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Let $X$ be a CW complex, $p:E\to X$ a covering map. Then $E$ has an induced CW complex structure defined as follows. If $\Phi:D^n\to X$ is a covering, it lifts to a map $D^n\to E$ (since $D^n$ is simply connected and we can apply the lifting criterion). These give the desired cell decomposition.

However, I don't know how to prove that $E$ has the weak topology induced by the aforementioned cell decomposition. Somehow I must use the fact that $p:E\to X$ is a covering and that $X$ has the weak topology, but I don't know how...

This question has been asked several times on this site, e.g., here, and here. But (surprisingly) none of the answers there give a proof of this...

Thanks in advance!

Eric Wofsey
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Yuxiao Xie
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1 Answers1

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Suppose $A\subseteq E$ is open in every cell of $E$ (i.e., $\Phi^{-1}(A)$ is open for each cell $\Phi:D^n\to E$). It suffices to show that $A\cap p^{-1}(U)$ is open for any open $U\subseteq X$ which is evenly covered by $p$, since such sets $p^{-1}(U)$ are an open cover of $E$.

Since $U$ is evenly covered by $p$, we may identify $p^{-1}(U)$ with a disjoint union of copies $U_i$ of $U$. Consider each such copy $U_i$ as sitting inside a copy $X_i$ of $X$. Consider a cell $\Phi:D^n\to X_i$ and a connected component $V$ of $\Phi^{-1}(U_i)$. Then there is a cell $\Phi':D_n\to E$ such that $\Phi'|_V=\Phi|_V$ (just let $\Phi'$ be the lift of $\Phi$ to $E$ that agrees with it on $V$). Since $A$ is open in every cell of $E$, we conclude that $V\cap \Phi^{-1}(A\cap U_i)=V\cap\Phi'^{-1}(A\cap U_i)$ is open. Since $V$ was an arbitrary connected component of $\Phi^{-1}(U_i)$, this means $\Phi^{-1}(A\cap U_i)$ is open.

Thus we have shown that $A\cap U_i$ is open in every cell of $X_i$. Since $X_i$ has the weak topology, this means $A\cap U_i$ is open in $X_i$, and thus open in $U_i$. Since $p^{-1}(U)$ is the disjoint union of the $U_i$, this means $A\cap p^{-1}(U)$ is open in $p^{-1}(U)$ and hence in $E$.

Eric Wofsey
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  • $\Phi'$ is a map from $D^n$ to $E$, $\Phi$ is a map from $D^n$ to $X_i$ which is in $X$, I don't see why $\Phi'$ and $\Phi$ restricts to $V$ is the same, since they map to different spaces. – CuriousAlpaca Oct 18 '20 at 02:07
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    $X_i$ is a homeomorphic copy of $X$ obtained by replacing each element of $U\subseteq X$ with its preimage in $U_i$. So, restricted to $V$, both $\Phi'$ and $\Phi$ map into $U_i$. – Eric Wofsey Oct 18 '20 at 03:03
  • Thank you, I get that now. But why do we need to lift $\Phi$ to $\Phi'$? $\Phi$ is a cell of $X_i$, so it must also be a cell of $E$? – CuriousAlpaca Oct 18 '20 at 04:59
  • $X_i$ is not contained in $E$; it's just an abstract copy of $X$ containing $U_i$. We can partition $p^{-1}(U)$ into copies of $U$ (the $U_i$) since we chose $U$ to be evenly covered, but we can't necessarily get copies of all of $X$ inside $E$. (But, crucially, since $D^n$ is simply connected, we can always lift maps out of $D^n$ up to $E$.) – Eric Wofsey Oct 18 '20 at 05:16
  • I see what you mean. So $E$ must contain at least 1 copy of $X$, since it is a covering space, but it may not necessarily contain multiple copies of $X$. Is this correct thinking? – CuriousAlpaca Oct 18 '20 at 05:54
  • $E$ need not contain any copy of $X$. For instance, $\mathbb{R}$ is a covering space of $S^1$ but does not contain any subspace homeomorphic to $S^1$. – Eric Wofsey Oct 18 '20 at 13:43
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    In your second sentence, why does it suffice to show that $A \cap p^{-1}(U)$ is open for any open $U \subseteq X$? How would this imply the weak topology? – Yuxi L Nov 28 '23 at 14:19