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Hatcher defined a covering space as follows:

$\textbf{Defn:}$ A covering space of a space $X$ is a space $\tilde{X}$ together with a map $p: \tilde{X} \to X$ satisfying the following conditions: There exists an open cover $\{ U_{\alpha}\}$ of $X$ such that for each $\alpha$, $p^{-1}(U_{\alpha})$is a disjoint union of open sets in $\tilde{X}$ each of which is mapped by $p$ homeomorphically onto $U_{\alpha}$

I don't understand how such a map can fail to be surjective. Any concrete examples??

Balloon
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user7090
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1 Answers1

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For instance, if $\tilde{X}$ is the empty set, and $p$ is the empty map. Then each $p^{-1}(X)$ is empty, so it is an (empty) disjoint union of open sets, "each" of which are mapped homeomorphically onto $X$ (vacuously, as there are no opens to check the condition on).

Elle Najt
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    And there are intermediate situations: if $X$ is not connected, one component can be completely outside of the image for this reason. – Mariano Suárez-Álvarez Oct 16 '15 at 01:26
  • This is not correct. The definitions asks that $p$ maps each disjoint open set onto $U_\alpha$. This cannot happen in your case, since the image of an empty set is empty. – Pedro Oct 22 '15 at 17:31
  • @PedroTamaroff I think it is correct. The open cover of $\emptyset$ is not the empty set ${ \emptyset }$, it is the empty collection of sets ${ }$. Each open $U \in { }$ satisfies any condition you want, since there are no opens to check. Do you disagree? – Elle Najt Oct 24 '15 at 15:31
  • @area The cover is of the base space, not the covering space. – Pedro Oct 24 '15 at 16:50
  • @PedroTamaroff I was being imprecise. You take the covering ${X}$ of the base space. What I mean is that $p^{-1}(X)$ is an empty disjoint union - it is $\cup_{U \in \emptyset} U$. Since there are no opens in this disjoint union to check, vacuously each maps onto $X$. – Elle Najt Oct 24 '15 at 17:06
  • No, they don't map onto $X$. The induced map must be an homeomorphism, and it clearly isn't in this case. – Pedro Oct 24 '15 at 17:32
  • @PedroTamaroff I think you are misunderstanding me. It is analogous to the following situation: Let A be a collection of some spaces. In this case, X. Then the empty set is always a disjoint union of spaces in A. How? It is the empty disjoint union. Each set in the disjoint union is homeomorphic to X, but this is only because there are no sets in the disjoint union. There are no induced maps that can fail to be a homeo, so all of them are homeomorphisms. This changes if you require that the preimage is a nonempty disjoint union of open sets in ... – Elle Najt Oct 24 '15 at 17:41
  • @PedroTamaroff the covering. But really, this is maximal pedantry, and completely a distraction from the beautiful ideas of covering space theory. – Elle Najt Oct 24 '15 at 17:42
  • The empty function is a function to $X$, but it is not an homeomorphism, because it is not onto, say. – Pedro Oct 24 '15 at 17:45
  • @PedroTamaroff I know that the empty function is not a homemorphism. That is not my point. – Elle Najt Oct 24 '15 at 17:59
  • @AreaMan OK, I see now what the misunderstanding is. You're allowing to take a union indexed by the empty set, and I wasn't. – Pedro Oct 31 '15 at 03:13
  • @PedroTamaroff Precisely the difference. I am pretty sure that taking empty unions is allowed under standard set theoretic rules... it is a pretty weird thing to do though. If you know some sheaf theory, you can use this trick to figure out what a sheaf of Abelian groups on the empty set has to be. – Elle Najt Nov 01 '15 at 15:04