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Let $q: X \to Y$ and $r: Y \to Z$ be covering maps. Let $p=r \circ q$. Show that if $r^{-1}(z)$ is finite $\forall z$, then $p$ is a covering map.

$\textbf{My Attempt:}$

Let $U$ be an arbitrary open set in $Z$. Then $p^{-1}(U)=q^{-1}(r^{-1}(U))$. Since $r$ is a covering map, we know that $r^{-1}(U)=\bigcup_{i=1}^{N} V_i$ where $V_i$ is open $\bigcap_{i=1}^{N} V_i = \emptyset$.

Is the following true, $q^{-1}(\bigcup_{i=1}^{N} V_i)=\bigcup_{i=1}^{N} q^{-1}(V_i)$ ?

Now since $q$ is covering for each $V_i$ we have that $q^{-1}(V_i)=\bigcup_{\alpha} K_{\alpha}^i$ where the $K_{\alpha}^{i}$ are all disjoint and open.

Then, $q^{-1}(\bigcup_{i=1}^{N} V_i)=\bigcup_{i=1}^{N}(\bigcup_{\alpha} K_{\alpha}^i)$.

So we just need to show that $\bigcap_{i=1}^{N}(\bigcup_{\alpha} K_{\alpha}^i)=\emptyset$ which is fairly straightforward.

I am not sure if this is the right way to go about this proof as I have not utilized the fact that the number of $V_i$ will be finite.

How do I use the fact that $r^{-1}(z)$ is finite?

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marked as duplicate by Stefan Hamcke, Najib Idrissi, Dan Rust, MathOverview, Kevin Carlson May 13 '14 at 18:45

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  • $\begingroup$ I would like to know the answers to your questions, I hope someone answers and explains very well why this happens! $\endgroup$ – user424241 Jan 29 '18 at 22:43