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Given a topological space $X$ with a base point $x_0$, which is locally path-connected, path-connected, and semi-locally simply connected.

So, by Hatcher's book, its universal covering can be constructed by $$\tilde{X}:=\{[\gamma]|~\gamma:I\to X;\ \gamma(0)=x_0 \}$$ with the projection map $$p:\tilde{X}\to X: \ \ [\gamma]\mapsto \gamma(1)$$ where $[\gamma]$ is the homotopy equivalence class of $\gamma$.

Then we have this is a universal covering space of $X$ with the fiber $p^{-1}(x)$ is isomorphic to the set of the homotopy equivalence classes of paths from $x_0$ to $x$.

For every covering space, we have the monodromy action, i.e. an action of $\pi_1(X,x_0)$ on the fiber $p^{-1}(x)$.

I was trying to write this action explicitly for the universal covering. Obviously, it should be $$[\alpha]\cdot[\gamma] := [\alpha\ast \gamma]$$ where $[\alpha]\in \pi_1(X,x_0)$, $[\gamma]$ is an element of the set of the homotopy equivalence classes of paths from $x_0$ to $x$, and $[\alpha\ast \gamma]$ is the homotopy equivalence classe of the comobined path $\alpha\ast \gamma$.

(We know $\mathrm{Aut}(\tilde{X}|X)\cong \pi_1(X,x_0)$. Can we write out the group action from this?)

However, is there a geometric interpretation? In wikipedia, it reads every monodromy action should be a lifting of a loop $$\gamma:I\to X$$ to a path $$\tilde{\gamma}:I\to \tilde{X}.$$ Since the lifting path is not necessary a loop, $\tilde{\gamma}(1)$ can be a different element in $p^{-1}(x)$. Then we defined an action of $\pi_1(X,x_0)$ on $p^{-1}(x)$.

But for covering space, it is hard for me to see this. Because the path in $\tilde{X}$ is hard to imagine.

Any insight and answer are welcome!


$X$ is semi-locally simply connected if for every point $x$, we can find a neighborhood $U$ such that the loop inside can be contracted to a point in $X$, not necessarily in $U$. In other words, we have trivial homomorphism $$\pi_1(U,x)\to \pi_1(X,x).$$

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    $\begingroup$ Visualizing this in the full abstract setting is not really going to work. However, for your understanding it might be useful to visualize this for specific examples. Try to visualize it for the universal covering space of the circle $S^1$ which is $\mathbb R$ and which has fundamental group $\mathbb Z$. Or even better, for the universal covering space of the torus $T^2 = S^1 \times S^1$ which is $\mathbb R^2$ and which has fundamental group $\mathbb Z^2$. Also try the universal cover of the figure 8 which is the infinite 4-valent tree and has fundamental group $F_2$, the free group of rank 2. $\endgroup$ – Lee Mosher Nov 5 '18 at 16:22

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