trivial) asymptotic homomorphisms to (weakly quasidiagonal)
extensions, i.e. we need to reverse the Connes-Higson construction.
This was already done in \cite{K-JT}, but at the cost of an additional
suspension. The key idea here is to use an alternative approach based on
'discretizing' the asymptotic homomorphisms. This was suggested by the
work of Mishchenko and Noor Mohammad, \cite{MN}, and Manuilov and
Mishchenko, \cite{MM}.



\section{Weakly quasidiagonal extensions.}

Throughout the paper $A$ and $B$ will denote separable $C^*$-algebras,
with $B$ stable.

An extension
\begin{equation}\label{qd0}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E   \ar[r]^p  & A  \ar[r] & 0 }
\end{xymatrix}
\end{equation}
of $A$ by $B$ is \it semi-split \rm when there is a completely positive
linear contraction $s : A \to E$ such that $p\circ s \ = \ \Id_A$.
Two semi-split extensions,
\begin{equation*}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E_1   \ar[r]  & A  \ar[r] & 0 }
\end{xymatrix}
\end{equation*}
and
\begin{equation*}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E_2   \ar[r]  & A  \ar[r] & 0  \ , }
\end{xymatrix}
\end{equation*}
are homotopic when there is a semi-split extension of $A$ by
$IB = C[0,1]\otimes B$ sitting in the middle of a commuting diagram
\begin{equation*}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E_1   \ar[r] & A \ar@{=}[d]  \ar[r] & 0 \\
0 \ar[r] & IB \ar[d]_-{\pi_1} \ar[u]^-{\pi_0} \ar[r] & E_3 \ar[d] \ar[u]
\ar[r]  & A  \ar[r] & 0 \\
0 \ar[r] & B \ar[r] & E_2  \ar[r]  & A \ar@{=}[u]  \ar[r] & 0  \ ,
}
\end{xymatrix}
\end{equation*}
where $\pi_0, \ \pi_1 : IB \to B$ denote evaluation at $0$ and $1$,
respectively. It follows from the work of Kasparov and others that the
homotopy classes of semi-split extensions form an abelian group
$\Ext^{-1}(A,B)$ which is isomorphic to $KK^1(A,B)$, cf. Theorem 3.3.14
of \cite{K-JT}. In the following we denote for any $C^*$-algebra $D$ by
$M(D)$ the multiplier algebra of $D$, by $Q(D)$ the generalized Calkin
algebra $Q(D) = M(D)/D$ and by $q_D$ the quotient map $q_D : M(D) \to Q(D)$.

\bigskip


\begin{defn}\label{qd1} An extension (\ref{qd0}) is \it weakly
quasidiagonal \rm when there is an increasing sequence
$P_1 \leq P_2 \leq P_3 \leq \cdots $ of projections in $M(E)$ such that
\begin{enumerate}
\item[1)] $P_n E \ \subseteq \ B$ for all $n \in \Bbb N$,
\item[2)] $\lim_{n \to \infty} P_n e - eP_n \ = \ 0$ for all $e \in E$,
\item[3)] $\lim_{n \to \infty} P_nb \ = \ b$ for all $b \in B$.
\end{enumerate}
\end{defn}


Following \cite{Sa} and \cite{BD} we call an extension (\ref{qd0}) \it
quasidiagonal \rm when it is weakly quasidiagonal, and the projections
$\{P_n\}$ required by Definition \ref{qd1} can be found in $B$. A
crucial difference between the notion of quasidiagonal and weakly
quasidiagonal extensions is that
weakly quasidiagonal extensions always exist. Indeed, the direct sum
extension
\begin{equation*}
\begin{xymatrix}{ 0 \ar[r] & B \ar[r] & B\oplus A \ar[r] & A\ar[r] & 0 }
\end{xymatrix}
\end{equation*}
is always weakly quasidiagonal, but only quasidiagonal when $B$ has an
approximate unit consisting of projections.



Let $\Ext^{-1}_q(A,B)$ and $Q(A,B)$ denote the sets in $\Ext^{-1}(A,B)$
consisting of the elements which can be represented by a weakly
quasidiagonal extension, and a quasidiagonal extension, respectively.
 $\Ext^{-1}_q(A,B)$ and $Q(A,B)$ are then always abelian sub-semigroups
in $\Ext^{-1}(A,B)$ (although $Q(A,B)$ can be empty), and
$\Ext^{-1}_q(A,B)$ is a subgroup when $A$ or $B$ is a suspension.
The same is true
about $Q(A,B)$ if the set is not empty (as it is automatically when
$B$ is a suspension). Clearly, $Q(A,B) \subseteq \Ext^{-1}_q(A,B)$.




\bigskip

From the work of Busby we know that the extensions of $A$ by $B$ can
be identified with $\Hom(A,Q(B))$. This will be done systematically
in the following and we need therefore the following characterization
of the weakly quasidiagonal extensions in terms of their Busby invariant.
The simple proof is left to the reader.


\begin{lemma}\label{char} Let $\varphi \in \Hom(A,Q(B))$, and let
$\lambda : A \to M(B)$ be a map such that $\varphi(a) \ = \ q_B \circ
\lambda(a) \ , \ a \in A$. Then $\varphi$ is weakly quasidiagonal if and
only if there is a sequence $P_1 \leq P_2 \leq P_3 \leq \cdots $ of
projections in $M(B)$ such that
\begin{enumerate}
\item[a)] $P_n\lambda(A) \ \subseteq \ B \ , \ n \in \Bbb N$
\item[b)] $\lim_{n \to \infty} P_n \lambda(a) - \lambda(a)P_n \ = \ 0$
for all $a \in A$,
\item[c)] $\lim_{n \to \infty}P_n b \ = \ b \ , \ b \in B$ .
\end{enumerate}
\end{lemma}





\bigskip




\begin{lemma}\label{func} Let $\varphi : A \to A_1$ and $\psi : B \to B_1$
be $*$-homomorphisms. Assume that $B$ and $B_1$ are stable. Then
$$
\varphi^*(\Ext^{-1}_q(A_1,B)) \ \subseteq \  \Ext^{-1}_q(A,B) \ ,
$$
and
$$
\psi_*(\Ext^{-1}_q(A,B)) \ \subseteq \  \Ext^{-1}_q(A,B_1) \ .
$$
\begin{proof} The assertion about $\varphi^*$ is trivial. To prove the
assertion about $\psi$, recall from \cite{T1} that $\psi$ is homotopic
to a quasi-unital $*$-homomorphism. So by homotopy invariance we may
assume that $\psi$ is quasi-unital, i.e. that there is a projection
$e \in M(B_1)$ such that $\overline{\psi(B)B_1} \ = \ eB_1$. Let
$\lambda : A \to M(B)$ be a completely positive map such that
$q_B \circ \lambda \in \Hom(A,Q(B))$. Let $\{P_n\}$ be a sequence of
projections in $M(B)$ meeting the conditions of Lemma \ref{char}. Then
$$
\psi_*[q_B \circ \lambda] \ = \ [q_{B_1} \circ \overline{\psi} \circ
\lambda] \ ,
$$
where $\overline{\psi} : M(B) \to M(B_1)$ is the unique extension of
$\psi$ which is strictly continuous on the unit ball, cf. \cite{T1}.
Note that $\overline{\psi}(1) \ = \ e$. Set $Q_n \ = \ \overline{\psi}(P_n)
+ (1- e)$. It is straightforward to check that this is a sequence of
projections in $M(B_1)$ satisfying the three conditions of Lemma \ref{char}
relative to $\overline{\psi}\circ \lambda$.
\end{proof}
\end{lemma}

\bigskip

For any $C^*$-algebra $D$ we let $s_D : D \to \mathcal K \otimes D$ be
the $*$-homomorphism given by $s_D(d) \ = \ e \otimes d$ for some minimal
non-zero projection in $\mathcal K$.


\begin{cor}
$$
{s_B}_* \ : \ \Ext^{-1}_q(A,B) \ \to  \ \Ext^{-1}_q(A,\mathcal K \otimes B)
$$
and
$$
s_A^* \ : \  \Ext^{-1}_q(\mathcal K \otimes A,B) \to  \Ext^{-1}_q(A,B)
$$
are both isomorphisms.
\begin{proof} The inverses of the isomorphisms ${s_B}_* \ : \
\Ext^{-1}(A,B) \ \to  \ \Ext^{-1}(A,\mathcal K \otimes B)$ and
$s_A^* \ : \  \Ext^{-1}(\mathcal K \otimes A,B) \to  \Ext^{-1}(A,B)$
are given by
$$
[\varphi]  \mapsto \ \mu_*[\varphi] \ \ \ ,  \ \ \ \varphi \in
\Hom(A,Q(\mathcal K \otimes B)) \ ,
$$
and
$$
[\psi]  \mapsto \ \mu_*[\Id_{\mathcal K} \otimes \psi] \ \ \ ,  \ \ \
\psi\in \Hom(A,Q(B)) \ ,
$$
respectively, where $\mu :   \mathcal K \otimes B \to B$ is a
$*$-isomorphism. So they both respect weak quasidiagonality.
\end{proof}
\end{cor}






\begin{prop}\label{qd2} Let
\begin{equation*}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E   \ar[r]  & A  \ar[r] & 0 }
\end{xymatrix}
\end{equation*}
be a weakly quasidiagonal extension. By applying the $K$-functor we obtain
a pure extension
\begin{equation*}
\begin{xymatrix}{
0 \ar[r] & K_*(B) \ar[r] & K_*(E)   \ar[r]  & K_*(A)  \ar[r] & 0 } \ .
\end{xymatrix}
\end{equation*}
\begin{proof} The proof of \cite{BD}, Theorem 8, can be taken over ad
verbatim.
\end{proof}
\end{prop}


\bigskip

For a convenient short introduction to pure extensions, see Section 2
of \cite{BD}, or \cite{R}, where the notion was first introduced in the
classification program.

\bigskip


It follows from Proposition \ref{qd2} that there is a map
$$
\tau \ : \ \Ext^{-1}_q(A,B) \ \to \ \Pext(K_*(A),K_*(B))  \ ,
$$
where
$$ \Pext(K_*(A),K_*(B)) \ = \  \Pext(K_0(A),K_0(B)) \oplus
\Pext(K_1(A),K_1(B)) \ .
$$
$\tau$ is natural with respect to the functoriality in both $A$ and $B$.













\section{From extensions to asymptotic homomorphisms.}

In \cite{CH} Connes and Higson used a procedure which out of a given
extension of the form (\ref{qd0}) produces an asymptotic homomorphism
$\varphi = (\varphi_t)_{t \in [0,\infty)} \ : \ SA \to B$. In \cite{H-LT}
it was shown that by applying their construction to a semi-split extension
the resulting asymptotic homomorphism is asymptotically equal to a
completely positive asymptotic homomorphism, i.e. an asymptotic
homomorphism, $\varphi = (\varphi_t)_{t \in [0,\infty)}  $, where the
individual maps, the $\varphi_t$'s, are all completely positive
contractions. By using universal properties of the functors involved,
it was concluded in \cite{H-LT} that
$$
KK(A,B) \ \simeq \ [[SA \otimes \mathcal K, SB \otimes \mathcal K]]_{cp} \ ,
$$
where $[[ \cdot \ , \ \cdot ]]_{cp}$ denotes the homotopy classes of
completely positive asymptotic homomorphisms. We need to know that this
isomorphism is actually effectuated by the Connes-Higson construction.
For this we need the following.


\begin{thm}\label{DL} (Dadarlat and Loring, \ \cite{DL1})  Let $B$ be a
stable $C^*$-algebra. The suspension map
$$
S \ : \ [[SA,B]]_{cp} \ \to \ [[S^2A,SB]]_{cp}
$$
is an isomorphism.
\begin{proof} This was proved \cite{DL1} for $E$-theory, $[[ \cdot , \
\cdot]]$. It is straightforward to check that their argument works equally
well with completely positive asymptotic homomorphisms.
\end{proof}
\end{thm}

\bigskip

\begin{thm}\label{ch1} Let $B$ be a stable $C^*$-algebra. Then the
Connes-Higson construction gives rise to an isomorphism
$$
CH \ : \ \Ext^{-1}(A,B) \ \to \ [[SA,B]]_{cp} \ .
$$
\begin{proof} From Section 4 of \cite{H-LT} we take the commuting diagram
\begin{equation}\label{ch2}
\begin{xymatrix}{
\Ext^{-1}(A,B)  \ar[r]^-S \ar[d]_-{CH}  &  \Ext^{-1}(SA,SB) \ar[d]^-{CH} \\
[[SA,B]]_{cp} \ar[ru]^-{M}  \ar[r]^-{-S} &  [[S^2A,SB]]_{cp} }
\end{xymatrix}
\end{equation}
>From the work of Kasparov we know that $S  \ : \ \Ext^{-1}(A,B) \to
\Ext^{-1}(SA,SB)$ is an isomorphism, cf. 6.7 of \cite{Sk}. Since the
lower suspension map is an isomorphism by Theorem \ref{DL}, it follows
that $CH : \Ext^{-1}(A,B) \to [[SA,B]]_{cp}$ is an isomorphism.
\end{proof}
\end{thm}

\bigskip

We are going to identify the image of the $\Ext^{-1}_q(A,B)$ under the
$CH$-map when $A$ is a suspension. The first step in this direction is
the following.





\begin{lemma}\label{ch3}
Let
\begin{equation}\label{ch5}
\begin{xymatrix}{
0 \ar[r] & B \ar[r] & E   \ar[r]^-p  & A  \ar[r] & 0 }
\end{xymatrix}
\end{equation}
be a weakly quasidiagonal semi-split extension. There is then a completely
positive asymptotic homomorphism
$\varphi = (\varphi_t)_{t \in [0,\infty)} \ : \ SA \to B$, which is
homotopic to the asymptotic homomorphism obtained by applying the
Connes-Higson construction to (\ref{ch5}), and satisfies that
$$
\lim_{n \to \infty} \varphi_n(a) \ = \ 0 \ \ \ , \ \ \ a \in SA \ .
$$
\begin{proof} Let $\{P_n\}$ be the sequence of projections in $M(E)$
meeting the conditions of Definition \ref{qd1}, and let $s : A \to E$
be a set-theoretic right-inverse for $p$. By convex interpolation we can
construct a norm-continuous path $V_t, \ t \in [0,\infty)$, of elements
$0 \leq V_t \leq 1$ in $M(E)$ such that $V_t E \subseteq B, \
 t \in [0,\infty)$, $\lim_{t \to \infty} V_te - eV_t \ = \ 0$ for all
$e \in E$, $\lim_{t \to \infty} V_tb \ = \ b, \ b \in B$, and
$V_n \ = \ P_n$ for all $n \in \Bbb N$. By using the same arguments as
in \cite{CH} we obtain an asymptotic homomorphism
$\varphi \ = \ (\varphi_t)_{t \in [0,\infty)} \ : \ SA \to B$ such that
$$
\lim_{t \to \infty} \|\varphi_t(f \otimes a) \ - \ f(V_t)s(a)\| \ = \ 0
$$
for all $f \in S, \ a \in A$. To see that $\varphi$ is homotopic to the
asymptotic homomorphism coming from a genuine Connes-Higson construction,
consider an approximate unit $u_t \in B, \ t \in [0, \infty)$, which
asymptotically commutes with $E$. A convex combination of $u_t$ and $V_t$
can then be used to obtain a homotopy between $\varphi$ and the result
of the proper Connes-Higson construction where one uses $\{u_t\}$. Note
also that
$$
f(V_n)s(a) \ = \ f(P_n)s(a) \ = \ 0
$$
for all $n \in \Bbb N, \ f \in S = C_0(0,1), \ a \in A$. It follows
readily that $\lim_{n \to \infty} \varphi_n(x) \ = \ 0$ for all $x \in SA$.
Since the extension (\ref{ch5}) is semi-split by assumption, we may assume
that $s$ is a completely positive contraction. By (the proof of) Lemma 4.1
of \cite{H-LT}, there is a complete positive asymptotic homomorphism
$\sigma = (\sigma_t)_{t \in [0,\infty)} : SA \to B$ such that
$\lim_{t \to \infty} \|\sigma_t(x) - \varphi_t(x)\| \ = \ 0$ for all
$x \in SA$.
\end{proof}
\end{lemma}





\section{Sequentially trivial asymptotic homomorphisms.}

Motivated by Lemma \ref{ch3} we make the following definition.

\begin{defn}\label{at} An asymptotic homomorphism $\varphi \ =
\ (\varphi_t)_{t \in [0,\infty)} : A \to B$ is said to be sequentially
trivial when
$$
\lim_{n \to \infty} \varphi_n(a) \ = \ 0
$$
for all $a \in A$.
\end{defn}


In this paper we shall only be concerned with sequentially trivial
asymptotic homomorphisms which are also completely positive. We denote
by $[[A,B]]^0_{cp}$ the elements of $[[A,B]]_{cp}$ which can be
represented by a  sequentially trivial completely positive asymptotic
homomorphism.


 The composition product of Connes and Higson gives us a product
$$
[[A,B]]_{cp} \times [[B,C]]_{cp} \ \ni \ (x,y) \ \mapsto \ y \bullet x
\ \in \ [[A,C]]_{cp} \ .
$$
The construction of this product and the proof of its associativity is
much simpler in the present case where we deal with asymptotic
homomorphisms which are also completely positive. So we will only
summarize the construction here.

\bigskip




 A \it parametrization \rm $r : [0,\infty) \to [0,\infty)$ is a continuous
function such that $\lim_{t \to \infty} r(t) = \infty$. Given two
parametrizations, $r$ and $s$, we write $r \preceq s$ when $r(t) \leq s(t)$
for all large enough $t$.

\begin{thm} \label{defprod} There is a map
{$$
 \ [[A,B]]_{cp} \ \times \ [[B,C]]_{cp} \ \ni \ (x,y) \ \mapsto \ y
\bullet x \ \in \ [[A,C]]_{cp}
$$}
with the following properties :


a) (Definition) :       When {$\varphi : A \to B$} and {$\psi : B \to C$}
are completely positive asymptotic homomorphisms there is a parametrization
$r : [0,\infty) \to [0,\infty)$ such that $\lambda \ = \ \{\psi_{s(t)}\circ
\varphi_t\}_{t \in [0,\infty)}$ is a completely positive asymptotic
homomorphism for every parametrization $s$ with $r \preceq s$, and
$$
[\psi] \bullet [\varphi] \ = \ [\lambda] \ .
$$



b) (Associativity) :
$$
z \bullet (y \bullet x) \ = \ (z \bullet y) \bullet x \ .
$$
\end{thm}

\bigskip



The map $\bullet$ will be called \it the composition product. \rm



\bigskip


\begin{lemma}\label{prod} a) Let $x \in [[A,B]]_{cp}^0, \ y
\in [[B,C]]_{cp}$. Then
$$
y \bullet x \ \in \ [[A,C]]_{cp}^0 \ .
$$


b)  Let $x \in [[A,B]]_{cp}, \ y \in [[B,C]]_{cp}^0$. Then
$$
y \bullet x \ \in \ [[A,C]]_{cp}^0 \ .
$$

\begin{proof} a) Let $\varphi \ = \ (\varphi_t)_{t \in [0,\infty)} \ :
\ A \to B$ and $\psi \ = \  (\psi_t)_{t \in [0,\infty)} \ : \ B \to C$
be completely positive asymptotic homomorphisms representing $x$ and $y$,
respectively. We may assume that $\varphi$ is sequentially trivial.
For any parametrization $r : [0,\infty) \to [0,\infty)$ we have that
$$
 \|\psi_{r(n)} \circ \varphi_n(a)\| \ \leq \ \| \varphi_n(a)\| \ .
$$
Hence $\lim_{n \to \infty} \psi_{r(n)} \circ \varphi_n(a) \ = \ 0$ since
$\lim_{n \to \infty} \varphi_n(a) \ = 0$.

b) Let $\varphi \ = \ (\varphi_t)_{t \in [0,\infty)} \ : \ A \to B$ and
$\psi \ = \  (\psi_t)_{t \in [0,\infty)} \ : \ B \to C$ be completely
positive asymptotic homomorphisms representing $x$ and $y$, respectively.
We may assume that $\psi$ is sequentially trivial. Let $r : [0,\infty)
\to [0,\infty)$ be a parametrization such that $(\psi_{s(t)} \circ
\varphi_t)_{t \in [0, \infty)}$ represents $y \bullet x$ for any
parametrization $r \preceq s$. Let $\{a_1,a_2, a_3, \cdots \}$ be a
dense sequence in $A$. Since $\psi$ is sequentially trivial we can find
an increasing sequence $m_1 < m_2 < m_3 < \cdots $ of natural numbers
such that $m_n \geq \Sup_{t \in [0,n+1]} r(t)$ and
$$
\|\psi_{k}(\varphi_n(a_i))\| \ \leq \ \frac{1}{n} \ \ , \ \ i = 1,2,
\cdots , n \ , \ k \geq m_n \ .
$$
Let $s : [0,\infty) \to [0,\infty)$ be a parametrization which is linear
on $[n-1,n]$ and has $s(n) = m_n$ for all $n \in \Bbb N$. Then
$(\psi_{s(t)}\circ \varphi_t)_{t \in [0,\infty)}$ represents
$y \bullet x$ (since $r \preceq s$) and is sequentially trivial.



\end{proof}
\end{lemma}


\bigskip


As a very special case of Lemma \ref{prod} we see that we can make
$[[A,B]]_{cp}^0$ functorial in both $A$ and $B$ by restricting the
functoriality of $[[A,B]]_{cp}$. It follows also that we have isomorphisms
$$
{s_A}^* \ : \ [[\mathcal K \otimes SA,B]]_{cp}^0 \ \to \  [[SA,B]]_{cp}^0
$$
and
$$
{s_B}^* \ : \  [[ SA,B]]_{cp}^0 \ \to \  [[SA,\mathcal K \otimes B ]]_{cp}^0
\ ,
$$
when $B$ is stable.


In addition we get the following :

\begin{lemma}\label{st1} Let $B$ be a stable $C^*$-algebra. Then the
suspension map $S \ : \ [[SA,B]]_{cp}^0 \ \to \ [[S^2A,SB]]^0_{cp}$ is
an isomorphism.
\begin{proof} By using the isomorphisms ${s_A}^*$ and ${s_{SA}}^*$ from
above, we may assume that $A$ is stable. We then see from \cite{DL1} that
the isomorphism $S \ : \ [[SA,B]]_{cp} \to [[S^2A,SB]]_{cp}$ of Theorem
\ref{DL} is given by
$$
x  \ \mapsto \ [\alpha] \bullet (Sx) \bullet [\beta]
$$
where $Sx \in [[S^3A,S^2B]]_{cp}$ is the suspension of $x \in
[[S^2A,SB]]_{cp}$, $\beta \in \Hom(SA,S^3A)$ and $\alpha : S^2B \to B$
is a completely positive asymptotic homomorphism. By Lemma \ref{prod},
this inverse map takes a sequentially trivial asymptotic homomorphism
to one which is also sequentially trivial.
\end{proof}
\end{lemma}



\section{From asymptotic homomorphisms to extensions.}


\begin{lemma}\label{ae1}
Let $\{\psi_t\}_{t \in [0,\infty)} : A \to B$ be a completely positive
asymptotic homomorphism. There is then a sequence
$0 = t_0 \leq t_1 \leq t_2 \leq \cdots $ in $[0,\infty)$ such that
\begin{enumerate}
\item[a)] $\lim_{n \to \infty} t_n \ = \ \infty \ . $
\item[b)] $\lim_{n \to \infty} \sup_{s \in [t_n,t_{n+1}]} \|\psi_{t_n}(a)
\ - \ \psi_s(a)\| \ = \ 0$ for all $a \in A$.
\end{enumerate}
\begin{proof} Let $a_1,a_2,a_3,\cdots $ be a dense sequence in $A$. Let
$n \in \Bbb N$. There is a $\delta_n > 0$ such that
\begin{equation*}
\|\psi_s(a_i) \ - \ \psi_t(a_i)\| \ \leq \ \frac{1}{n}
\end{equation*}
for all $s,t \in [n-1,n]$ with $|s-t| < \delta_n$ and all
$i=1,2,\cdots , n$. Choose $m_n \in \Bbb N$ so large that
$1/m_n \ < \ \delta_n$, and set
$$
s^n_i \ = \ n-1 + \frac{i}{m_n} \ , \ \ i = 0,1,2, \cdots , m_n \ .
$$
Let $t_0 < t_1 < t_2 < \cdots $ be the sequence given by reading the list
\begin{equation*}
\begin{split}
&s^1_0 \ , \ s^1_1 \ , \ s^1_2 \ , \ s^1_3 \ , \ \cdots \ , s^1_{m_1}  \\
&s^2_0 \ , \ s^2_1 \ , \ s^2_2 \ , \ s^2_3 \ , \ \cdots \ , s^2_{m_2}  \\
&s^3_0 \ , \ s^3_1 \ , \ s^3_2 \ , \ s^3_3 \ , \ \cdots \ , s^3_{m_3}  \\
&\vdots \ \ \ \ \ \vdots \ \ \ \ \ \vdots \ \ \ \ \vdots \ \ \ \ \
\vdots \ \ \ \ \vdots \\
\end{split}
\end{equation*}
Then $\lim_{n \to \infty} t_n =\infty$ and $\lim_{n \to \infty}
\sup_{s \in [t_n,t_{n+1}]} \|\psi_{t_n}(a_i) - \psi_{s}(a_i)\| \ = \ 0$
for all $i$. Since $\{a_i\}$ is dense in $A$ and the $\psi_s$'s are
uniformly bounded b) follows immediately.

\end{proof}
\end{lemma}

\bigskip

The sequence of maps $\varphi_{t_n} : A \to B \ , \ n = 0,1,2,3,\cdots$,
from Lemma \ref{ae1}  will be called a \it discretization \rm of
$\{\varphi\}_{t \in [0,\infty)}$.


\bigskip
Let $H_B \ = \ l_2(\Bbb Z) \otimes B$ be the Hilbert $B$-module
consisting of bi-infinite sequence
$( \cdots b_{-3}, \ b_{-2}, \ b_{-1}, \ b_{0}, \ b_1, \ b_2, \ b_3, \cdots )$
in $B$ such that $\sum_{i=-\infty}^{\infty} b_i^*b_i$ converges in $B$.
We identify $M(\mathcal K \otimes B)$ and $ \mathcal K \otimes B$ with
the adjoinable and the compact operators on $H_B$, respectively.
For any given norm-bounded sequence $(x_i)_{i = 0}^{\infty}$ in $M(B)$
we can define an element
$$
D[(x_i)] \ \in \ M(\mathcal K \otimes B)
$$
by
$$
D[(x_i)](b_j)_k \ = \ \cases x_kb_k  \ \ , & k \geq 0 \\ 0 \ \ , & k < 0
\endcases \ \ \ \ \ \ , \ \ (b_j) \in H_B \ .
$$
Note that $D[(x_i)] \in \mathcal K \otimes B$ if and only if $x_i \in B$
for all $i$, and $\lim_{i \to \infty} \|x_i\| \ = \ 0$.
\bigskip

Let $\varphi_{t_n} : A \to B \ , \ n = 0,1,2,3,\cdots$, be a
discretization of the completely positive asymptotic homomorphism
$\varphi = (\varphi_t)_{t \in [0,\infty)} : A \to B$. We can then define
a map $\bold\Phi^0_{(t_i)} : A \to M( \mathcal K \otimes B)$ by
$$
\bold\Phi^0_{(t_i)}(a) \ = \ D[\varphi_{t_i}(a)] \ .
$$
It is easy to see that $\bold\Phi^0_{(t_n)}$ is a $*$-homomorphism modulo
$ \mathcal K \otimes B$, and we denote the resulting extension of $A$ by
$ \mathcal K \otimes B$ by $\bold\Phi_{(t_n)} \in \Hom(A,Q( \mathcal K
\otimes B))$.  The twosided shift
$$
V_0(b_j) \ = \ (b_{j+1}) \ \ , \ \ (b_j) \in H_B \ ,
$$
is a unitary in $M(\mathcal K \otimes B)$, and we have that
$$
V_0\bold\Phi^0_{(t_n)}(a) \ - \ \bold\Phi^0_{(t_n)}(a)V_0 \in \mathcal K
\otimes B
$$
for all $a \in A$ because of condition b) in Lemma \ref{ae1}. It follows
that
\begin{equation}\label{aee1}
V\bold\Phi_{(t_n)}(a) \ = \ \bold\Phi_{(t_n)}(a) V
\end{equation}
in $Q(\mathcal K \otimes B )$ for all $a \in A$, where $V$ denotes the
image of $V_0$ in $Q(\mathcal K \otimes B)$. It follows from (\ref{aee1}),
and the fact that $V$ is unitary that we may define an extension
$\bold\Psi_{(t_n)} \in \Hom(C(\Bbb T) \otimes A , Q( \mathcal K
\otimes B))$ such that
\begin{equation*}
\bold\Psi_{(t_n)}(f \otimes a) \ = \ f(V) \bold\Phi_{(t_n)}(a) \ \ ,
\ \ f \in C(\Bbb T) \ , \ a \in A \ .
\end{equation*}

\bigskip



\begin{lemma}\label{ae2} $\bold\Psi_{(t_n)} \in Ext^{-1}(C(\Bbb T)
\otimes A, \mathcal K \otimes B)$.
\begin{proof} By Theorem 6 of \cite{A} (combined with Lemma 3.2.8 of
\cite{K-JT}) it suffices to find a sequence $\chi_m : C(\Bbb T)
\otimes A \to M(\mathcal K \otimes B)$ of completely positive contractions
such that
\eq{\label{aee2}
\lim_{m \to \infty} q_{\mathcal K \otimes B} \circ \chi_m(f) \ = \
\bold\Psi_{(t_n)}(f) \ \ , \ \ f \in C(\Bbb T) \otimes A \ . }
For each $m \in \Bbb N$ we choose a partition of unity $\{h^m_i \ : i = 1,
\cdots, m\}$ in $C(\Bbb T)$ such that the maximal diameter of the support
of the $h^m_i$'s tend to $0$ as $m$ tends to infinity. For each $m,i$,
take an element $x^m_i$ in the support of $h^m_i$. Define $\chi_m :
C(\Bbb T) \otimes A \to M(\mathcal K \otimes B)$ by
 $$
\chi_m(f) \ = \ \sum_{i=1}^m \sqrt{h^m_i}(V_0) \bold\Phi^0_{(t_n)}(f(x^m_i))
\sqrt{h^m_i}(V_0) \ .
$$
Then $\chi_m$ is a completely positive contraction. For every
$g \in C(\Bbb T)$ and $a \in A$,
\eQ{
 q_{\mathcal K \otimes B} \circ \chi_m(g \otimes a) \ = \
[\sum_{i=1}^m g(x^m_i)h^m_i](V)  \bold\Phi_{(t_n)}(a) \ . }
Hence
$$
\lim_{m \to \infty} \| q_{\mathcal K \otimes B} \circ \chi_m(g \otimes a)
\ - \  \bold\Psi_{(t_n)}(g \otimes a)\| \ = \ 0
$$
since $\lim_{m \to \infty} \|\sum_{i=1}^m g(x^m_i)h^m_i \ - \ g\| \ = \ 0$
in $C(\Bbb T)$. Since all maps in sight are linear contractions,
(\ref{aee2}) follows.
\end{proof}
\end{lemma}

\bigskip


\begin{lemma}\label{ae3}  Up to homotopy (by  semi-split extensions)
$\bold\Psi_{(t_n)} $ is independent of the chosen discretization.
\begin{proof}  Let $\{\varphi_{s_n}\}$ be a discretization of
$\{\varphi_{t_n}\}$ with the property that $\{s_n\}$ is a refinement
of $\{t_n\}$, meaning that $\{t_n\} \subseteq \{s_n\}$
(counting multiplicity). It is not difficult to see that two given
discretizations have a common refinement in this sense, so to prove
the lemma it suffices to prove that $\bold\Psi_{(s_n)}$ is homotopic
to $\bold\Psi_{(t_n)}$.

Let $x^0_1 \leq x^0_2 \leq\cdots \leq x^0_{m_0}$ be the elements from
$\{s_n\}$ which lie between the first occurence of $t_0$ and the last
occurence of $t_{1}$. For $n \geq 1$, let $x^n_1 \leq x^n_2 \leq\cdots
\leq x^n_{m_n}$ be the elements from $\{s_n\}$ which lie between the
last occurence of $t_n$ and the last occurence of $t_{n+1}$. For each
$t \in [0,1]$, each $n$ and each $i = 1,2, \cdots , m_n$, set
$$
y^{n,i}_t \ = \ tx^n_i + (1-t)t_{n+1} \ .
$$
Let $s^t_n, n = 0,1,2, \cdots $, denote the sequence obtained by reading
the list
\begin{equation*}
\begin{split}
&t_0 ,y^{0,1}_t , y^{0,2}_t, \cdots , y^{0,m_0}_t, \\
&t_1 , y^{1,1}_t , y^{1,2}_t, \cdots , y^{1,m_1}_t, \\
&t_2 , y^{2,1}_t , y^{2,2}_t, \cdots , y^{2,m_2}_t, \\
& \vdots  \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \vdots
\end{split}
\end{equation*}
Then $[0,1] \ni t \mapsto \bold\Phi^0_{(s^t_n)}$ is path of completely
positive contractions such that $ t \mapsto  \bold\Phi^0_{(s^t_n)}(a)$
is normcontinuous for all $a \in A$,
$\bold\Phi^0_{(s^t_n)}(a)\bold\Phi^0_{(s^t_n)}(b) \ - \
\bold\Phi^0_{(s^t_n)}(ab) \ \in \ \mathcal K \otimes B$ for all
$a,b \in A$, $ \bold\Phi^0_{(s^0_n)} \ = \  \bold\Phi^0_{(s_n)}$ and
$ \bold\Phi^0_{(s^1_n)} \ = \  \bold\Phi^0_{(t^{\prime}_n)}$, where
$\{t^{\prime}_n\}$ is the sequence
\begin{equation*}
t_0, \underset{m_0 \
\text{repetitions}}{\underbrace{t_1, t_1, \cdots , t_1}},
t_1, \underset{m_1 \ \text{repetitions}}{\underbrace{t_2,t_2, \cdots ,
t_2}},t_2, \underset{m_2 \ \text{repetitions}}{\underbrace{t_3,t_3,
\cdots , t_3}},t_3,t_4,t_4, \dots
\end{equation*}
 By using convex combinations of $t_1$ and $t_2$, we construct in the
same way a path $[0,1] \ni t \mapsto \bold\Phi^0_{(s^{1,t}_n)}$ of
completely positive contractions such that  $ t \mapsto
\bold\Phi^0_{(s^{1,t}_n)}(a)$ is norm-continuous for all $a \in A$,
$\bold\Phi^0_{(s^{1,t}_n)}(a)\bold\Phi^0_{(s^{1,t}_n)}(b) \ - \
\bold\Phi^0_{(s^{1,t}_n)}(ab) \ \in \ \mathcal K \otimes B$ for all
$a,b \in A$, $ \bold\Phi^0_{(s^{1,0}_n)} \ = \
\bold\Phi^0_{(t^{\prime}_n)}$ and $ \bold\Phi^0_{(s^{1,1}_n)} \ = \
\bold\Phi^0_{(t^{\prime \prime}_n)}$, where $\{t^{\prime \prime }_n\}$
is the sequence
\begin{equation*}
t_0, t_1, \underset{m_0 + m_1 +1 \
\text{repetitions}}{\underbrace{t_2,t_2,t_2, \cdots , t_2}},
\underset{m_2 \ \text{repetitions}}{\underbrace{t_3,t_3, \cdots ,
t_3}},t_3, \underset{m_3 \ \text{repetitions}}{\underbrace{t_4,t_4,
\dots ,t_4}},t_4, t_5 \cdots
\end{equation*}
By proceding in this way, using convex combinations of $t_2$ and $t_3$,
and then $t_3$ and $t_4$ etc., we get a path $[0,1[ \ \ni \ t \ \mapsto
\   \bold\Psi_t$, of completely positive contractions such that
$ t \mapsto  \bold\Psi_t(a)$ is norm-continuous for all $a \in A$,
$\bold\Psi_t(a)\bold\Psi_t(b) \ - \ \bold\Psi_t(ab) \ \in \ \mathcal K
\otimes B$ for all $a,b \in A$, $ \bold\Psi_0 \ = \  \bold\Phi^0_{(s_n)}$
and $ \bold\Psi_{1 - \frac{1}{k}} \ = \  \bold\Phi^0_{(w^k_n)}, k \geq 3$,
where $\{w^k_n\}$ is the sequence
\begin{equation*}
t_0, t_1, \cdots , t_{k-2}, \underset{m_0 + m_1 + \cdots + m_{k-2} +1 \
\text{repetitions}}{\underbrace{t_{k-1},t_{k-1}, \cdots , t_{k-1}}},
\underset{m_k \ \text{repetitions}}{\underbrace{t_{k},t_{k}, \cdots ,
t_{k}}},t_{k}, \underset{m_{k+1} \
\text{repetitions}}{\underbrace{t_{k+1},t_{k+1}, \dots ,t_{k+1}}},t_{k+1},
\cdots
\end{equation*}
Set $\bold\Psi_1 \ = \  \bold\Phi^0_{(t_n)}$. For each $a \in A$, define
$\bold\Psi(a) \in M(I\mathcal K \otimes B)$ by
$$
(\bold\Psi(a)f)(t) \ = \ \bold\Psi_t(a)f(t) \ \ ,  \ \ t \in [0,1] \ ,
\ f \in I\mathcal K \otimes B \ .
$$
It is easy to see that $q_{I \mathcal K \otimes B} \circ \bold\Psi$ is a
$*$-homomorphism. If $T_0$ denotes the canonical image of $V_0$ in
$M(I\mathcal K \otimes B)$, we have that
$$
\bold\Psi(a)T_0 \ - \ T_0\bold\Psi(a) \ \in \ I \mathcal K \otimes B
$$
for all $a \in A$. This follows from the construction of $\bold\Psi$ and
condition b) of Lemma \ref{ae1}. Hence $\bold\Psi$ gives rise to an
extension of $C(\Bbb T) \otimes A$ by $I \mathcal K \otimes B$, and
the argument from the proof of Lemma \ref{ae2} shows that this extension
is semi-split. It is then clear that it gives a homotopy connecting
$\bold\Psi_{(s_n)}$ to $\bold\Psi_{(t_n)}$.
\end{proof}
\end{lemma}


\bigskip


By Lemma \ref{ae3} $[\bold\Psi_{(t_n)}] \in \Ext^{-1}(C(\Bbb T) \otimes A,
\mathcal K \otimes B )$ is independent of the sequence $\{t_n\}$ we choose
to discretize $\{\varphi_t\}$ with, and consequently we denote this element
of the extension group by $\bold\Psi_{\varphi}$.


\bigskip
\begin{lemma}\label{ae4} Let $\{\varphi^1_t\}_{t \in [0,\infty)}, \
\{\varphi^2_t\}_{t \in [0,\infty)} \ : A \to B$ be completely positive
asymptotic homomorphism which are homotopic. It follows that
$$
\bold\Psi_{\varphi^1} \ = \ \bold\Psi_{\varphi^2}
$$
in $\Ext^{-1}(C(\Bbb T) \otimes A, \mathcal K \otimes B)$.
\begin{proof} Let $\{\psi_t\}_{t \in [0,\infty)} : A \to IB$ be a
completely positive asymptotic homomorphism given rise to a homotopy
between $\{\varphi^1_t\}$ and $\{\varphi^2_t\}$. Let $\{\psi_{t_n}\}$
be a discretization of $\{\psi_t\}$. Since a refinement of the sequence
$\{t_n\}$ will still be a discretization we may assume that
$\{\varphi^1_{t_n}\}$ and $\{\varphi^2_{t_n}\}$ are discretizations of
$\{\varphi^1_t\}$ and $\{\varphi^2_t\}$, respectively. It is then obvious
that $\bold\Psi_{\psi} \in \Ext^{-1}(C(\Bbb T)\otimes A, I\mathcal K
\otimes B)$ defines a homotopy connecting $\bold\Psi_{\varphi^1}$ and
$\bold\Psi_{\varphi^2}$.
\end{proof}
\end{lemma}


\bigskip

It follows that we get a well-defined map
$$
E_0 : [[A,B]]_{cp} \ \to \ \Ext^{-1}(C(\Bbb T) \otimes A, \mathcal K
\otimes B) \ ,
$$
given by $E_0[\varphi] \ = \ \bold\Psi_{\varphi}$. There is an obvious
map $\Ext^{-1}(C(\Bbb T) \otimes A,  \mathcal K \otimes B)  \ \to \
\Ext^{-1}(SA, \mathcal K \otimes B)$ induced by the embedding
$S \subset C(\Bbb T)$, and by composing with $E_0$ we get a map
$$
E : [[A,B]]_{cp} \ \to \ \Ext^{-1}(SA, B \otimes \mathcal K) \ .
$$





To compare the map $E$ with other general constructions we remind the
reader of the existence of an asymptotic homomorphism
$$
\chi \ = \ \{\chi_t\}_{t \in [0,\infty)} \ : \ SC(\Bbb T) \ \to \
\mathcal K \
$$
coming, via the Connes-Higson construction, from the Toeplitz extension :
\begin{equation*}
\begin{xymatrix}{
0 \ar[r] & {\mathcal K} \ar[r] &  T \ar[r]  &  C(\Bbb T) \ar[r] & 0 \ . }
\end{xymatrix}
\end{equation*}
To describe it, choose a sequence of continuous functions
$\kappa_n : [0, \infty) \to [0,1], \ n =0,1,2,3, \cdots$, such that

\eq{\label{aee20} \lim_{t \to \infty} \kappa_n(t) \ = \ 1 \ , \ n
\in \Bbb N \ , }
\eq{\label{aee5}
\lim_{n \to \infty} \kappa_n(t) \ = \ 0 \ \ ,  \ \ t \in [0,\infty)\  , }
and
\eq{\label{aee6} \lim_{n \to \infty} \sup_{t \in [0,\infty)}
\|\kappa_{n+1}(t) - \kappa_n(t)\| \ = \ 0 \ . }
One way of constructing such a sequence of functions is to set
$a_n \ = \ \sum_{i=1}^n \frac{1}{i}$ and let $\kappa_n, n \geq 1$,
be the function
$$
\kappa_n(t) \ \ = \ \ \cases 0 \ \ , \ \ & t \in [0,a_n] \\  t - a_n \ \ ,
\ \ & t \in [a_n,a_n+1] \\ 1 \ \ ,  \ \ & t \geq a_n+ 1 \endcases
$$

But the actual choice is not important. Let
$s : C(\Bbb T) \to M(\mathcal K)$ be a completely positive map such that
$q_{\mathcal K} \circ s(f) \ = \ f(T)$ for all $f \in C(\Bbb T)$, where
$T$ is the image of the one-sided shift on $l^2(\Bbb N)$. There is then a
completely positive asymptotic homomorphism, $\chi = (\chi_t)_{t \in
[0,\infty)} : SC(\Bbb T) \to \ \mathcal K$ such that
$$
\lim_{t \to \infty} \|\chi_t(f \otimes g) \ - \  diag(f(\kappa_1(t)),
f(\kappa_2(t)),  f(\kappa_3(t)), \cdots \cdots ) s(g)\| \ = \ 0
$$
for all $f \in S, \ g \in C(\Bbb T)$, cf. Lemma 4.1 in \cite{H-LT}.
(Note that we are viewing $\mathcal K$ as the compact operators on
$l_2(\Bbb N)$, rather than $l_2(\Bbb Z)$, as above.) By restricting $\chi$
to $S^2 \subset SC(\Bbb T)$, we get a completely positive asymptotic
homomorphism $\chi_0 \ : \ S^2 \to \mathcal K$ which gives rise to
Bott-periodicity in $E$-theory.

\bigskip

It is important to observe that we can replace the sequence $\kappa_1,
\kappa_2, \kappa_3, \cdots$ in the definition of $\chi$ and $\chi_0$
by any other sequence $\kappa^{\prime}_1, \kappa^{\prime}_2,
\kappa^{\prime}_3, \cdots $ of continuous functions which satisfy
(\ref{aee20}),(\ref{aee5}) og  (\ref{aee6}). Such a different choice
will not change the classes of $\chi$ and $\chi_0$ in $[[SC(\Bbb T),
\mathcal K]]_{cp}$ and $[[S^2, \mathcal K]]_{cp}$, respectively.

\bigskip








\begin{lemma}\label{ae20}   The diagram
\begin{equation*}
\begin{xymatrix}{
\Ext^{-1}(C(\Bbb T) \otimes A,  \mathcal K \otimes B) \ar[dr]^-{CH} &  \\
[[A,B]]_{cp} \ar[u]^-{E_0}  \ar[r]_-{[\varphi] \mapsto [\chi \otimes
\varphi]} &  [[SC(\Bbb T) \otimes A, \mathcal K \otimes B]]_{cp} \  }
\end{xymatrix}
\end{equation*}
commutes.
\begin{proof} Let $\varphi = \{\varphi_t\}_{t \in [0,\infty)} : A  \to B$
be a completely positive asymptotic homomorphism. For any  discretization
$\{\varphi_{t_n}\}$ of $\{\varphi_t\}_{t \in [0,\infty)}$, and any sequence
of functions $(\kappa_n)$ such that  (\ref{aee20}), (\ref{aee5}) and
(\ref{aee6}) hold, set $\varphi_{t_n} = \kappa_n = 0$ for negative integers.
Let $z^j, \ j \geq 0$, denote the polynomial $z \mapsto z^j$.
$CH \circ E_0[\varphi] \in [[SC(\Bbb T) \otimes  A, \mathcal K
\otimes B]]_{cp}$ is then represented by an asymptotic homomorphism
$\psi$ such that
\begin{equation*}
\lim_{t \to \infty} \|\psi_t(f \otimes z^j \otimes a) \ - \
D[f(\kappa_i(t))\varphi_{t_i}(a)]V_0^j\| \ = \ 0 \  ,
\end{equation*}
for all $f \in S$, $  a \in A$ and $j \in \Bbb N$. On the other hand
$[\chi \otimes \varphi]$ is represented by an asymptotic homomorphism
$\lambda$ such that
\begin{equation*}
\lim_{t \to \infty}  \|\lambda_t(f \otimes z^j \otimes a) \ - \
D[f(\kappa_i(t))\varphi_{t}(a)] (PV_0)^j)\| \ = \ 0 \  ,
\end{equation*}
for all $f \in S$, $ a \in A$ and $j \geq 0$. $P$ denotes here the
orthogonal projection $P : l_2(\Bbb Z)\otimes B \to l_2(\Bbb N)\otimes B$.
Note that
\eQ{\spl{
&\lim_{t \to \infty} \|D[f(\kappa_i(t))\varphi_{t_i}(a)]V_0^j \ - \
D[f(\kappa_i(t))\varphi_{t_i}(a) (PV_0)^j \| \\
& = \ \lim_{t \to \infty}  \|D[f(\kappa_i(t))\varphi_{t_i}(a)]PV_0^j \ - \
D[f(\kappa_i(t))\varphi_{t_i}(a)] (PV_0)^j) \ = \ 0 \ . }}
Hence
$$
\lim_{t \to \infty} \|\psi_t(f \otimes z^j \otimes a) \ - \
D[f(\kappa_i(t))\varphi_{t_i}(a)](PV_0)^j\| \ = 0
$$
for all $f, a$ and $j$. By using the freedom in the choice of the
$\kappa_i$'s we can arrange that there is a sequence $0 < m_1 < m_2 <
\cdots $ in $\Bbb N$ such that
\begin{equation}\label{aee7}
\kappa_i(t) \ = 1  \ \ \forall t \in  [t_j,t_{j+1}] , \ i = 1,2,
\cdots , m_j
\end{equation}
and
\begin{equation}\label{aee8}
\kappa_i(t) \ = 0 \ \ \forall t \in  [t_j,t_{j+1}] , \ i \geq m_{j+1} \
\end{equation}
Define a new sequence $s_1 \leq s_2 \leq s_3 \cdots $ in $[0,\infty)$
such that
$$
s_i = 0 \ , \ 0 \leq i < m_1 \ ,
$$
$$
s_{m_1} \ = \ s_{m_1+1} \ = \cdots \ = s_{m_2-1} \ = \  t_1 \ ,
$$
$$
s_{m_2} \ = \ s_{m_2+1} \ = \cdots \ = s_{m_3-1} \ = \  t_2 \ ,
$$
$$
s_{m_3} \ = \ s_{m_3+1} \ = \cdots \ = s_{m_4-1} \ = \  t_3 \ ,
$$
and so on. Then $\{\varphi_{s_n}\}$ is also a discretization of $\varphi$,
and we may therefore assume that
\begin{equation*}
\lim_{t \to \infty} \|\psi_t(f \otimes z^j \otimes a) \ - \
  D[f(\kappa_i(t))\varphi_{s_i}(a))] (PV_0)^j \| \ = \ 0
\end{equation*}
for all $f \in S, \ a \in A$ and $j \in \Bbb N$. After these changes we
have that
\begin{equation*}
\begin{split}
& \lim_{t \to \infty} \|\psi_t(f \otimes z^j \otimes a) \ - \
\lambda_t(f \otimes z^j \otimes a) \| \\
& \leq \ \lim_{t \to \infty} \sup_n \|f(\kappa_n(t))\varphi_{s_{n}}(a) \ - \
f(\kappa_n(t))\varphi_{t}(a)\| \\
&  = \ 0 \ ,
\end{split}
\end{equation*}
when $f \in S$, $a \in A$ and $j \in \Bbb N$. Since elements of the form
$f \otimes z^j \otimes a$ generate $SC(\Bbb T) \otimes A$ as a
$C^*$-algebra, it follows that $\lim_{t \to \infty} \|\psi_t(z) -
\lambda_t(z)\| \ = \ 0$ for all $z \in SC(\Bbb T) \otimes A$. Hence
$[\psi] = [\lambda]$ in $[[SC(\Bbb T) \otimes A, \mathcal K
\otimes B]]_{cp}$.
\end{proof}
\end{lemma}


\bigskip

It follows of course that also the diagram
\begin{equation}\label{ae14}
\begin{xymatrix}{
\Ext^{-1}(SA,  \mathcal K \otimes B) \ar[dr]^-{CH} &  \\
[[A,B]]_{cp} \ar[u]^-{E}  \ar[r]_-{[\varphi] \mapsto [\chi_0 \otimes
\varphi]} &  [[S^2A, \mathcal K \otimes B]]_{cp} \  }
\end{xymatrix}
\end{equation}
commutes, and it is this diagram we shall actually use in the following.

\bigskip

\begin{lemma}\label{ae5} $E_0([[A,B]]^0_{cp}) \ \subseteq \
\Ext_q^{-1}(C(\Bbb T) \otimes A, \mathcal K \otimes B)$.
\begin{proof} Let $\varphi \ = \ (\varphi_t)_{t \in [0,\infty)} \ : \
A \to B$ be a sequentially trivial completely positive asymptotic
homomorphism. It is easy to see that there is a completely positive
asymptotic homomorphism $ (\psi_t)_{t \in [0,\infty)}$ such that
$\psi_n = 0$ for all $n \in \Bbb N$ and $\lim_{t \to \infty} \|\psi_t(a) -
\varphi_t(a)\| \ = \ 0$ for all $a \in A$. We may therefore assume that
$\varphi_n \ = \ 0$ for all $n \in \Bbb N$. We can then find a
discretization $(\varphi_{t_n})$ of $\varphi$ such that there is an
increasing sequence $m_1 < m_2 < m_3 < \cdots $ in $\Bbb N$ with the
property that $m_{i+1} > m_i$ and
$$
t_j = t_{m_i+1}\in \Bbb N \ , \ j = m_i +1, m_i+2, m_i + 3, \cdots ,
m_i + i \ , \
$$
for all $i \in \Bbb N$. For each $k \in \Bbb N$, define a sequence
$(p^k_i)_{i=0}^{\infty}$ in $M(B)$ by
$$
p^k_i \ = \ \cases 1 \ , \ \ & 0 \leq i \leq k \\ 0 \ , \ \ & i > k \ .
\endcases
$$
Let $F : l_2(\Bbb Z) \otimes B \to  l_2(\Bbb Z) \otimes B$ be the
projection
$$
F(b_j)_k \ = \ \cases b_k \ , \ \ & k < 0 \\ 0 , \ \ & k \geq 0
\endcases \ \ \ \ \ , \ \ \ (b_j) \in H_B \ .
$$
Then $P_k \ = \ F + D[(p^k_i)]$ is a projection in $M(\mathcal K
\otimes B)$ for all $k \in \Bbb N$. Note that
\eq{
P_{m_k}\bold\Phi_{(t_n)}^0(a)V_0^j \ = \ \bold\Phi_{(t_n)}(a)^0V_0^jP_{m_k}
\ }
for all $a \in A$ and all $-k <  j <  k$.  It follows that
$$
\lim_{k \to \infty} P_{m_k} z \ - \  zP_{m_k} \ = \ 0
$$
for all $z \in q_{\mathcal K \otimes B}^{-1}(\bold\Psi_{(t_n)}(C(\Bbb T)
\otimes A))$. Since we clearly have that $P_{m_i} \leq P_{m_{i+1}}$,
$\lim_{k \to \infty} P_{m_k}b \ = \ b$ for all $b \in B$, and $P_{m_i}
z \in \mathcal K \otimes B$ for all $i \in \Bbb N$ and all $z \in
q_{\mathcal K \otimes B}^{-1}(\bold\Psi_{(t_n)}(C(\Bbb T) \otimes A))$,
it follows that $\bold\Psi_{\varphi} \ \in \ \Ext_q^{-1}(C(\Bbb T)
\otimes A,\mathcal K \otimes B)$.


\end{proof}
\end{lemma}




\bigskip




\begin{thm}\label{ae7} Assume that $B$ is stable. The maps
$ s_B^* \circ E_0 \ : \ [[SA,B]]_{cp}^0 \to \ \Ext^{-1}_q(S^2A,B)$ and
$CH : \Ext^{-1}_q(S^2A,B) \to [[S^3A,B]]_{cp}^0$ are isomorphisms.
\begin{proof} By using the stabilizing isomorphisms, ${s_A}^* \ : \
\Ext^{-1}_q(\mathcal K \otimes A, B) \ \to  \ \Ext^{-1}_q(\mathcal K
\otimes A, B)$, ${s_{SA}}^* \ : \ [[S \mathcal K\otimes A,B]]_{cp}^0 \
\to \  [[S A,B]]_{cp}^0$ and $ {s_{S^3A}}^* \ : \ [[S^3 \mathcal K
\otimes A,B]]_{cp}^0 \ \to \  [[S^3 A,B]]_{cp}^0$, we may assume that $A$
is stable. By combining the diagram (\ref{ae14}) with the diagram from
Section 4 of \cite{H-LT} we get the following commuting diagram
\begin{equation}\label{d2}
\begin{xymatrix}{
{} &  \Ext^{-1}_q(S^2A, \mathcal K \otimes B) \ar[d]^-{CH} \ar[r]^{S} &
\Ext^{-1}(S^3A, S(\mathcal K \otimes B)) \\
[[SA,B]]_{cp}^0 \ar[r]^-{\lambda} \ar[ur]^-{E_0} &  [[S^3A,\mathcal K
\otimes B]]_{cp}^0   \ar[ur]^-{M} & {}   , }
\end{xymatrix}
\end{equation}
where $\lambda[\varphi] \ = \ [\chi_0 \otimes \varphi]$. Since $A$ is
stable the inverse of
\begin{equation*}
\begin{xymatrix}{
 [[SA,B]]_{cp}  \ar[r]^-{\lambda} &   [[S^3A,\mathcal K \otimes B]]_{cp} }
\end{xymatrix}
\end{equation*} is given by $[\varphi] \to [ \psi_1 \circ \varphi \circ
\psi_2]$, where $\psi_1 :
\mathcal K \otimes B \to B$ and $\psi_2 : SA \to S^3A$ are both
$*$-homomorphisms. This inverse
clearly takes $ [[S^3A,\mathcal K \otimes B]]_{cp}^0 $ into
$[[SA,B]]_{cp}^0$, so we see
that the $\lambda$ of the above diagram is an isomorphism. Since the
suspension map $S$ in the diagram is injective
(being the restriction to $\Ext^{-1}_q(S^2A, \mathcal K \otimes B)$ of
an isomorphism), we conclude from the diagram
that both the CH-map and the $E_0$-map are isomorphisms.


\end{proof}
\end{thm}


\bigskip
\begin{cor}\label{ae8} Assume that $B$ is stable and has an approximate
unit consisting of projections. Then $Q(S^2A,B) \ = \
\Ext^{-1}_q(S^2A,B)$.
\begin{proof} Let $q_1 \leq q_2 \leq q_3 \leq \cdots $ be an approximate
unit for $B$ consisting of projections. If we let $n_1 < n_2 < n_3 <
\cdots $ be a sufficiently rapidly increasing sequence, and we if use
the sequence
$$
p^k_i \ = \ \cases q_{n_k} \ , \ \ & 0 \leq i \leq k \\ 0 \ , \ \ & i >
k \  \endcases
$$
in the proof of Lemma \ref{ae5}, we see that $E_0([[SA,B]]_{cp}^0)
\subseteq Q(S^2A,\mathcal K \otimes B)$. It follows then from Theorem
\ref{ae7} that $Q(S^2A,\mathcal K \otimes B) \ = \ \Ext^{-1}_q(S^2A,
\mathcal K \otimes B)$. Since $B$ is stable, $\mathcal K \otimes B \
\simeq \ B$, and the conclusion follows.
\end{proof}
\end{cor}

\bigskip


\begin{thm}\label{ae9} Assume that $B$ is stable. The map $CH \ : \
\Ext^{-1}_q(SA,B) \ \to \ [[S^2A, B]]_{cp}^0$ is an isomorphism. If
$B$ has an approximate unit consisting of projections we have in addition
that $Q(SA,B) \ = \ \Ext^{-1}_q(SA,B)$.
\begin{proof} We may assume that $A$ is stable. There is then a
$*$-homomorphism $\psi : SA \to S^3A$ which is invertible in $KK$.
There is therefore a commuting diagram
\begin{equation*}
\begin{xymatrix}{
\Ext^{-1}_q(S^3A,B) \ar[d]_-{CH} \ar[r]^-{\psi^*} &  \Ext^{-1}_q(SA,B)
\ar[d]^-{CH} \\
[[S^4A,B]]_{cp}^0 \ar[r]^-{\psi^*}  &  [[S^2A,B]]_{cp}^0 \ . }
\end{xymatrix}
\end{equation*}
By Theorem \ref{ae7} the $CH$-map to the left is an isomorphism, and it
follows from Theorem \ref{DL}, that
$$
\psi^* \ : \ [[S^4A,B]]_{cp} \to  [[S^2A,B]]_{cp}
$$
is an isomorphism. The inverse is obtained by taking the composition
product with an element of $[[S^4A,S^2A]]_{cp}$, so it follows from
Lemma \ref{prod}  that also the $\psi^*$ in the diagram is an isomorphism.
Therefore the right $CH$-map is surjective. By Theorem \ref{ch1} this map
is also injective, and hence an isomorphism. Since $\psi^*(Q(S^3A,B))
\subseteq Q(SA,B)$, and $Q(S^3A,B) \ = \ \Ext^{-1}_q(S^3A,B)$ when $B$
has an approximate unit consisting of projections, by Corollary \ref{ae8},
it follows that $Q(SA,B) \ = \ \Ext^{-1}_q(SA,B)$ in this case.
\end{proof}
\end{thm}


\bigskip


\section{Putting the pieces together}


Let $A$ be a $C^*$-algebra in $\mathcal N$ - the class of $C^*$-algebras
for which the UCT holds (with any $B$). Then the map
$$
\tau \ : \ \Ext^{-1}_q(A,B) \ \to \ \Pext(K_*(A), K_{*}(B))
$$
is injective. Indeed, when $\delta \ : \  \Pext(K_*(A), K_{*}(B)) \to \
\Ext^{-1}(A,B)$ is the injection from the UCT we have that $\delta \circ
\tau$ is the inclusion $\Ext^{-1}_q(A,B) \ \subseteq \ \Ext^{-1}(A,B)$.


\begin{lemma}\label{pp1} Let $A$ be a unital abelian $C^*$-algebra, and
assume that $B$ has an approximate unit consisting of projections. Then
$$
\tau \ : \ \Ext^{-1}_q(A,B) \ \to \ \Pext(K_*(A), K_{*}(B))
$$
and
$$
\tau \ : \  \Ext^{-1}_q(SA,SB) \ \to \  \Pext(K_*(SA), K_{*}(SB)) \ \simeq
\ \Pext(K_*(A), K_{*}(B))
$$
are isomorphisms.
\begin{proof} Being abelian $A$ is automatically quasidiagonal relative
to $B$. It follows therefore from Theorem 1.4 of [S] that already the
restriction of $\tau$ to $Q(A,B)$ is surjective. Since the suspension of
extensions gives us a commuting diagram
\begin{equation*}
\begin{xymatrix}{
\Ext^{-1}_q(A,B) \ar[r]^-{\tau}   \ar[d]_-{S} &  \Pext(K_*(A), K_{*}(B))
\ar@{=}[d] \\
\Ext^{-1}_q(SA,SB) \ar[r]^-{\tau} &  \Pext(K_*(SA), K_{*}(SB))  \ , }
\end{xymatrix}
\end{equation*}
where we now know that the upper $\tau$-map is an isomorphism, and the
lower $\tau$-map is injective, it follows that the lower $\tau$-map is
also an isomorphism.




\end{proof}
\end{lemma}


\begin{thm}\label{main} Let $A$ and $B$ be separable $C^*$-algebras.
Assume that $A$ is $KK$-equivalent to a separable abelian $C^*$-algebra.
It follows that
$$
\Ext^{-1}_q(SA,\mathcal K \otimes B) \ \simeq \ [[S^2A, \mathcal K
\otimes B]]^0_{cp} \ \simeq \   \Pext(K_*(A), K_{*-1}(B)) \ .
$$
The isomorphisms are natural in $A$ and $B$.



\begin{proof}




By combining Lemma \ref{pp1} with Theorem \ref{ae9} we see that there
is an isomorphism
$$
[[SA, \mathcal K \otimes B]]^0_{cp} \ \simeq \ [[S^2A, S\mathcal K
\otimes B]]^0_{cp} \ \simeq \ \Ext^{-1}_q(SA,S\mathcal K \otimes B) \
\simeq \  \Pext(K_*(A), K_{*}(B)) \ ,
$$
when $A$ is a unital abelian $C^*$-algebra and $B$ has an approximate
unit consisting of projections. The combined map,
$$
[[SA, \mathcal K \otimes B]]^0_{cp} \ \to \   \Pext(K_*(A), K_{*}(B)) \ ,
$$
can be defined for arbitrary (separable) $C^*$-algebras, $A$ and $B$,
and is natural in both variables, not only with respect to
$*$-homomorphisms but in fact with respect to the pairings with
$KK$-theory. The last assertion may be deduced from Theorem 4.9 of
\cite{T2}. Hence, by adjoining units and using the split-exactness of
all functors involved, we see first that the map is an isomorphism for
any (separable) abelian $A$ and any (separable) $B$, and then that
it is also an isomorphism for any separable $B$, and any separable
$A$ which is $KK$-equivalent to a separable abelian $C^*$-algebra.
So for any such $A$ and $B$ we have that
$$
\Ext^{-1}_q(SA,\mathcal K \otimes B) \ \simeq \ [[S^2A, \mathcal K
\otimes B]]^0_{cp} \ \simeq \   \Pext(K_*(A), K_{*-1}(B))
$$
by Theorem \ref{ae9}  .
\end{proof}
\end{thm}








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\end{document}
