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I have read on several places that given a (say path connected) topological space $X$ and its universal covering $\tilde{X}\stackrel{p}\rightarrow X$, there is an isomorphism

$$\mathrm{Deck}(\tilde{X}/X) \simeq \pi_1(X, x_0).$$

Here $\mathrm{Deck}(\tilde{X}/X)$ denotes the group of deck transformations of the universal cover and $x_0$ is some basepoint. (Maybe I am omitting some assumptions, but I am interested in this mainly for path connected smooth manifolds, which should be "nice enough" topological spaces.)

However, I was unable to find any explicit description of the isomorphism. What I have in mind is the following:

Given a loop $\gamma$ in $X$, what is the corresponding deck transformation? In other words, how does the fundamental group $\pi_1(X, x_0)$ act on the covering space $\tilde{X}$?

That is, I am interested in the "geometric picture" behind the isomorphism. I understand there is some choice involved, something like fixing some preimage $\tilde{x}_0 \in p^{-1}(x_0)$, I can imagine lifting the loop uniquely modulo this choice, however I cannot see how to obtain a homemorphism of $\tilde{X}$ using this lifted path (is it some use of the universal property of $\tilde{X}\rightarrow X$, perhaps?).

Thanks in advance for any help.

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  • $\begingroup$ what exactly is your problem? Given a loop $\gamma$ based at a point $x_0$ on $X$, you can lift it to a path $\tilde{\gamma}$ on the universal cover, the corresponding deck transform would send $\tilde\gamma(0)$ to $\tilde\gamma(1)$, by homotopy lifting property this is well-defined. You can do this for every point of $x \in X$ and get a well-defined bijection $\tilde X \to \tilde X$. I guess in order to show that it is a homeomorphism, you need to use that $\tilde X \to X$ is a local homeomorphism. $\endgroup$ – Ivan Bodhidharma Sep 22 '15 at 10:35
  • $\begingroup$ @IvanBodhidharma: I guess I do not understand the part "You can do this at every point $x \in X$" - the loop $\gamma$ is fixed, so how can I obtain, say, prescription where some point $\tilde{x}$ should be mapped if $p(\tilde{x})$ is not in the image of $\gamma$, i.e. "does not have anything to do with $\gamma$"? $\endgroup$ – Pavel Čoupek Sep 22 '15 at 12:13
  • $\begingroup$ I see. Actually, a deck transform is deformed by how it acts on a fibre, by uniqueness of path lifting (because of they way I phrased things in my previous comment, it is not obvious). In order to see how $\gamma \in \pi_1(X, x_0)$ acts on $\tilde X$ you first look at the action on the fibre over $x_0$ by lifting the loop to paths that start in preimages of $x_0$ then use the unique path lifting to extend this to all of $\tilde X$ $\endgroup$ – Ivan Bodhidharma Sep 22 '15 at 13:50
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Here is how it goes.

Let $B$, be a space nice enough to have a (simply connected) universal cover, say $B$ is connected, locally connected and semi-locally simply connected. Let $(X,x_0)\to (B,b_0)$ be its universal cover.

Take a loop $\gamma: (S^1,1)\to (B,b_0)$ then you can lift $\gamma$ to a path $\overline{\gamma}: I\to X$ that projects to $\gamma$. Now $\overline{\gamma}(1)$ is an element of $X_{b_0}$. You can use then the following theorem.

Let $(Y,y_0)\to (B,b_0)$ be a (path) onnected and locally path connected space over $B$ and $(X,x_0)\to (B,b_0)$ is a cover of $B$, then a lift of $(Y,y_0)\to (B,b_0)$ to $(Y,y_0)\to (X,x_0)$ exists iff the image of $\pi_1(Y,y_0)$ inside $\pi_1(B,b_0)$ is contained in the image of $\pi_1(X,x_0)$ inside $\pi_1(B,b_0)$

Use the previous theorem with $(Y,y_0)=(X,\overline{\gamma}(1) )$. This tells you that there exists a covering map $X\to X$ sending $x_0$ to $\gamma(1)$.

It is easy to see that this map depends only on the homotopy class of $\gamma$ using the following result

Let $(X,x_0)$ be a cover of $(B,b_0)$ and $Y$ be a connected space over $B$. If two liftings of $Y\to B$ to $Y\to X$ coincide at some $y_0$ in $Y$, the they're equal.

This tells you that if $\overline{\gamma}(1)=\overline{\tau}(1)$ then the two morphisms $X\to X$ you get, coincide. Moreover, using the inverse of $\gamma$, you see that the morphisms $X\to X$ you get are automorphisms.

This gives you a well defined map $\pi_1(B,b_0)\to \text{Aut}_B(X)$. Using what I said before, it is easy to see that it is an isomorphism.

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  • $\begingroup$ I agree with the answer of A. Rod, but also feel it is preferable to work using the algebraic model of covering maps in terms of covering morphisms of groupoids., see [2]:math.stackexchange.com/questions/1438169/… $\endgroup$ – Ronnie Brown Sep 23 '15 at 11:31

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