Exercise 12.9 book of Manetti (Topology) I am trying to prove the following exercise:
Let $p:E\rightarrow X$ be a covering space. Prove that for every $Y$ connected and every continuos map $f:Y\rightarrow X$,
$$f^*p:\{(e,y)\in E\times Y|p(e)=f(y)\}\rightarrow Y,\, f^*p(e,y)=y,$$
is a covering space.
I think, I already proved that $f^*p$ is continuos and surjective. But I can't still find admisible open sets.

I will write $E\times_X Y$ for $\{(e, y) \in E\times Y \mid p(e)=f(y)\}$. Let $f^*p : E\times_X Y \to Y$ be the projection on the $Y$ coordinate.
Given any $y \in Y$. Let $U$ be the evenly covered neighborhood of $f(y)$ in $X$. Then $f^{-1}(U)$ is a neighborhood of $y$. I claim that this $f^{-1}(U)$ is the evenly covered neighborhood of $y$. Note,
$(p^*f)^{-1}(f^{-1}(U)) = \{ (e, y) \in E\times_X Y \mid y \in f^{-1}(U) \} = \{(e, y) \in E \times Y \mid p(e) = f(y) \in U\}$. But since $U$ is evenly covered by $p$, $p^{-1}(U) = \bigsqcup_\alpha V_\alpha$, such that, $p|_{V_\alpha}$ is a homeomorphism. So,
$$(p^*f)^{-1}(f^{-1}(U)) = \bigsqcup_\alpha \{ (e, y) \in E\times Y \mid p(e) = f(y), e \in V_\alpha \} =: \bigsqcup_\alpha W_\alpha$$
Note that, $W_\alpha$ is open in $E\times_X Y$ since $W_\alpha = (V_\alpha\times Y) \cap (E\times_X Y)$. Now, you just need to check that $p|_{W_\alpha}$ is a homeomorphism, which should be easy.
