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\begin{document}
\title[On homotopical non-invertibility of $C^*$-extensions]
{On homotopical non-invertibility of $C^*$-extensions}
%----------Author 1
\author{Vladimir Manuilov}
\address{%
Dept. of Mechanics and Mathematics\\ Moscow State University\\
Leninskie Gory, Moscow\\ 119992 Russia}
\email{manuilov@mech.math.msu.su}
\thanks{The first named author was partially supported
by RFFI grant 05-01-00923.}
%----------Author 2
%\author{Klaus Thomsen}
%\address{%
%IMF, Department of Mathematics\\ Ny Munkegade, 8000\\Aarhus C,
%Denmark}
%\email{matkt@imf.au.dk}
%----------classification, keywords, date
\subjclass[2000]{Primary 46L80; Secondary 19K33 46L05}
\keywords{$C^*$-algebra, $C^*$-extension, homotopy}
%\date{January 1, 2004}
%----------additions
%\dedicatory{To my boss}
%%% ----------------------------------------------------------------------
\begin{abstract}
We have presented recently an example of a $C^*$-extension, which
is not invertible in the semigroup of homotopy classes of
$C^*$-extensions. Here we reveal the cause for existence of
homotopy non-invertible $C^*$-extensions: it is related to
non-exact $C^*$-algebras and to possibility to distinguish
different tensor $C^*$-norms by $K$-theory. We construct a special
$C^*$-algebra, $K$-theory of which hosts an obstruction for
homotopical non-invertibility, and show that this obstruction for
our example does not vanish.
\end{abstract}
%%% ----------------------------------------------------------------------
\maketitle
%%% ----------------------------------------------------------------------
%\tableofcontents
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
A $C^*$-extension is a short exact sequence
\begin{equation}\label{e}
\begin{xymatrix}{
0\ar[r]&B\ar[r]&E\ar[r]&A\ar[r]&0
}\end{xymatrix}
\end{equation}
of $C^*$-algebras. The classical Brown--Douglas--Fillmore theory
\cite{BDF} classifies $C^*$-extensions of $A$ by $B$, i.e.
different $E$'s in (\ref{e}), up to stable unitary equivalence
modulo split $C^*$-extensions and results in a functor
$\Ext(A,B)$, which turns out to be an abelian semigroup in the
case when $B$ is stable, i.e. $B\cong B\otimes\K$, where $\K$
denotes the $C^*$-algebra of compact operators.
The Brown--Douglas--Fillmore theory works nicely for
$C^*$-extensions of {\it nuclear} $C^*$-algebras because of the two
features:
\begin{enumerate}
\item
$\Ext(A,B)$ is a group, i.e. each $C^*$-extension is invertible;
\item
$\Ext(A,B)$ is homotopy invariant.
\end{enumerate}
Both of them do not hold in general! The first example of a
non-invertible $C^*$-extension was constructed by J.~Anderson
\cite{Anderson}, then a lot of other examples followed,
\cite{Wass,Wass1,Kirchberg,Ozawa,Haagerup}. Moreover, the example
by E.~Kirchberg, \cite{Kirchberg}, gives a non-invertible (hence
non-trivial) $C^*$-extension of a {\it contractible} (i.e.
homotopy equivalent to 0) $C^*$-algebra, thus showing that $\Ext$
is not homotopy invariant in general.
Besides the stable unitary equivalence, there is another natural
equivalence relation for $C^*$-extensions --- the homotopy
equivalence. It gives another functor, $\Ext_h(A,B)$, of homotopy
classes of $C^*$-extensions. As the homotopy equivalence is weaker
than the stable unitary equivalence modulo split $C^*$-extensions,
so the identity map gives rise to a surjective semigroup
homomorphism $\Ext(A,B)\to\Ext_h(A,B)$. Since $\Ext_h$ is patently
homotopy invariant, it was interesting to know, if
non-invertibility of $C^*$-extensions persists on the homotopy
level, in other words, if $\Ext_h$ is a group. In \cite{MT8} we
provided the first example of a non-invertible element of
$\Ext_h$.
The aim of this paper is to reveal the cause for existence of
homotopy non-invertible $C^*$-extensions (more exactly, one of the
causes, the only one known to us yet).
It was first understood by S.~Wassermann \cite{Wass2} that
non-invertibility (in $\Ext$) of $C^*$-extensions can be revealed
by using non-exact $C^*$-algebras. Recall that a $C^*$-algebra $D$
is exact if, after taking the minimal tensor $C^*$-product of $D$
by the $C^*$-algebras from (\ref{e}), the sequence
$$
\begin{xymatrix}{
0\ar[r]&B\otimes_{\min}D\ar[r]&E\otimes_{\min}D\ar[r]&A\otimes_{\min}D\ar[r]&0
}\end{xymatrix}
$$
is exact as well. Wassermann's idea was that if the $C^*$-norm on
the algebraic tensor product $A\odot D$ inherited from
$E\otimes_{\min}D/B\otimes_{\min}D$ is strictly greater than the
minimal tensor $C^*$-norm then this prevents (\ref{e}) from being
invertible. This is not enough for homotopy non-invertibility, but
here the following idea of G.~Skandalis (cf.
\cite{Higson-Skandalis}) can help: the difference between tensor
$C^*$-norms on $A\odot D$ can sometimes be detected on the level
of $K$-theory.
The next ingredient is to find an appropriate $C^*$-algebra $D$
and then to construct an ideal $R_{B,D}$ in the corona
$C^*$-algebra $Q(B\otimes_{\min}D)$ with two properties: firstly,
$R_{B,D}$ should have a non-trivial $K_0$-group and, secondly, the
images, under the Busby invariant of the extension (\ref{e}), of
all elements of $E\otimes_{\min}D/B\otimes_{\min}D$ that vanish in
$A\otimes_{\min}D$, should lie in $R_{B,D}$. Then, in order to
produce a homotopy non-invertible $C^*$-extension, it suffices to
find a projection in $E\otimes_{\min}D/B\otimes_{\min}D$ that
vanishes in $A\otimes_{\min}D$ and which is mapped by the Busby
invariant of the extension (\ref{e}) into a projection in
$R_{B,D}$, the $K$-theory class of which is non-trivial.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Basic definitions}
For a $C^*$-algebra $B$, let $M(B)$ denote the multiplier
$C^*$-algebra of $B$, which can be defined as the maximal
$C^*$-algebra that contains $B$ as an essential ideal. Let
$Q(B)=M(B)/B$ be the corresponding corona $C^*$-algebra and let
$q:M(B)\to Q(B)$ be the quotient $*$-homomorphism. Recall that any
extension (\ref{e}) gives rise in a natural way to a
$*$-homomorphism $E\to M(B)$, which passes to quotients: $A\to
Q(B)$. The latter is called the Busby invariant of the extension
(\ref{e}). It is well-known that, up to reasonable equivalence, a
$C^*$-extension can be recovered from its Busby invariant, so we
will not distinguish $C^*$-extensions and their Busby invariants.
Let $IB=C[0,1]\otimes B$ (we use notation $\otimes$ for the $C^*$-algebra
tensor product when it is unique, i.e. when there is only one $C^*$-norm
on the algebraic tensor product). The evaluation map $\ev_s:IB\to B$,
$s\in[0,1]$, gives rise to the map $\widehat{\ev}_s:Q(IB)\to Q(B)$. Two
extensions of $A$ by $B$ are called {\it homotopic} if there exists a
$*$-homomorphism $\Phi:A\to Q(IB)$ such that $\widehat{\ev}_0\circ\Phi$
and $\widehat{\ev}_1\circ\Phi$ coincide with the Busby invariants of the
two $C^*$-extensions. The set $\Ext_h(A,B)$ is defined as the set of
homotopy classes of $C^*$-extensions of $A$ by $B$.
If $B$ is stable, $B\cong B\otimes\K$, then the set $\Ext_h(A,B)$
has a natural semigroup structure. Stability of $B$ implies
existence of two elements $V_1,V_2\in M(B)$ with the properties
$$
V_1^*V_1=1_{M(B)}, \ \ \ V_2^*V_2=1_{M(B)}, \ \ \
V_1V_1^*+V_2V_2^*=1_{M(B)}.
$$
Let $\tau,\tau':A\to Q(B)$ be the Busby invariants of two extensions. Then
the sum of these two extensions is defined to have the Busby invariant
given by
$$
\widetilde{\tau}(a)=\Ad_{v_1}(\tau(a))+\Ad_{v_2}(\tau'(a)), \ \
a\in A,
$$
where $v_i=q(V_i)\in Q(B)$, $i=1,2$, and $\Ad_v(x)=vxv^*$. It is
well-known that this definition, up to homotopy, does not depend
on choice of $V_1$ and $V_2$. More detailed information on
$C^*$-extensions can be found in \cite{Blackadar}.
Besides the minimal and the maximal tensor $C^*$-norm on algebraic
tensor products we use one more tensor $C^*$-norm, the one defined
by the extension (\ref{e}). Assume that there is a unique tensor
$C^*$-norm on the algebraic tensor product $B\odot D$. Then,
thanks to the exactness of the maximal tensor product,
$E\otimes_{\min}D/B\otimes D$ is a quotient of
$A\otimes_{\max}D=E\otimes_{\max}D/B\otimes D$. On the other hand,
$A\otimes_{\min}D$ is the quotient of $E\otimes_{\min}D/B\otimes
D$. Therefore $A\odot D$ is a dense subspace in
$E\otimes_{\min}D/B\otimes D$. We denote the norm on $A\odot B$
inherited from $E\otimes_{\min}D/B\otimes D$ by $\|\cdot\|_E$.
Since this norm is a cross-norm, we may view
$E\otimes_{\min}D/B\otimes D$ as a tensor product of $A$ and $D$
and write $E\otimes_{\min}D/B\otimes D=A\otimes_E D$. If
$\|\cdot\|_\alpha$ and $\|\cdot\|_\beta$ are tensor $C^*$-norms on
$A\odot D$ and if $\|x\|_\alpha\geq\|x\|_\beta$ for any $x\in
A\odot D$ then we denote the canonical surjective $*$-homomorphism
by $q_{\alpha,\beta}:A\otimes_{\alpha}D\to A\otimes_{\beta}D$. For
more information on tensor products of $C^*$-algebras we refer to
\cite{Wass-book}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{A canonical $*$-homomorphism induced by extensions}
Let $D_k$, $k\in\mathbb N$, be a sequence of nuclear
$C^*$-algebras and let $\prod_{k\in\mathbb N}D_k$ be a
$C^*$-algebra of norm-bounded sequences $(d_1,d_2,\ldots)$,
$d_k\in D_k$. Denote by $p_k:\prod_{k\in\mathbb N}D_k\to D_k$ the
canonical projection. Let $D\subset \prod_{k\in\mathbb N}D_k$ be a
$C^*$-algebra such that $(p_k)|_D$ is surjective for each
$k\in\mathbb N$. Without abuse of notation we denote the
restriction $(p_k)|_D$ also by $p_k$. We also assume that there is
only one tensor $C^*$-norm on the algebraic tensor product $B\odot
D$. The case we are most interested is when the $C^*$-algebras
$D_k=M_{n_k}$ are the matrix algebras of increasing dimension
$n_k$ and the `big' $C^*$-algebra $D$ is not exact.
Denote by $J_{A,D}$ the kernel of the canonical surjection
$$
\begin{xymatrix}{
q_{\max,\min}:A\otimes_{\max}D\ar[r]& A\otimes_{\min}D.
}\end{xymatrix}
$$
Let $\tau:A\to Q(B)$ be the Busby invariant of the $C^*$-extension
(\ref{e}) and let $i:Q(B)\to Q(B\otimes D)$ and $j:D\to Q(B\otimes
D)$ be the canonical $*$-homomorphisms. Then $i\circ\tau(A)$ and
$j(D)$ commute in $Q(B\otimes D)$, hence there is a well-defined
$*$-homomorphism
$$
\begin{xymatrix}{
\tau_D:A\otimes_{\max}D\ar[r]& Q(B\otimes D)
}\end{xymatrix}
$$
due to the universal property of the maximal tensor product of
$C^*$-algebras. Another description of the map $\tau_D$ is given
by composing the $*$-homomorphism $q_{\max,E}:A\otimes_{\max}D\to
A\otimes_E D$ with the Busby invariant of the $C^*$-extension
$$
\begin{xymatrix}{
0\ar[r]&B\otimes D\ar[r]& E\otimes_{\min}D\ar[r]&A\otimes_E
D\ar[r]&0.
}\end{xymatrix}
$$
Consider the $*$-homomorphism
$$
\begin{xymatrix}{
\id_A\otimes p_k:A\otimes_{\max}D\ar[r]&A\otimes D_k.
}\end{xymatrix}
$$
\begin{lem}\label{Lemma1}
Let $x\in A\otimes_{\max}D$. Then $x\in J_{A,D}$ if and only if \
$\id_A\otimes p_k(x)=0$ for all $k\in\mathbb N$.
\end{lem}
\begin{proof}
This follows from injectivity of the canonical map
$A\otimes_{\min}D\to \prod_{k\in\mathbb N}(A\otimes D_k)$ and from
commutativity of the diagram
$$
\begin{xymatrix}{
A\otimes_{\max}D\ar[rr]\ar[drr]_-{\prod_{k\in\mathbb
N}\id_A\otimes p_k}&& A\otimes_{\min}D\ar[d]\\
&&\prod\nolimits_{k\in\mathbb N}(A\otimes D_k).
}\end{xymatrix}
$$
\end{proof}
Let
$$
\begin{xymatrix}{
\chi:Q(B\otimes_{\min}D)\ar[r]& M(B\otimes D)
}\end{xymatrix}
$$
be a continuous selfadjoint homogeneous section satisfying
$\|\chi(y)\|<2\|y\|$ for any $y\in Q(B\otimes D)$ (such sections
exist by the Bartle--Graves theorem, \cite{Bartle-Graves,Loring}).
Since the map $\id_B\otimes p_k:B\otimes D\to B\otimes D_k$ is
surjective, it extends to a surjective $*$-homomorphism
\begin{equation}\label{id_p}
\begin{xymatrix}{
\overline{\id_B\otimes p_k}:M(B\otimes D)\ar[r]& M(B\otimes D_k).
}\end{xymatrix}
\end{equation}
\begin{lem}\label{Lemma2}
If $x\in J_{A,D}$ then
$
\overline{\id_B\otimes p_k}\circ\chi\circ\tau_D(x)\in B\otimes D_k
$
for any $k\in\mathbb N$.
\end{lem}
\begin{proof}
Let
$$
\begin{xymatrix}{
\chi_k:Q(B\otimes D_k)\ar[r]& M(B\otimes D_k), \ \ k\in\mathbb N,
}\end{xymatrix}
$$
be continuous selfadjoint homogeneous sections satisfying
$\|\chi_k(z)\|<2\|z\|$ for any $z\in Q(B\otimes D_k)$. Denote by
$$
\begin{xymatrix}{
\tau\widehat{\otimes}\id_{D_k}:A\otimes D_k\ar[r]& Q(B\otimes D_k)
}\end{xymatrix}
$$
the $*$-homomorphism obtained by composing
$\tau\otimes\id_{D_k}:A\otimes D_k\to Q(B)\otimes D_k$ with the
canonical embedding $Q(B)\otimes D_k\subset Q(B\otimes D_k)$.
The $*$-homomorphism (\ref{id_p}) passes to quotients:
$$
\begin{xymatrix}{
\widehat{\id_B\otimes p_k}: Q(B\otimes D)\ar[r]& Q(B\otimes D_k)
}\end{xymatrix}
$$
and by checking on simple tensors one finds that
$$
\widehat{\id_B\otimes p_k}\circ\tau_D=(\widehat{\id_B\otimes
p_k})\circ(\id_A\otimes p_k),
$$
which implies that
\begin{equation}\label{2}
\overline{\id_B\otimes
p_k}\circ\chi\circ\tau_D(x)-\chi_k\circ(\widehat{\id_B\otimes
p_k})\circ(\id_A\otimes p_k)(x)\in B\otimes D_k
\end{equation}
for any $k\in\mathbb N$ and for any $x\in A\otimes_{\max}D$. By
Lemma \ref{Lemma1}, if $x\in J_{A,D}$ then $\id_A\otimes p_k(x)=0$
for any $k\in\mathbb N$, so due to homogeneity of $\chi_k$ it
follows from (\ref{2}) that $\overline{\id_B\otimes
p_k}\circ\chi\circ\tau_D(x)\in B\otimes D_k$ for any $k\in\mathbb
N$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Yet another short exact sequence}
Consider the $*$-homomorphism
\begin{equation}\label{P}
\begin{xymatrix}{
P=\prod\nolimits_{k\in\mathbb N}\overline{\id_B\otimes p_k}:
M(B\otimes D)\ar[r]& \prod\nolimits_{k\in\mathbb N}M(B\otimes D_k)
}\end{xymatrix}
\end{equation}
and denote by $N_{B,D}$ the preimage of $P$ of the ideal
$\prod_{k\in\mathbb N}B\otimes D_k\subset \prod_{k\in\mathbb
N}M(B\otimes D_k)$,
$$
N_{B,D}=P^{-1}\Bigl(\prod\nolimits_{k\in\mathbb N}B\otimes
D_k\Bigr)\subset M(B\otimes D).
$$
Since $B\otimes D\subset N_{B,D}$ is an ideal in $M(B\otimes D)$,
is it an ideal in $N_{B,D}$ as well. Denote the corresponding
quotient $C^*$-algebra $N_{B,D}/B\otimes D$ by $R_{B,D}$.
\begin{lem}\label{Lemma3}
$R_{B,D}$ is an ideal in $Q(B\otimes D)$.
\end{lem}
\begin{proof}
First notice that $\prod_{k\in\mathbb N}B\otimes D_k$ is an ideal
in $\prod_{k\in\mathbb N}M(B\otimes D_k)$. Notice also that the
map $P$ (\ref{P}) is injective. This implies that $N_{B,D}$ is an
ideal in $M(B\otimes D)$, hence $R_{B,D}$ is an ideal in
$Q(B\otimes D)$.
\end{proof}
Thus we get a short exact sequence
$$
\begin{xymatrix}{
0\ar[r]&R_{B,D}\ar[r]&Q(B\otimes D)\ar[r]&M(B\otimes
D)/N_{B,D}\ar[r]&0.
}\end{xymatrix}
$$
\begin{prop}
For any $*$-homomorphism $\tau:A\to Q(B)$ one has
$\tau_D(J_{A,D})\subset R_{B,D}$. In other words, any
$C^*$-extension (\ref{e}) gives rise to a morphism of extensions
$$
\begin{xymatrix}{
0\ar[r]&J_{A,D}\ar[r]\ar[d]&A\otimes_{\max}D\ar[r]\ar[d]^-{\tau_D}&A\otimes_{\min}D\ar[r]\ar[d]&0\\
0\ar[r]&R_{B,D}\ar[r]&Q(B\otimes D)\ar[r]&M(B\otimes
D)/N_{B,D}\ar[r]&0.
}\end{xymatrix}
$$
\end{prop}
\begin{proof}
This directly follows from the definition of $R_{B,D}$ and from
Lemma \ref{Lemma2}.
\end{proof}
The ideal $R_{B,D}$ is useful because of its non-trivial
$K$-theory.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some $K$-theory calculations}
Let $\nu=\{n_k\}_{k\in\mathbb N}$ be an increasing sequence of
positive integers. Let $\mathbb Z^\infty=\prod_{k\in\mathbb
N}\mathbb Z$ be the abelian group of all (not necessarily bounded)
infinite sequences of integers and let $Z_\nu\subset\mathbb
Z^\infty$ be a subgroup of all sequences $(m_k)_{k\in\mathbb N}$,
$m_k\in\mathbb Z$, satisfying
$$
\sup_{k\in\mathbb N}\frac{|m_k|}{n_k}<\infty.
$$
From now on we set $D_k=M_{n_k}$ and $D=\prod_{k\in\mathbb N}D_k$.
We also assume that $B$ is either $\K$ or $I\K=\K\otimes C[0,1]$.
\begin{prop}\label{P1}
For $B$ and $D$ as above,
\begin{enumerate}
\item
One has $K_0(B\otimes D)\cong Z_\nu$ and $K_1(B\otimes D)=0$;
\item
There is a split surjection $T:K_0(N_{B,D})\to\mathbb Z^\infty$
such that
\begin{enumerate}
\item
it maps the positive cone $K_0(N_{B,D})_+$ of $K_0(N_{B,D})$ into
the positive cone $(\mathbb Z^\infty)_+$ of $\mathbb Z^\infty$;
\item
the composition $T\circ i_*$ coincides with the isomorphism
$K_0(B\otimes D)\cong Z_\nu$, where $i_*:K_0(B\otimes D)\to
K_0(N_{B,D})$ is induced by the inclusion $B\otimes D\subset
N_{B,D}$.
\end{enumerate}
\end{enumerate}
\end{prop}
\begin{proof}
The $K$-groups of $B\otimes D$ are easy to calculate due to
stability of $K$-theory: one has $K_*(B\otimes D)\cong K_*(D)$.
The latter was calculated e.g. in \cite{Dadarlat-Eilers}. So let
us turn to the group $K_0(N_{B,D})$.
Let $i_k : D_k \to D$ be the natural right-inverse of $p_k:D \to
D_k$. and let $V_1,V_2, V_3, \dots $ be a sequence of isometries
in $M(B)$ such that $\sum_{k=1}^{\infty} V_kV_k^* = 1$ with
convergence in the strict topology. Then
$$
\varphi_k = \Ad_{V_k} \otimes i_k : \ B \otimes D_k \to B \otimes
D
$$
is a $*$-homomorphism. Let $q_k = V_kV_k^* \otimes i_k\left(
1_{D_k}\right) \in M(B \otimes D)$.
\begin{lem}\label{Obs1}
Each $\varphi_k$ is quasi-unital, in particular,
$$
\overline{\varphi_k\left(B \otimes D_k\right) B\otimes D} = q_kB
\otimes D.
$$
\end{lem}
\begin{proof}
The inclusion $\subseteq $ is obvious. To prove the reversed
inclusion it suffices to show that $q_k \left( b \otimes d\right)
\in \overline{\varphi_k\left(B \otimes D_k\right)B \otimes D}$ for
every $b \in B$ and every $d = \left(d_1,d_2, d_3, \dots \right)
\in D$. Write $b = b_1b_2$, $b_1,b_2\in B$, and $d_k = xy$,
$x,y\in D_k$. Then
\begin{eqnarray*}
q_k\left(b \otimes d\right)& = &V_kV_k^*b \otimes \left(
0,0,0,..., 0,d_k,0,...... \right) \\ & = & \varphi_k\left( V_k^*
b_1 \otimes x\right)\left(V_kb_2 \otimes \left(0,0,\dots, 0,y,0,0,
\dots\right)\right)\\ & \in & \varphi_k\left(B \otimes D_k\right)B
\otimes D.
\end{eqnarray*}
\end{proof}
Since $B \otimes D_k$ and $B \otimes D$ are both $\sigma$-unital
there is, for each $k$, a $*$-homomorphism $\overline{\varphi}_k:
M\left(B \otimes D_k\right) \to M\left(B \otimes D\right)$ which
extends $\varphi_k$ and is strictly continuous on the unit ball
(see e.g. \cite{K-Jensen-Thomsen}, Corollary 1.1.15).
\begin{lem}\label{Obs2}
For each $x = \left(x_1,x_2, x_3, \dots \right) \in
\prod_{k=1}^{\infty} M\left( B \otimes D_k\right)$, the sum
$$
\sum\nolimits_{k=1}^{\infty} \overline{\varphi}_k\left(x_k\right)
$$
converges in the strict topology of $M\left(B \otimes D\right)$.
\end{lem}
\begin{proof}
Let $z \in B \otimes D$ and $\epsilon > 0$ be arbitrary. It
suffices to show that there is a $K \in \mathbb N$ such that
$$
\left\| \sum\nolimits_{j=N}^M
\overline{\varphi}_j\left(x_j\right)z \right\| \leq \epsilon \|x\|
$$
when $M > N \geq K$. Note first that
$\overline{\varphi}_k\left(1_{D_k}\right) = q_k$ so that
$\overline{\varphi}_i\left(x_i\right)\overline{\varphi}_j\left(x_j\right)
=
\overline{\varphi}_j\left(x_j\right)\overline{\varphi}_i\left(x_i\right)
= 0$ when $i \neq j$. It follows that
$$
\left\| \sum\nolimits_{j=N}^M \overline{\varphi}_j\left(x_j\right)
\right\| \leq \|x\|
$$
for all $M > N$. Set $P_n = \sum_{j=1}^n V_jV_j^*$ and note, by
checking on simple tensors, that $\lim_{n \to \infty}
\left\|\left(P_n \otimes 1_D\right)z - z\right\| = 0$. Choose $K
\in \mathbb N$ such that $\left\|\left(P_K \otimes 1_D\right)z -
z\right\| \leq \epsilon$. Then
$$
\left\| \sum\nolimits_{j =N}^M
\overline{\varphi}_j\left(x_j\right)z\right\| \leq \left\|
\sum\nolimits_{j =N}^M
\overline{\varphi}_j\left(x_j\right)P_Kz\right\| + \|x\| \epsilon
= \|x\|\epsilon
$$
when $M > N \geq K$.
\end{proof}
It follows from Lemma \ref{Obs2} that we can define a
$*$-homomorphism $\Phi : \prod_{k=1}^{\infty} M\left( B \otimes
D_k\right) \to M\left(B \otimes D\right)$ such that
$$
\Phi (m) = \sum\nolimits_{k=1}^{\infty}
\overline{\varphi}_k\left(m_k\right)
$$
where $m = \left(m_1,m_2, m_3, \dots \right) \in
\prod_{k=1}^{\infty} M\left( B \otimes D_k\right)$.
\begin{lem}\label{Obs3}
One has
\begin{enumerate}
\item
$
P \circ \Phi (m) = \left( \Ad_{V_1\otimes 1_{D_1}}(m_1),
\Ad_{V_2\otimes 1_{D_2}}(m_2), \Ad_{V_3\otimes
1_{D_3}}(m_3), \dots \right)
$ \newline
for all $m = \left(m_1,m_2, m_3, \dots \right)$, where $P$ is the
map defined by (\ref{P});
\item
$
\Phi(\prod_{k\in\mathbb N}B\otimes D_k)\subset N_{B,D}.
$
\end{enumerate}
\end{lem}
\begin{proof}
To prove the first assertion, it suffices to show that
$$
\overline{\id_B \otimes p_k} \circ \Phi \left(m_1,m_2, m_3, \dots
\right) = \Ad_{V_k \otimes 1_{D_k}}\left(m_k\right)
$$
for each $k$. To this end let $x \in B \otimes D_k$. Then
\begin{equation*}
\begin{split}
& \overline{\id_B \otimes p_k} \circ \Phi \left(m_1,m_2, m_3,
\dots \right)x \\ & = \overline{\id_B \otimes p_k} \left(\Phi
\left(m_1,m_2, m_3, \dots \right) \left( \id_B \otimes
i_k\right)(x)\right) \\ & = \overline{\id_B \otimes p_k}\left(
\overline{\varphi}_k\left(m_k\right)\left( \id_B \otimes
i_k\right(x)\right) \\ & = \overline{\id_B \otimes p_k} \circ
\overline{\varphi}_k\left(m_k\right)x = (*)
\end{split}
\end{equation*}
When $b \in B, d \in D_k$ we find that
$$
\overline{\id_B \otimes p_k}\circ \overline{\varphi}_k \left( b
\otimes d\right) = \left(\id_B \otimes p_k\right) \circ
\varphi_k\left( b \otimes d\right) = V_kbV_k^* \otimes d .
$$
It follows then by strict continuity of both $\Ad_{V_k \otimes
1_{D_k}}$ and $\overline{\id_B \otimes p_k}\circ
\overline{\varphi}_k$ that $(*) = \Ad_{V_k \otimes
1_{D_k}}\left(m_k\right)x$, completing the proof of the first
assertion of Lemma \ref{Obs3}. The second assertion is obvious.
\end{proof}
\begin{lem}\label{Obs4}
$P_* : K_*\left(N_{B,D}\right) \to K_*\left( \prod_{k=1}^{\infty}
B \otimes D_k\right)$ is a split surjection.
\end{lem}
\begin{proof}
Set $W_k = V_k \otimes 1_{D_k} \in M\left(B \otimes D_k\right)$.
It follows from Lemma \ref{Obs3} that it suffices to show that the
endomorphism $\psi : \prod_{k=1}^{\infty} B \otimes D_k \to
\prod_{k=1}^{\infty} B \otimes D_k$ given by
$$
\psi\left(x_1,x_2,x_3, \dots\right) =
\left(\Ad_{W_1}(x),\Ad_{W_2}(x),\Ad_{W_3}(x), \dots \right)
$$
induces the identity map on K-theory. This follows from standard
arguments by using that $\psi = \Ad_W$, where $W = \left(W_1,W_2,
W_3, \dots \right)$ is an isometry in $M\left(
\prod_{k=1}^{\infty} B \otimes D_k \right)$.
\end{proof}
\begin{lem}\label{Obs5}
$
K_0\left(\prod_{k=1}^{\infty} B \otimes D_k\right) \simeq
\prod_{k=1}^{\infty} K_0\left(B \otimes D_k\right)
$
under the obvious map.
\end{lem}
\begin{proof}
Let
$$
\begin{xymatrix}{
\iota : K_0\left(\prod\nolimits_{k=1}^{\infty} B \otimes
D_k\right) \ar[r]& \prod\nolimits_{k=1}^{\infty} K_0\left( B
\otimes D_k\right)
}\end{xymatrix}
$$
be the map we are considering.
Surjectivity: Let $(x_k) \in \prod_{k=1}^{\infty} K_0\left(B
\otimes D_k\right)$. Since $B$ is stable there are projections
$p_k,q_k \in B\otimes D_k$ such that
$$
x_k = \left[p_k\right] - \left[q_k\right]
$$
for all $k$. Then $p = \left(p_1,p_2,p_3, \dots \right), \ q =
\left(q_1,q_2, \dots \right)$ are projections in
$\prod_{k=1}^{\infty} B \otimes D_k$ and
$\iota\left(\left[p\right] - \left[q\right] \right) = (x_k)$.
Injectivity: Let $x,y \in K_0\left(\prod_{k=1}^{\infty} B \otimes
D_k\right)$ be such that $\iota(x) = \iota(y)$. Note that
$\prod_{k=1}^{\infty} B \otimes D_k$ has an approximate unit of
projections since $B$ does (obviously not a \emph{countable}
approximate unit, but that is irrelevant). It follows therefore
that there is $m \in \mathbb N$ and projections $p,q,p',q' \in
M_m\left( \prod_{k=1}^{\infty} B \otimes D_k\right) =
\prod_{k=1}^{\infty} M_m\left(B \otimes D_k\right)$ such that
$$
x = [p] -[q], \ \ \ y = [p'] - [q'] .
$$
(cf. Proposition 5.5.5 in \cite{Blackadar}). Write $p =
\left(p_1,p_2,p_3, \dots \right), \ q = \left(q_1,q_2,q_3, \dots
\right), \ p' = \left(p'_1,p'_2,p'_3, \dots \right), \ q'
=\left(q'_1,q'_2, q'_3, \dots \right)$, where
$$
p_k,q_k,p'_k,q'_k \in M_m\left( B \otimes D_k\right) .
$$
Then $\left[p_k\right] - \left[q_k\right] = \left[p'_k\right] -
\left[q'_k\right]$ for each $k$ since $\iota(x) = \iota(y)$. Since
$B$ is stable there are projections $r_k \in B \otimes D_k$ such
that $p_k \oplus q'_k \oplus r_k = V_kV_k^*$ and $V_k^*V_k = p'_k
\oplus q_k \oplus r_k$ in $M_{2m+1}\left(B \otimes D_k\right)$ for
some partial isometry $V_k \in M_{2m+1}\left( B \otimes
D_k\right)$. Set $r = \left(r_1,r_2, r_3, \dots \right) \in
\prod_{k=1}^{\infty} B \otimes D_k , \ V = \left(V_1,V_2, \dots
\right) \in M_{2m+1} \left( \prod_{k=1}^{\infty} B \otimes
D_k\right)$ and note that
$$
p \oplus q' \oplus r = VV^*, \ \ \ V^*V = p'\oplus q \oplus r
$$
in $M_{2m+1} \left( \prod_{k=1}^{\infty} B \otimes D_k\right)$.
\end{proof}
Note that our proof holds not only for our $B$, but for a general
stable $C^*$-algebra with an approximate unit of projections. It
does not hold in complete generality, cf. \cite{Dadarlat-Eilers}.
Now we can finish the proof of Proposition \ref{P1}. Let $\tr_k$
be the standard trace on $B \otimes D_k$ normalized by
$\tr_k(e)=1$, where $e\in B\otimes D_k$ is a minimal projection.
As this trace defines an isomorphism $(\tr_k)_*:K_0(B \otimes
D_k)\to\mathbb Z$, the composition $\prod_{k\in\mathbb
N}(\tr_k)_*\circ\iota:K_0(\prod_{k=1}^{\infty} B \otimes
D_k)\to\mathbb Z^\infty$ is also an isomorphism. Put
$T=\prod_{k\in\mathbb N}(\tr_k)_*\circ\iota\circ P_*$. Then, by
Lemmas \ref{Obs4} and \ref{Obs5}, $T$ is split surjective. It is
obvious that $T$ respects positivity and that its restriction onto
$K_0(B\otimes D)$ is an isomorphism onto $Z_\nu$.
\end{proof}
\begin{cor}
The quotient map $\widehat{T}:K_0(R_{B,D})\to\mathbb
Z^\infty/Z_\nu$ induced by $T$ is split surjection and maps the
positive cone $K_0(R_{B,D})_+$ of $K_0(R_{B,D})$ to the positive
cone $(\mathbb Z^\infty/Z_\nu)_+$ of $\mathbb Z^\infty/Z_\nu$.
\end{cor}
\begin{proof}
Consider the six-term exact sequence
\begin{equation}\label{seq}
\begin{xymatrix}{
K_0(B\otimes D)\ar[r]&K_0(N_{B,D})\ar[r]&K_0(R_{B,D})\ar[d]\\
K_1(R_{B,D})\ar[u]^-{\partial}&\ldots\ar[l]&K_1(B\otimes D)\ar[l]
}\end{xymatrix}
\end{equation}
By Proposition \ref{P1}, the map $K_0(B\otimes D)\to K_0(N_{B,D})$
induced by the inclusion is injective, hence the boundary map
$\partial$ is zero. One also has $K_1(B\otimes D)=0$. Therefore,
the sequence (\ref{seq}) is a short exact sequence of
$K_0$-groups, hence $K_0(R_{B,D})=K_0(N_{B,D})/Z_\nu$ is mapped
surjectively onto $\mathbb Z^\infty/Z_\nu$. Positivity of
$\widehat{T}$ is obvious.
\end{proof}
Thus we see that $K_0(R_{B,D})$ is non-trivial. Our argument also
shows that the cases $B=\K$ and $B=I\K$ give the same
$K_0$-groups. More exactly, let $\ev_s:I\K\to\K$ be the evaluation
map at the point $s\in[0,1]$. It induces the maps
$\overline{\ev}_s:N_{I\K,D}\to N_{\K,D}$ and
$\widehat{\ev}_s:R_{I\K,D}\to R_{\K,D}$.
\begin{cor}
The composition of $(\widehat{\ev}_s)_*:K_0(R_{I\K,D})\to
K_0(R_{\K,D})$ with the quotient map
$\widehat{T}:K_0(R_{\K,D})\to\mathbb Z^\infty/Z_\nu$ is an
isomorphism independent of $s$.
\end{cor}
\begin{rem}\label{rem1}
The positive cone $(\mathbb Z^\infty/Z_\nu)_+$ satisfies the
following property: if $\xi,\eta\in (\mathbb Z^\infty/Z_\nu)_+$
and $\xi\neq 0$ then $\xi+\eta\neq 0$.
\end{rem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Homotopical non-invertibility: a sufficient condition}
The key ingredient to produce homotopically non-invertible
extensions is existence of a non-trivial projection $p\in
J_{A,D}$.
\begin{thm}\label{T1}
Let $\tau:A\to Q(\K)$ be a $*$-homomorphism. Suppose there exists
a projection $p\in J_{A,D}\subset A\otimes_{\max}D$ such that the
class $[\tau_D(p)]\in K_0(R_{\K,D})_+$ is mapped by $\widehat{T}$
to a non-trivial element in $(\mathbb Z^\infty/Z_\nu)_+$. Then the
class of $\tau$ in $\Ext_h(A,\K)$ is not invertible.
\end{thm}
\begin{proof}
Suppose that $\tau$ is homotopically invertible. Then there exists
an extension $\tau':A\to Q(\K)$ such that their sum
$\widetilde{\tau}:A\to Q(\K)$ given by
$\widetilde{\tau}(a)=v_1\tau(a)v_1^*+v_2\tau'(a)v_2^*$, $a\in A$,
is homotopic to the zero $*$-homomorphism via a homotopy
$\Phi:A\to Q(I\K)$. Then $\widetilde{\tau}_D:A\otimes_{\max}D\to
Q(\K\otimes D)$ is also homotopic to zero via the homotopy
$\Phi_D:A\otimes_{\max}D\to Q(I\K\otimes D)$ and $\Phi_D(p)\in
R_{I\K,D}\subset Q(I\K\otimes D)$. Then
$$
0=\widehat{T}[\widehat{\ev}_0\circ\Phi_D(p)]=\widehat{T}[\widehat{\ev}_1\circ\Phi_D(p)]=
\widehat{T}[\widetilde{\tau}_D(p)],
$$
hence
$$
\widehat{T}[\tau_D(p)]+\widehat{T}[\tau'_D(p)]=0
$$
in $Z_\infty/Z_\nu$. But $\widehat{T}[\tau'_D(p)]$ is non-negative
and $\widehat{T}[\tau_D(p)]$ is both non-negative and non-trivial,
hence we get a contradiction, cf. Remark \ref{rem1}.
\end{proof}
So, in order to produce examples of homotopically non-invertible
$C^*$-extensions, one has to expose a projection satisfying the
conditions of Theorem \ref{T1}. We do that in the next section.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Homotopical non-invertibility of some $C^*$-extensions related to property T groups}
Let $G$ be an infinite countable discrete group with the property
T of Kazhdan, \cite{Kazhdan,HV}, which means that the trivial
representation is isolated in the space of all representations of
$G$. We assume that $G$ has an infinite sequence $\pi_k$,
$k\in\mathbb N$, of inequivalent finite-dimensional irreducible
unitary representations. Denote by $n_k$ the dimension of the
representation $\pi_k$ and fix the sequence
$\nu=(n_k)_{k\in\mathbb N}$. Let $\bar{\pi}_k$ be the
contragredient representation of $\pi_k$. The latter means that
there is an antilinear involution $J$ on the Hilbert space
$H_{\pi_k}$ of the representation $\pi_k$ and
$\bar{\pi}_k=J\pi_kJ$. Put $D_k=\mathbb
B(H_{\bar{\pi}_k})=M_{n_k}$ and $D=\prod_{k\in\mathbb N}D_k$. Let
$C^*(G)$ be the (full) group $C^*$-algebra of $G$. Then the
representation $\bar{\pi}_k$ defines a $*$-homomorphism $C^*(G)\to
D_k$.
By \cite{Va}, $C^*(G)$ contains a copy of $\mathbb C$ generated by
the minimal projection $p_0$ corresponding to the trivial
representation of $G$, as a direct summand. We identify elements
of $G$ with their images in $C^*(G)$. Let $S\subset G$ be a finite
symmetric set of generators of $G$ (recall that countable property
T groups are always finitely generated, cf. \cite{HV}, Theorem
1.10). Put
\begin{equation}\label{x}
x=\frac{1}{|S|}\sum\nolimits_{g\in S}g\in C^*(G).
\end{equation}
Since $S$ is symmetric (i.e. for each $g\in S$, $S$ contains
$g^{-1}$), $x$ is selfadjoint. So, $\Sp(x)\subset[-1,1]$. Property
T means that there is a spectral gap between $1\in\Sp{x}$ and the
remaining part of the spectrum: there exists some $\delta>0$ such
that $\Sp(x)\subset[-1,1-\delta]\cup\{1\}$. The projection $p_0$
is the spectral projection of $x$ corresponding to the point 1
\cite{HRV}.
Let $E$ be the $C^*$-algebra on the Hilbert space
$H=\oplus_{k\in\mathbb N}H_{\pi_k}$ generated by all operators of
the form $\oplus_{k\in\mathbb N}\pi_k(g)$, $g\in G$, and by all
compact operators $\K(H)$ on $H$. Put $A=E/\K$. Then the
$C^*$-extension $0\to\K\to E\to A\to 0$ is the $C^*$-extension
considered by S. Wassermann in \cite{Wass1}. We need the following
modification of this $C^*$-extension \cite{MT8}.
Fix a sequence $\mu=(m_k)_{k\in\mathbb N}$ of positive integers
with $\lim_{k\to\infty}\frac{m_k}{n_k}=\infty$. Let $\pi_k^{m_k}$
be the direct sum of $m_k$ copies of the representation $\pi_k$.
Define $E_\mu$ as the $C^*$-algebra generated by all operators of
the form $\oplus_{k\in\mathbb N}\pi_k^{m_k}(g)$, $g\in G$, and by
all compact operators on $H_\mu=\oplus_{k\in\mathbb
N}H_{\pi_k}^{m_k}$. Note that $E_\mu/\K\cong A$, because both
$E/\K$ and $E_\mu/\K$ can be obtained from the group ring $\mathbb
C[G]$ by completing it with respect to the same $C^*$-algebra
seminorm $\|\cdot\|=\lim\sup_{k\to\infty}\|\pi_k(\cdot)\|$.
If we denote the inclusion $E_\mu\subset\B(H_\mu)$ by
$\overline{\tau}^\mu$ then, passing to the quotients, the
composition $\tau^\mu=q\circ\overline{\tau}^\mu:A\to Q(\K)$ is the
Busby invariant of the extension
\begin{equation}\label{e-nu}
\begin{xymatrix}{
0\ar[r]&\K\ar[r]&E_\mu\ar[r]&A\ar[r]&0.
}\end{xymatrix}
\end{equation}
Let $\sigma:C^*(G)\to E_\mu$ be the $*$-homomorphism defined by
$\sigma(g)=\oplus_{k\in\mathbb N}\pi_k^{m_k}(g)$, $g\in G$.
Composing it with the quotient map, we get a surjective
$*$-homomorphism $\widehat{\sigma}:C^*(G)\to A$. Consider also
another $*$-homomorphism $\oplus_{k\in\mathbb
N}\bar{\pi}_k:C^*(G)\to D$.
Let $\underline{\Delta}:G\to G\times G$ be the diagonal
homomorphism. It induces the $*$-homomorphism $\Delta:C^*(G)\to
C^*(G)\otimes_{\max}C^*(G)$. Composing it with the map
$$
\begin{xymatrix}{
\widehat{\sigma}\otimes(\oplus_{k\in\mathbb N}\bar{\pi}_k):
C^*(G)\otimes_{\max}C^*(G)\ar[r]& A\otimes_{\max}D
}\end{xymatrix}
$$
we obtain a $*$-homomorphism
$$
\begin{xymatrix}{
\Delta\circ(\widehat{\sigma}\otimes(\oplus_{k\in\mathbb
N}\bar{\pi}_k)): C^*(G)\ar[r]& A\otimes_{\max}D.
}\end{xymatrix}
$$
Put
$$
p=\Delta\circ(\widehat{\sigma}\otimes(\oplus_{k\in\mathbb
N}\bar{\pi}_k))(p_0)\in A\otimes_{\max}D.
$$
\begin{lem}\label{proj}
The projection $p$ satisfies the following conditions:
\begin{enumerate}
\item
$p\in J_{A,D}$;
\item
$\widehat{T}[\tau^\mu_D(p)]=[\mu+Z_\nu]\neq 0$ in $\mathbb
Z^\infty/Z_\nu$.
\end{enumerate}
\end{lem}
\begin{proof}
To show that $p\in J_{A,D}$ one has to check that
$q_{\max,\min}(p)=0$ in $A\otimes_{\min}D$. The latter was proved
by S.~Wassermann in \cite{Wass1}, cf. \cite{MT8}.
Let us calculate $\widehat{T}[\tau^{\mu}_D(p)]$. The map
$\tau^{\mu}_D$ factorizes through the Busby invariant
$$
\begin{xymatrix}{
A\otimes_{E_\mu}D\ar[r]& Q(\K\otimes D)
}\end{xymatrix}
$$
of the $C^*$-extension
$$
\begin{xymatrix}{
0\ar[r]&\K\otimes
D\ar[r]&E_\mu\otimes_{\min}D\ar[r]&A\otimes_{E_\mu}D\ar[r]&0,
}\end{xymatrix}
$$
which is obtained from the inclusion $E_\mu\subset M(\K\otimes D)$
by passing to quotients. Therefore, it suffices to calculate
$T[\bar{p}]\in \mathbb Z^\infty$, where
$$
\bar{p}=(\sigma\otimes (\oplus_{k\in\mathbb
N}\bar{\pi}_k))(\Delta(p_0))\in E_\mu\otimes_{\min}D
$$
is the projection corresponding to the trivial representation of
$G$.
Consider the map $\id_{E_\mu}\otimes p_k:E_\mu\otimes_{\min}D\to
E_\mu\otimes D_k$. Then $T[\bar{p}]=(T_k)_{k\in\mathbb N}$, where
$T_k=\tr(\id_{E_\mu}\otimes p_k(\bar{p}))$. By the spectral
theorem, $\id_{E_\mu}\otimes p_k(\bar{p})$ equals the spectral
projection of the image
$$
\frac{1}{S}\sum\nolimits_{g\in
S}\sigma(g)\otimes\bar{\pi}_k(g)\in\B(H_\mu)\otimes D_k
$$
of $x$ (\ref{x}) corresponding to the (still isolated) point 1.
So, it remains to calculate the dimensions of the trivial
representation as a subrepresentation of
$\sigma\otimes\bar{\pi}_k$. But $\sigma=\oplus_{k\in\mathbb
N}\pi_k^{m_k}$ and it is known, \cite{Wass1}, that the
representation $\pi_l\otimes\bar{\pi}_k$ does not contain the
trivial representation if $l\neq k$ and $\pi_k\otimes\bar{\pi}_k$
contains exactly one one-dimensional trivial representation.
Hence, $T_k=m_k$ and $\widehat{T}[\tau^\mu_D(p)]=[\mu+Z_\nu]\neq
0$.
\end{proof}
\begin{cor}\label{non-invert}
The $C^*$-extension (\ref{e-nu}) is not invertible in
$\Ext_h(A,\K)$.
\end{cor}
\begin{cor}
The semigroup $\Ext_h(A,\K)$ contains a copy of the semigroup
$(Z^\infty/Z_\nu)_+$.
\end{cor}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Concluding remarks}
\begin{rem}
It is not necessary to have a projection with the properties as in
Theorem \ref{T1} in an ideal of $A\otimes_{\max}D$. A smaller
tensor product suffices. Let $\|\cdot\|_\alpha$ be a tensor
$C^*$-norm on $A\odot D$ such that
$\|\cdot\|_E\leq\|\cdot\|_\alpha$ for {\it any} extension $E$ of
$A$. Then one can use a projection in $A\otimes_\alpha D$
satisfying the properties of Lemma \ref{proj} to prove homotopical
non-invertibility of $C^*$-extensions of $A$ in the same way.
Using this tensor $C^*$-product one can prove homotopical
non-invertibility of similar $C^*$-extensions arising from
property $\tau$ groups (\cite{Lubozky}; this property is weaker
than the property T and means that the trivial representation is
isolated from finitedimensional representations only), when the
projection corresponding to the trivial representation does not
lift to a projection in $A\otimes_{\max}D$.
\end{rem}
\begin{rem}
A. Connes and N. Higson, the founders of $E$-theory, have
expressed in \cite{CH} their opinion that $E$-theory is the
quotient, modulo homotopy, of the theory of $C^*$-extensions. Now
we know that the deficiency of non-invertibility for
$C^*$-extensions persists at the level of homotopy. As $E$-theory
always has a group structure, the Connes--Higson construction,
which defines a map $\Ext_h(A,B)\to E_1(A,B)$, cannot be faithful.
So the relation between two theories is more complicated. Our
conjecture is that $E$-theory coincides with the group of
invertible elements of $\Ext_h(A,B)$.
\end{rem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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