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Let $p:Y\to X$ be a covering space and $p^{-1}(x)$ countable for every $x\in X$.

Task: Let $X$ be a smooth manifold. Show, that $Y$ has the structure of a smooth manifold, regarding this $p$ is smooth.

Hello,

I have a question to this task. We showed here () a statement, which is needed for this task. To show, that $Y$ is a smooth manifold it has to be a second-countable Hausdorff space with a smooth structure. With the task from the other thread it is clear, that $Y$ is second-countable. So I need to show, that $Y$ is a Hausdorff space with a maximal atlas.

But I do not know, how to show that. And than I need to show, that $p$ is smooth.

Do you have any tips how to start? Thanks in advance.

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  • $\begingroup$ There are problems with the grammar of the problem, but it looks like you're explicitly told that $p$ is smooth. $\endgroup$ – anomaly Jun 30 '16 at 15:19
  • $\begingroup$ I am sorry for mistakes with the grammer. $p$ needs to be smooth under the smooth strucur of $Y$. So we need to show, that $p$ is smooth. It is not assumed. $\endgroup$ – MrTopology Jun 30 '16 at 15:23
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Separated, let $x,y\in Y$ if $p(x)\neq p(y)$, since $X$ is separated, there exists open subsets $p(x)\in U_x, p(y)\in U_y$ of $X$ such that $U_x\cap U_y$ is empty, $p^{-1}(U_x)\cap p^{-1}(U_y)$ is also empty and $x=\in p^{-1}(U_x), y\in p^{-1}(U_y)$. If $p(x)=p(y)$, since $p$ is a covering, there exists an open subset $p(x)\in U$ such that $p^{-1}(U)=\bigcup_iU_i$ and the restriction of $p:U_i\rightarrow U$ is an homeomorhism and $U_i\cap U_j$ is empty if $i\neq j$. Let $x\in U_{i_x}, y\in U_{i_y}$, $p(x)=p(y)$ implies that $y$ is not in $U_{i_x}$ since the restriction of $p$ to $U_{i_x}$ is injective.

Manifold: You can suppose that $p(U_{i_x})$ is a chart by taking an open subset,$(p(U_{i_x},\phi_x)$ consider the atlas on $Y$ defined by $(U_{i_x},\phi_x\circ p)$.

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  • $\begingroup$ Thank you, I have a question to your hint about the manifold. Do you mean a random open subset, or $(p(U_{i_x},\phi_x))$ is supposed to be this subset. The latter would seem odd to me. $\endgroup$ – MrTopology Jun 30 '16 at 16:38

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