# Definition of homotopy between Kasparov modules

I'm trying to understand the definition of homotopy between Kasparov modules as presented in Blackadar's book on K-theory for operator algebras. $A,B$ will be C*-algebras, while $E$ will denote a Hilbert $B$-module. We define homotopy:

$Let\ i=0,1\ and\ let\ f_{i}\ denote\ the\ respective\ evaluation\ morphism\ from\ IB:=C([0,1],B)\ to\ B.\ Two\ Kasparov\ A-B\ modules\ (E_{i},\phi_{i},F_{i})\ are\ homotopic\ if\ there\ exists\ an\ A-IB\ module\ (E,\phi,F)\ such\ that\ (E \otimes_{f_{i}}B,f_{i}\circ\phi,f_{i,*}F)\simeq^{unitary}(E,\phi_{i},F_{i}).$

Here $E\otimes_{f_{i}}B$ denotes the graded tensor product with respect to $f_{i}$. My troubles are with the definition of the third Kasparov module. In the definition of a Kasparov module we need $f_{i}\circ\phi$ to be a graded *-homomorphism from $A$ to $\mathbb{B}(E\otimes_{f_{i}}B)$. However, $\phi:A\rightarrow\mathbb{B}(E)$ while $f_{i}:IB\rightarrow B$, so obvious composition doesn't work. One could try to simply demand $f_{i}\circ\phi:=\phi\otimes 1$, but this is independent of $f_{i}$ and hence most likely the false definition. Similarly, I don't see where we naturally apply the evaluation morphism with an element of $\mathbb{B}(E)$, for example $F$, to turn it into an element of $\mathbb{B}(E\otimes_{f_{i}}B)$.

Thank you for your help!

## 1 Answer

The notation is a little confusing, yes. In fact, you are right that you could just take $f_i \circ \phi := \phi \otimes 1$. This looks like it is independent of $f_i$, but it's not: $f_i$ is hidden in the definition of the tensor product. If you want to clarify this, write it as $\phi\otimes_{f_i}1$.

Effectively, you should see $E$ as a family of $A$-$B$-bimodules indexed by $i \in [0,1]$. The tensor product $E\otimes_{f_i}B$ is the fibre at $i$. The representation $\phi\otimes_{f_i}1$ is the restriction of the $A$ action to this fibre.

• You are of course right that $f_{i}$ plays a role in the definition of the tensor product. I guess we then write $f_{i,*}F:=F\otimes_{f_{i}} 1$ as well? – David Hornshaw Sep 25 '14 at 10:32