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Let $Z_m$ act on $S^1$ by multiplication with $e^{2\pi ki/m}$ for $k \in Z_m$. Let $X = S^1 / Z_m$ be the orbit space of this action. Then we have a universal cover $q:S^1 \rightarrow X$ given by the canonical projection. To determine the fundamental group of $X$ we can consider the group of deck transformations $Aut(q)$ of $q$, since we know that $\Pi_1(X) = Aut(q)$.

$Aut(q)$ is by definition the group of all homeomorphisms $f: S^1 \rightarrow S^1$ such that $qf=q$. We see that all rotations by degree $e^{2\pi ki/m}$ for $k \in Z_m$ are such homeos. But why are there no other ones?

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    $\begingroup$ Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $\pi_1 (X) \cong \text{Aut}(q)$. In fact, it is correct that $\text{Aut}(q) \cong \mathbb{Z}/m\mathbb{Z}$, but $\pi_1 (X) \cong \mathbb{Z}$, since $X$ is homeomorphic to $S^1$. $\endgroup$ – Sameer Kailasa Jan 2 at 19:43
  • $\begingroup$ yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$ $\endgroup$ – CHwC Jan 2 at 19:52
  • $\begingroup$ I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations $\endgroup$ – CHwC Jan 2 at 19:53
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$\pi_1(X)$ is not meant to be equal to ${\rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.

[An example of a universal cover of $X$ would be the map $q_{\rm univ} : \mathbb R \to X$, sending $x \mapsto x {\rm \ mod \ } \tfrac{2\pi}{m}$. This is a universal cover because $\mathbb R$ is simply connected. The deck transformation group for this universal cover is ${\rm Aut}(q_{\rm univ}) = \mathbb Z$. Since $X$ itself is a small circle, $\pi_1(X) = \mathbb Z$, which agrees with ${\rm Aut}(q_{\rm univ})$.]

As for your original covering $q : S^1 \to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_\star (\pi_1(S^1)) $ is the image of $\pi_1(S^1)$ under the group homomorphism $q_\star : \pi_1(S^1) \to \pi_1(X)$, then ${\rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_\star (\pi_1(S^1))$ in $\pi_1(X)$. Now $\pi_1(X) = \mathbb Z$, and $H$ is the subgroup $m\mathbb Z$, which is a normal subgroup, so the theorem tells us that ${\rm Aut}(q) = \mathbb Z_m$. And this agrees perfectly with your counting.

Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x \in X$ consists of $m $ points. Given a point $x \in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.

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  • $\begingroup$ okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations? $\endgroup$ – CHwC Jan 2 at 19:50
  • $\begingroup$ when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally $\endgroup$ – CHwC Jan 2 at 19:51
  • $\begingroup$ @CHwC See edit... $\endgroup$ – Kenny Wong Jan 2 at 20:02
  • $\begingroup$ why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing $\endgroup$ – CHwC Jan 2 at 21:35
  • $\begingroup$ @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : \widetilde X \to X$ and a map $f : Y \to X$ with two lifts $\widetilde f_1, \widetilde f_2 : Y \to \widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = \widetilde X$ and $f = q$, then $\widetilde f_1, \widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere. $\endgroup$ – Kenny Wong Jan 2 at 21:40

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