# 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 $$\mathbb{B}(E\otimes_{f_{i}}B)$$.

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? Sep 25, 2014 at 10:32