I am reading proposition 17.2.3 Blackadar: If $ \varepsilon =(E, \varphi, T) \in \mathbb{D}(A, B)$, then $(E, \varphi, T)$ is homotopic to the $0$-module.
I have some questions about the proof. We must find an element $ \bar{\varepsilon} = (\bar{E}, \bar{\varphi}, \bar{T})$ in $\mathbb{D}(A, B[0,1])$ such that $ev_{0,*}(\bar{\varepsilon})=(\bar{E}\otimes_{ev_{o}}B, \bar{\varphi}\otimes_{ev_{o}}id_{b}, \bar{T}\otimes_{ev_{o}}idB)$ is isomorphic to 0, and $ ev_{1,*}(\bar{\varepsilon})$ is isomorphic to $\varepsilon$,where $ev_{t}:B[0,1] \rightarrow B$, is the evaluation map
Because we need a Hilbert $ B[0,1]$-module, we can consider $E(0,1]= \{ e:[0,1] \rightarrow E : e \ is \ contionuous \ and \ e(0)=0 \}$, which is a Hibert $ B[0,1]$-module. we define a $A$-action on $E(0,1]$ by $(a · e)(t) = a(e(t))$ for all $a \in A, e \in E(0, 1] \ and \ t \in [0, 1]$. Define the operator Define $\bar{T} \in L(E(0, 1]), e → Toe$.
Could you please explain how $ev_{0,*} = 0$ and $ev_{1,*} =\varepsilon.$ I think maby we can say that for example for $E(0,1] \otimes_{ev_{0}}B$, if we consider $e(t) \beta\otimes_{ev_{0}}b= e(0)\otimes\beta(0)b =0$, where $\beta \in B[0,1].$
But how it gives us $E(0,1] \otimes_{ev_{1}}B = E$?