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!