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Let $X$ be a topological space that has a universal cover $p:\tilde{X}\to X$. Fix a basepoint $\tilde{x}_0\in \tilde{X}$ and let $x_0=p(\tilde{x}_0)$. Then the group $G$ of deck transformations of $\tilde{X}$ is isomorphic to $\pi_1(X,x_0)$: for a loop $\gamma$, let $\tilde{\gamma}$ be the lift of $\gamma$ such that $\tilde{\gamma}(0)=\tilde{x}_0$. Then there is a unique deck transformation $\tau_{\gamma}$ such that $\tau_{\gamma}(\tilde{x}_0)=\tilde{\gamma}(1)$. The map $[\gamma]\to \tau_{\gamma}$ gives an isomorphism $\pi_1(X,x_0)\to G$. Via this isomorphism, we see that the group $\pi_1(X,x_0)$ acts on $\tilde{X}$. Does this action depend on the choice of $\tilde{x}_0 \in p^{-1}(x_0)$? For example, in the case of $\Bbb R\to S^1$, the action does not depend on the choice of a basepoint of $\Bbb R$. So I'm curious about the general case.

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  • $\begingroup$ You have to make precise what you want to know. If $\tilde x_1 \in \tilde X$ and $x_1 = p(\tilde x_1) \ne x_0$, then $\pi_1(X,x_0) \ne \pi_1(X,x_1)$ and your question does not make much sense because two distinct groups cannot have an identical action (or you must explain what this means). So do you consider points in the same fiber? $\endgroup$
    – Paul Frost
    Oct 18, 2021 at 8:25
  • $\begingroup$ @PaulFrost Yes, I do. I have made an edit $\endgroup$
    – user302934
    Oct 18, 2021 at 13:03

1 Answer 1

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Note that there is no canonical isomorphism between $\pi_1(X,x_0)$ and $\pi_1(X, x_1)$ for $x_0,x_1\in X$, as it depends on the homotopy class of the paths from $x_0$ to $x_1$. So we know the answer must be false, but it's worse than that.

To define the action of $\pi_1(X, x_0)$ on $\tilde{X}$, one needs to pick a lifting $\tilde{x_0}$ of $x_0$, therefore the action of $\pi_1(X, x_0)$ is not canonically defined. So the problem is ill-posed.

The reason that it all works out for $\mathbb R/S^1$ is $\pi_1(S^1)$ is abelian, so the group action doesn't really depend on the choice of the base point lifting.

To give a slightly different perspective, the Deck transformation group $G$ by defintion acts on $\tilde{X}$ canonically. For any point $\tilde{x}\in \tilde{X}$, we can define an isomorphism $\rho_{\tilde{x}}: G\rightarrow \pi_1(X,x)$ by $\rho(g) = [\text{projection of any path from } \tilde{x} \text{ to } g(\tilde{x})]$, but this depends on the choice of $\tilde{x}$, not just its projection $x$. In Hatcher's book, he carefully denoted $G(\tilde{X})\approx \pi_1(X)$ instead of " $\cong$".

However, the action is canonical and therefore natural for $\tilde{x}\in\tilde{X}$. More precisely, we can define the functor from the fundamental groupoid to the category of group actions on $\tilde{X}/X$ by sending $\tilde{x}\in \tilde{X}$ to the group action of $\pi_1(X, x)$ on $\tilde{X}/X$ through the lifting $x\rightarrow\tilde{x}$.

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  • $\begingroup$ For a fixed $x_0\in X$, and two liftings $\tilde{x}_0, \tilde{x}_0'$, is there an example of $X$ that the two actions of $\pi_1(X,x_0)$ on $\tilde{X}$ are different? $\endgroup$
    – user302934
    Oct 26, 2021 at 18:18
  • $\begingroup$ We can give an abstract argument, following a nontrivial conjugation $g\rightarrow h^{-1}gh$ of $\pi_1$. To give a concrete one: Let $X$ be the wedge sum of two circles, and pick the intersection of the two circles as the base point, call it $O$. $\tilde{X}$ is well-known as the Cayley graph of $F_2$, so each arch is labeled by one of $a, b, a^{-1}, b^{-1}$. Now we lift $O$ to $A$, then the path $ab$ in $\pi_1(X, O)$ will carry $A$ to $B$ following the arc labelled by $a$, then to $C$ by $b$. However, if we lift $O$ to $B$, then $ab$ will carry $B$ following $b$ first, instead of $a$. $\endgroup$ Oct 26, 2021 at 19:04

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