Restriction of a covering map to a subspace Let $p:X\rightarrow Y$ be a covering map and let $Y_0 \subset Y$. Show that $p|:p^{-1}(Y_0)\rightarrow Y_0$ is a covering map.
Hint: Show first that if $V\subset Y$ is well-covered by $p$, then $Y_0\cap V$ is well-covered by $p|:p^{-1}(Y_0)\rightarrow Y_0$.
All I've got so far is the definition of a covering map written out. I can't even seem to use the hint given.
 A: Since $p$ is a covering map, there exists an open cover $\{U_{\alpha}\}$ of $X$ by evenly covered sets $U_{\alpha}$. That is, for all $\alpha$, $p^{-1}(U_{\alpha}) = \bigsqcup_{i \in I} V^i_{\alpha}$ for some index set $I$, and the restriction $p:V^i_{\alpha} \to U_{\alpha}$ for any particular $i$ is a homeomorphism. Since the $U_{\alpha}$ cover $X$, $\{U_{\alpha} \cap Y_0\}$ is an open cover of $Y_0$. Additionally, for all $\alpha$, $p^{-1}(U_{\alpha} \cap Y_0) = p^{-1}(U_{\alpha}) \cap p^{-1}(Y_0) = \bigsqcup_{i \in I} V^i_{\alpha} \cap p^{-1}(Y_0)$. $p:V^i_{\alpha} \to U_{\alpha}$ is a homeomorphism, so the restriction $p: V_{\alpha}^i \cap p^{-1}(Y_0) \to p(V_{\alpha}^i \cap p^{-1}(Y_0))$ is also a homeomorphism. But $p(V_{\alpha}^i \cap p^{-1}(Y_0)) = U_{\alpha} \cap Y_0$ since a homeomorphism is in particular bijective, so the $U_{\alpha} \cap Y_0$ are evenly covered. Hence $Y_0$ has an open cover by evenly covered sets, so the restriction $p:p^{-1}(Y_0) \to Y_0$ is a covering map.
