# The fundamental group of an $S^1$-quotient

Let $M$ be a compact manifold with an $\mathbb S^1$-action that fixes a point on $M$. Is it correct that $\pi_1(M/S^1)=\pi_1(M)$?

I believe this is correct and is a corollary of some well-known statement.

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Suppose $S^1$ acts on $S^2$ by rotation around one axis. This action commutes with the antipodal map and hence gives an action on $\mathbb{R}P^2$. But $\mathbb{R}P^2/S^1\cong [0,1]$ and hence this cannot be true.
Thank you Henrik! Do I understand at least correctly that $\pi_1(M)$ is at worse a finite cyclic extension on $\pi_1(M/S^1)$? – aglearner Jul 23 '13 at 12:49
I tried to look at the $S^1$ space $\mathbb{R}P^2\times \mathbb{R}P^2$ with the diagonal action. It has a map to the old quotient. Let us look how preimages look like. On any point in the interior of the interval it is just $\mathbb{R}P^2$, on the boundary points it is $D^2/\pm \cong D^2$ and $[0,1]$. Thus it should be a homotopy pushout of those spaces and hence its fundamental group should be the pushout of $1 \leftarrow \mathbb{Z}/2 \rightarrow 1$ and hence it should be trivial. – HenrikRüping Jul 23 '13 at 13:22