In Milnor-Stasheff's book about characteristic classes there is an appendix about Chern-Weil theory. Suppose that $E \to M$ is a (complex) vector bundle with connection $\nabla$: this connection can be extended to act on one forms with values in $E$ and one defines $K=K_{\nabla}=\nabla^2$ to be a curvature of this connection. It is shown that $K$ is $C^{\infty}(M)$ linear: it is locally determined by a matrix of two-forms $\Omega=(\Omega_{ij})_{i,j}$ (if $(s_i)_i$ is a local basis of sections of $E$ then $K(s_i)=\sum_j\Omega_{ij} \otimes s_j$). Given an invariant polynomial $P:M_n(\mathbb{C}) \to \mathbb{C}$ one can (correctly) define $P(K)$ locally as $P(\Omega)$. It is shown that $dP(\Omega)=0$ i.e. this is cloes form and therefore it defines a class in de-Rham cohomology.
Claim. This class does not depend on the choice of connection.
Proof. Let $\nabla_0$ and $\nabla_1$ be two connections on $E$. Consider $M \times \mathbb{R} \ni (x,t) \mapsto x \in M$ and consider $E'$ to be a pullback bundle and denote $\nabla_0',\nabla_1'$ the pullback connections. Consider $\nabla=t\nabla_1'+(1-t)\nabla_0'$: then $P(K_{\nabla})$ is a de Rham cocyle on $M \times \mathbb{R}$.
Now consider $i_{\epsilon}:M \ni x \mapsto (x,\epsilon) \in M \times \mathbb{R}, \epsilon=0,1$. Then the induced connection $i_{\epsilon}^*\nabla$ may be identified with $\nabla_{\epsilon}$
It seems to me that $i_{\epsilon}^*\nabla$ may be rather identified with $t\nabla_1+(1-t)\nabla_0$, not with $\nabla_{\epsilon}$. Am I right? If not, why? If yes, how to proceed further?
(EDIT: according to comments below this issue is clarified.)
Therefore $i_{\epsilon}^*(P(K_{\nabla}))=P(K_{i^*_{\epsilon}\nabla})=P(K_{\nabla_{\epsilon}})$. As $i_0$ and $i_1$ are homotopic, so are $i_0^*$ and $i_1^*$ therefore the class of $i_0^*(P(K_{\nabla}))$ is the same as the class of $i_1^*(P(K_{\nabla}))$.
Why the first equality, namely $i_{\epsilon}^*(P(K_{\nabla}))=P(K_{i^*_{\epsilon}\nabla})$ is true? Could someone explain it?