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$\newcommand{\Aut}{\operatorname{Aut}} $ I would need your help regarding covering spaces.

I'm going to try to summarize what seems important and I'm going to ask you some questions regarding parts that I don't really get (in fact, I think that I have trouble to connect everything).


$(1)$ Let $p : E \rightarrow B$ be a covering space. There is a right group action of the fundamental group $\pi_{1}(B, b_{0})$ on the fiber $p^{- 1}(b_{0})$ defined in the following way :

$$\pi_{1}(B, b_{0}) \times p^{- 1}(b_{0}) \rightarrow p^{- 1}(b_{0}), ([\gamma], x) \mapsto \widetilde{\gamma}(1)$$

where $\widetilde{\gamma}$ is the unique lift of $\gamma$ such that $\widetilde{\gamma}(0) = x$.

The two mains theorems which are related to this action are the following :

  • Let $B$ be a path-connected, locally path-connected and semi-locally simply connected topological space and $b_{0} \in B$. Then, for every subgroup $H$ of the fundamental group $\pi_{1}(B, b_{0})$, there exists a covering space $p : (E, x_{0}) \rightarrow (B, b_{0})$ such that $E$ is path-connected and $p_{\ast}(\pi_{1}(E, x_{0})) = H$ (in particular, if $H = \{[e_{b_{0}}]\}$, then we obtain 'the' universal cover).

  • Let $B$ be a path-connected, locally path-connected and semi-locally simply connected topological space and $b_{0} \in B$. Then, the following application is a bijection :

$$\{\text{Equivalence classes of covering spaces from $(E, x_{0})$ to $(B,b_{0})$, $E$ path-connected}\}$$ $$\rightarrow \{\text{Subgroups of $\pi_{1}(B, b_{0})$}\},$$ $$[p] \mapsto p_{\ast}(\pi_{1}(E, x_{0}))$$ (if we get rid of base points in the total space $E$, then we obtain a bijection between the equivalence classes and the conjugacy classes of subgroups).

$(2)$ Let $p : E \rightarrow B$ be a covering space. There is a left group action of the group $\Aut(p)$ (of automorphisms of covering maps) on the fiber $p^{- 1}(b_{0})$ defined in the following way :

$$\Aut(p) \times p^{- 1}(b_{0}) \rightarrow p^{- 1}(b_{0}), (f, x) \mapsto f(x)$$

I noticed one important theorem :

Let $p : E \rightarrow B$ be a regular covering space, $b_{0} \in B$ and $x_{0} \in p^{- 1}(b_{0})$. Moreover, let $E$ be locally path-connected. Then $\Aut(p)$ is ismorphic (as a group) to the quotient group $\pi_{1}(B, b_{0}) \backslash p_{\ast}(\pi_{1}(E, x_{0}))$. Also, if $E$ simply connected, then $\Aut(p)$ is ismorphic to $\pi_{1}(B, b_{0})$.

$(3)$ If we consider any topological space $X$ and a topological group $G$ acting on it (by a covering space action), then :

  • $pr : X \rightarrow X/G$ is a covering space ;
  • If $X$ is path-connected, then $\Aut(p)$ is isomorphic (as a group) to $G$ ;
  • If $X$ is path-connected and locally path-connected, then the quotient group $\pi_{1}(X/G) \backslash p_{\ast}(\pi_{1}(X))$ is isomorphic to $G$ ;
  • If $X$ is simply connected, then $\pi_{1}(X/G)$ is isomorphic to $G$ ;

First of all, could you tell me if something is wrong in what I said ? Second, these are my main 'doubts' and questions where I need your help :

  • I cannot see the link between the two actions on the fiber.

  • As I understood it, approach $(1)$ is useful when you start from the base space (i.e., when you start from "below"), you already know its fundamental group and you want to find every possible covering spaces of this base space.

  • Concerning approach $(2)$, I understood that it is useful when you start with a covering space (i.e., when you start from "above"), but I don't really get it...

  • Concerning approach $(3)$, I don't really understand it. I mean, I just have the feeling that your start from an arbitrary space, an arbitrary covering space action and you want to find the fundamental group of another space. If I consider for example $S^{1}$, I already know that I will have to consider it as $\mathbb{R} \backslash \mathbb{Z}$, where $\mathbb{Z}$ acts by translation on $\mathbb{R}$, but that's because I already know the fundamental group of $S^{1}$ (so I know that I don't do it in the good way). If I would not know its fundamental group, I clearly would not be able to see how to use approach $(3)$ to find it (Why do I have to choose $\mathbb{R}$ ? Maybe there is another covering space action than the one of $\mathbb{Z}$ ?).

I know it's a bit long, but I took some time in order to try to pose coreectly the problem.

Thank you !

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  • $\begingroup$ I like the idea of "algebraic model" and for covering spaces we have the model of covering morphism of groupoids, explained in my book "Topology and groupoids". . $\endgroup$ – Ronnie Brown Jun 2 at 10:43
  • $\begingroup$ For you first point, have you seen a proof of the important theorem you stated? The construction of the isomorphism might help you to understand the correspondence between the two actions. For the next two points I don't really understand what you are looking for. Do you want examples of applications of approach (1) and (2) ? $\endgroup$ – Adam Chalumeau Jun 3 at 20:16
  • $\begingroup$ Thank you for your answers. Some time ago actually... I should take a look again. If we choose for example the circle $S^{1}$, I'm going to try and tell me please if something is wrong. $\endgroup$ – deeppinkwater Jun 3 at 20:30
  • $\begingroup$ Approach 1 : I know that the fundamental group is isomorphic $Z$, so I can look at the subgroups of $Z$ to find every covering spaces. $\endgroup$ – deeppinkwater Jun 3 at 20:31
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    $\begingroup$ A relates answer is at math.stackexchange.com/questions/135457/… $\endgroup$ – Ronnie Brown Jun 4 at 14:21

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