If $$p: Y \to X$$ a covering space and $$f: Z \to X$$ a function, we define:
$$f^* Y=[(y,z):p(y)=f(z) ] \subset Y \times Z.$$
I have to find out if the function: $$q: f^* Y \to Z, (y,z) \to z, $$ is a covering space. A definition of a covering space is : A surjective continuous map $p:E\rightarrow B$ of spaces $E$ and $B$ is a covering map (and $E$ a covering space of $B$) if for every $b\in B$ there is an open neighborhood $U$ of $b$, a discrete space $S$ and a homeomorphism $\varphi:p^{-1}(U)\rightarrow U\times S$ such that $p|_{p^{-1}(U)}=\varphi\circ\pi_U$ ($\pi_U$ the projection map). How can i handle my issue, I get lost
Fix $z \in Z$. Choose an evenly-covered neighborhood $V \subset X$ of $f(z)$. (i.e. $V$ satisfies the definition you gave for the covering space $p\colon Y \to X$ for the point $f(z)$.). Write $U = f^{-1}(V)$. By continuity this is an open neighbourhood of $z$. Consider $q^{-1}(U)$. A pair $(y,z)$ is in $q^{-1}(U)$ exactly when $p(y) = f(z) \in V$. Since $V$ is evenly-covered by $p$, take your discrete set $S$ and homeomorphism $\varphi\colon p^{-1}(V) \to V \times S$. Notice that $q^{-1}(U) = (p^{-1}(V)\times U)\cap f^\ast Y = \{(y,z) \colon y \in p^{-1}(V),\, z \in U \mid p(y) = f(z)\}$.
Consider the map $\psi \colon q^{-1}(U) \to U\times S$ given by $\psi(y,z) = (z,\pi_S\circ\varphi(y))$. This map is obviously continuous, with continuous inverse given by $(z,s) \mapsto (\varphi^{-1}(f(z),s),z)$. Surjectivity and continuity of $q$ are obvious, so we've proven that $q \colon f^\ast Y \to Z$ is a covering map.