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Jul 23, 2013 at 13:22 comment added HenrikRüping 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.
Jul 23, 2013 at 12:49 comment added aglearner Thank you Henrik! Do I understand at least correctly that $\pi_1(M)$ is at worse a finite cyclic extension on $\pi_1(M/S^1)$?
Jul 23, 2013 at 12:47 vote accept aglearner
Jul 23, 2013 at 12:33 history answered HenrikRüping CC BY-SA 3.0