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$E,B$ are topological spaces and let's say that $p:E\to B$ is a covering map.

Is $p$ open?

i tried to show it as follows:

let $U$ be an open set in $E$, and now for every $x\in U$, $p(x)\in B$. $p$ is a covering map so there is a open neighborhood of $p(x)$, call it $V$ that is evenly coverd by $p$. e.g there is a partition of $p^{-1}(V)$ of open disjoint subsets $\{V_\alpha\}$ that covers $E$. and $p|_{V_\alpha}$ is homeomorphism between $V_\alpha$ and $V$.

so for some $\alpha$, $x\in V_\alpha$. so $V_\alpha \bigcap U$ is an open neighborhood of $x$ and $V_\alpha \bigcap U\subset V_\alpha$ and it is open in $V_\alpha$ , and since $p|_{V_\alpha}$ is homeomorphism between $V_\alpha$ and $V$ we get that $p(V_\alpha \bigcap U)$ is an open neighborhood of $p(x)$ and it is a subset of $p(U)$. then $p(U)$ is open.

is it right?

Hans
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sha
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    Note that it suffices to show $p(U)$ is open for every $U\in\mathcal{B}$, where $\mathcal{B}$ is any basis for the topology. The claim follows now from the fact that there is a basis whose elements are all mapped diffeomorphically by $p$. – Amitai Yuval Sep 14 '14 at 12:55
  • yeah i see, for every element of $B$ (the covered space) we take an open neighborhood that is evenly covered. and all the partitions for all the elements crate that basis. right? – sha Sep 14 '14 at 13:01
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    @AmitaiYuval diffeomorphism has nothing to do with it, as we are in general spaces... – Henno Brandsma Sep 14 '14 at 13:14
  • does my proof stands? – sha Sep 14 '14 at 13:18
  • @HennoBrandsma You're right, of course, thanks for the correction. I meant homeomorphically instead of diffeomorphically. (too much differential geometry, I guess...) – Amitai Yuval Sep 14 '14 at 14:16
  • @AmitaiYuval, I’m unfamiliar with the claim about the existence of such a basis. How does one construct it? – Jos van Nieuwman Sep 11 '20 at 17:52

1 Answers1

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Your argument is quite correct, but start with let $y = p(x)$ be a point of $p[U]$ where $x \in U$, and then proceed (you start with an arbitrary element of $p[U]$ and show its an interior point, not an element of $U$, though here the difference is slight...).

This generalises to local homeomorphisms $f: X \rightarrow Y$, where every $x$ has an open neighbourhood $U_x$ such that $f|_{U_x}$ is a homeomorphism between $U_x$ and the open set $f[U_x]$. The argument is almost the same.

Henno Brandsma
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  • Just to make this answer extra clear. Note that openness in arbitrary topological spaces, just like metric spaces, can still be intuitively understood as: start with an arbitrary element of $X$ and show it has an open neighborhood around it in $X$. (in comparison, in metric spaces, such a neighborhood is an open ball). In this particular question, we can also exploit the fact that an open map $X \to Y$ is equivalent to $f^{-1}(B^{\mathrm{o}}) \subset ({f^{-1}(B)})^{\mathrm{o}} $for any $B\subset Y $for any general spaces. – Dinoman Nov 08 '22 at 16:41