I'm having trouble proving the following result of algebraic topology.
Let $p:X\rightarrow B$ be a covering and let $f:Y\rightarrow B$ be a continuous map. Define the subspace $Z=\{(y,x)\in Y\times X:f(y)=p(x)\}$. Let $q:Z\rightarrow Y$ be the restriction of the natural projection $Y\times X\rightarrow Y:(x,y)\mapsto y$. Prove $q:Z\rightarrow Y$ is a covering.
I've tried to use the information regarding $f(y)=p(x)$, but I simply cannot make this work.
Well, let me be the first to tell you that this is a very special case of a pullback, in this case of covers (but more generally bundles.)
Hint: take a point $y \in Y$, and consider $f(y)$. Take a neighborhood $U$ around it so that there is a homeomorphism $p^{-1}(U) \cong U \times F$, and consider the preimage of this under $f$, i.e: $f^{-1}(U)$. Let $\pi:Z \to X$ be the projection.
Then over this point, we have $$q^{-1}(f^{-1}(U))=\{(x,y_0) \mid p(x)=f(y_0)=f(y)\}=\pi^{-1}(p^{-1}(U)).$$
From this, can you show the desired property? A different way of "saying" all of this is replacing $p^{-1}(U) \cong U \times F$ with $p^{-1}(U)= \coprod U_i$ and they are mapped homeomorphically onto $U$ by $p$. In this case, you also get a clean expression.