# This projection is a covering map?

If $\tilde X$ is the product of $X$ with a discrete space, the projection $\tilde X \to X$ is a covering map.

This question seems really easy, but as I'm a beginner there are some things a little bit confusing.

In order to solve this question, let $Y$ be a set with the discrete topology, so we have to prove that $X\times Y\to X$ is a covering map, to do so, let $x$ be a point in $X$, then take any open set $U$ which $x$ is contained in to be our evenly covered open set by $p$, thus $p^{-1}(U)=U\times Y$.

Am I on the right way? I'm stuck here.

Thanks

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Take any point $x\in X$ and note that for an neighborhood $U$ of $x$ one has that $p^{-1}(U)=\bigcup_{d\in D}\{d\}\times U$ where $D$ is your discrete space. Now, each of the $\{d\}\times X$ is open, and $p\mid_{\{d\}\times X}$ is a homeomorphism. Thus you're done!
Why $p^{-1}(U)=\bigcup_{d\in D}\{d\}\times U$? why it isn't $p^{-1}(U)=Y\times U$? thank you for your answer :) –  user42912 Dec 9 '12 at 23:42