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Given a covering map (continuous, surjective, closed, irreducible, compact) $f:X \to Y$, and a dense subset $D$ of $Y$, is the preimage of $D$ by $f$ a dense subset of $X$?

Edit: The question is missing context, but not details. Quite frankly, it was just a minor question pertaining to some research I'm doing -- and I thought I would give StackExchange a try. My original thoughts were about the absolute of a Hausdorff space (X being the Iliadis absolute of Y), but I thought phrasing the question more generally would be better. The question is true, btw. Thanks.

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  • $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ Nov 15, 2013 at 10:31
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    $\begingroup$ The question is missing context, but not details. Quite frankly, it was just a minor question pertaining to some research I'm doing -- and I thought I would give StackExchange a try. My original thoughts were about the absolute of a Hausdorff space ($X$ being the Iliadis absolute of $Y$), but I thought phrasing the question more generally would be better. The question is true, btw. Thanks. $\endgroup$
    – ihaphleas
    Nov 15, 2013 at 12:48
  • $\begingroup$ Ok. Please include this on the body of the question. This will prevent the downvotes and votes to close (I didn't). $\endgroup$ Nov 15, 2013 at 12:59

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If $D$ is dense in $Y$, then $f^{-1}(D)$ will be dense in $X$. The only property which is really used here is that $f$ is an open map. Take an open set $U$ of $X$, then its image $f(U)$ is open and thus intersects $D$. There is thus point $x\in U$ such that $f(x)\in D$, and that means $x\in U\cap f^{-1}(D)$.

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