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Is the pullback of a covering map $p:E\rightarrow B$ along a continuous map $f:C\rightarrow B$ a covering map?

J.J.
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    ..and I guess that (at least for nice topological spaces, where coverings are usually used) the pullback covering is the covering of $C$ corresponding to the kernel of $\pi_1(f) ,: \pi_1(C) \to \pi_1(B)$.. – fritz Feb 14 '17 at 15:12

2 Answers2

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Yes. Let $E^\prime=E\times_BC$ be the pullback with $p^\prime:E^\prime\rightarrow C$ the projection and $f^\prime:E^\prime\rightarrow E$ the other projection. Suppose $U\subseteq B$ is evenly covered by $p$, meaning that $p^{-1}(U)=\coprod_{i\in I}V_i$ with the $V_i$ open subsets of $E$, each of which is mapped homeomorphically onto $U$ by $p$. Then I claim that $f^{-1}(U)$, an open subset of $C$, is evenly covered by $p^\prime$. In fact, we have $(p^\prime)^{-1}(f^{-1}(U))=(f^\prime)^{-1}(p^{-1}(U))=\coprod_{i\in I}(f^\prime)^{-1}(V_i)$. So we need to check that each $(f^\prime)^{-1}(V_i)$ is mapped homeomorphically onto $f^{-1}(U)$ by $p^\prime$. But the restriction $p^\prime:(f^\prime)^{-1}(V_i)\rightarrow f^{-1}(U)$ is the pullback of the restriction $p:V_i\rightarrow U$ along $f:f^{-1}(U)\rightarrow U$, so because the restriction of $p$ is a homeomorphism, so is the restriction of $p^\prime$.

Keenan Kidwell
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    For further readers: homeomorphism comes from universal property of pullback along arrows $id_C|{f^{-1}(U)}$ and $(p|{V_i})^{-1} \circ f|_{f^{-1}(U)}$. – Stephen Dedalus Feb 26 '15 at 15:11
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    Why the pullback of the restriction $p : V_i \to U$ along $f:f^{-1}(U) \to U$is also homeomorphism? – ChoMedit Mar 27 '18 at 11:24
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    @ChoMedit We can see that explicitly: The total space for the pullback of $p\colon V_i\to U$ consists of pairs $(x,v)$ such that $x\in f^{-1}(U), v\in V_i$ and $f(x)=p(v)$. Since $p|{V_i}$ is invertible, we can write $v=(p|{V_i}^{-1})\circ f(x)$. Now, the projection map of the pullback works as $(x,(p|_{V_i}^{-1})\circ f(x))\mapsto x$, and it is clear that it is continuous and invertible, with the inverse continuous as well (because the inverse is given by continuous functions in each coordinate). – mathreader Jul 25 '22 at 20:38
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Yes. If $c \in C$, choose a n.h. $U$ of $f(c)$ such that the preimage $p^{-1}(U)$ of is a union of open subsets each mapping isomorphically to $U$. Then $f^{-1}(U)$ is a n.h. of $c$ with the analogous property (since the preimage of the pull-back over $f^{-1}(U)$ is the same as the pull-back of $p^{-1}(U) \to U$ via $f^{-1}(U) \to U$).

Matt E
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