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!