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Suppose $q : E \to X$ is a covering map, show that if $X$ is a $n$-manifold, then $E$ is too.

My Attempted Proof:

To show $E$ is an $n$-manifold, every $e \in E$ needs to have a neighbourhood $\Gamma$ hoemomorphic to an open subset of $\mathbb{R}^n$. To that end pick $e \in E$, we prove the existence of such a neighbourhood $\Gamma$.

We have that $q(e) = x$ for some $x \in X$. Since $X$ is a $n$-manifold, let $U$ be a neighbourhood of $x$ homeomorphic to an open subset $\widetilde{U}$ in $\mathbb{R}^n$ and let $\phi$ denote this homeomorphism.

Since $q$ is a covering map, $x$ also has an evenly covered neighbourhood $U'$ in $X$, such that $q^{-1}[U'] = \bigcup_{i \in I} V_i$, where each $V_i$ contains exactly one $e_i \in q^{-1}(x)$ and each $V_i$ is a connected open set of $E$, that is mapped homeomorphically onto $U'$, by the restriction $q|_{V_i}$. Note that we have $e_j = e$ for some $j \in I$. Fix this $j \in I$.

Put $W = U' \cap U$. We have that $W$ is nonempty, since it contains $x$ and it is open, being the intersection of two open sets of $X$, moreover it is also an open set of $U$. Then $\phi|_{W} : W \to \phi[W]$ is a homeomorphism (being the restriction of the homeomorphism $\phi : U \to \widetilde{U}$) and $\phi[W]$ is open in $\widetilde{U}$ and hence $\mathbb{R}^n$, since an open set of an open subset (endowed with the subspace topology) of a parent topoological space, is open in the parent topological space.

Then we have $q^{-1}[W] \subseteq \bigcup_{i \in I}V_i$. Now $W$ is an open set of $U'$ and that $V_j$ is mapped homeomorphically onto $U'$. Let $\varphi$ be the restriction of $q|_{V_{i}}$ to $(q|_{V_i})^{-1}[W]$, and put $\Gamma = (q|_{V_i})^{-1}[W] \subseteq V_j$, then $\varphi : \Gamma \to W$ is a homeomorphism (being the restriction of the homeomorphism $p|_{V_i}$ to $\Gamma$).

We thus have $\Gamma \cong W \cong \phi[W]$, so $\Gamma$ contains $e$ and it is an open set of $E$, as desired. $\square$

Is this proof correct? Is it rigorous enough? Can it be improved or proved in a shorter way?

José Carlos Santos
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Perturbative
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  • Does your definition of $n$-manifold include the Hausdorff condition? Because if so, then the statement you're trying to prove is false. This answer describes an action of $\mathbb Z$ on the manifold $X = \mathbb R^2\smallsetminus{(0,0)}$ such that the quotient map $X\to X/\mathbb Z$ is a covering map, but $X/\mathbb Z$ is not Hausdorff. – Jack Lee Nov 30 '17 at 17:50
  • @JackLee Professor Lee, this is problem 11.1 (b) in your book Introduction to Topological Manifolds, the previous problem 11.1(a) deals with the Haursdoff condition and shows that for a coveirng map $q : E \to X$, if $X$ is Haursdoff, then $E$ is too. Since $X$ is a $n$-manifold (whose definition I'm using from your book), it is Haursdoff, so $E$ must be Haursdoff too. (As a side note Introduction to Topological Manifolds is probably my favorite math book I've ever read, thank you for writing such a wonderful book!) – Perturbative Nov 30 '17 at 18:43
  • Aack! I misread what you wrote -- for some reason, I thought you were asking to show that if $E$ is a manifold, then $X$ is too. Forget my previous comment. (And thank you for the very kind compliment!) – Jack Lee Nov 30 '17 at 18:46

1 Answers1

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Yes, it is correct. I suppose that I would have started with a neighborhood $U$ of $x$ wich not only is homeomorphic to $\mathbb{R}^n$ but which is also an evenly covered neighbourhood of $x$. Therefore, there would be no need to consider the intersection of two neighborhhods. I would also avoid using the symbols $\phi$ and $\varphi$ in the same text, since they are two different ways of representing the same greek letter.

I don't know which definition of manifold you're using, but usually there is an extra requirement for the global topology of the manifold, such as being paracompact. It is not obvious (not to me at least) that $X$ being paracompact implies that $E$ is paracompact.

José Carlos Santos
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  • Hmm, how do you know that the neighborhood $U$ of $x$ which is homeomorphic to an open subset of $\mathbb{R}^n$ is also an evenly covered neighbourhood of $x$? Because not all neighbourhoods of $x$ are evenly covered, nor are all neighbourhoods of $x$ homeomorphic to an open subset of $\mathbb{R}^n$. (Also the definition of a topological manifold that I'm using doesn't assume paracompactness) – Perturbative Nov 30 '17 at 09:53
  • @Perturbative The answer to your question is that I don't know. But you could have started with a neighborhood $U^\star$ that is evenly covered and that to work with a neighborhood $U\subset U^\star$ which was homeomorphic to $\mathbb{R}^n$. Then $U$ would be evenly covered too. – José Carlos Santos Nov 30 '17 at 09:57
  • To expand on the previous comment, it is true that every simply connected open subset of $X$ is evenly covered, although it is something that has to be proved; the proof uses lifting lemmas. However, you don't need all of that for this answer to work. All you need is that $x$ has a neighborhood basis of open subsets $(U_i)$ each homeomorphic to $\mathbb{R}^n$. For once you have that then, if $U^\star$ is any evenly covered open neighborhood of $x$, there exists $i$ such that $U_i \subset U^\star$, and therefore $U_i$ is evenly covered. – Lee Mosher Nov 30 '17 at 16:46