Definition of homotopy between Kasparov modules

Question

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!

Answer 1

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.