Fundamental group of a manifold with an $S^1$-action Let $M$ be a compact connected manifold with an $S^1$-action. Suppose that $S^1$ has a fixed point in $M$. Is it true that $\pi_1(M)=\pi_1(M/S^1)$?
I is there some reference or a short proof of this fact?
PS. I am sorry for amending the question. In reality I only want to know that the kernel of the map $\pi_1(M)\to \pi_1(M/S^1)$ is finite. Is at least this fact true?
 A: In general, the answer to the original question is no.
In fact, $S^1$ acts on $\mathbf{RP}^2$ in such a way that that the action has a fixed point. The quotient space is homeomorphic to a closed interval, hence $\pi_1(\mathbf{RP}^2/S^1)=\{1\}$ whereas $\pi_1(\mathbf{RP}^2) = \mathbf{Z} / 2 \mathbf{Z}$. 
The description of the action is the following. We see $\mathbf{RP}^2$ as the quotient of $S^2$ by the antipodal map $x \mapsto -x$. Then the action of $S^1$ on $S^2$ given by rotations around the vertical axis descends to $\mathbf{RP}^2$. All the points have trivial stabilizer, except the point  corresponding to the class of the two poles (which is fixed) and the points corresponding to classes on the equatorial circle (whose stabilizers have order $2$).
Hovewer, the answer is yes when $M$ is simply connected. This is a consequence of a more general result of Armstrong concerning actions of compact Lie groups on simply connected spaces, see the paper
M-A. Armstrong, Calculating the fundamental group of an orbit space, Proceedings of the American Mathematical Society 84 (1982), 267-271, in particular Example 4. 
A: The answer to the amended question still seems to be NO For a source of counterexamples, check out Frank Raymond's 1968 paper on circle actions on 3-manifolds.
(it seems that the fundamental group of the quotient is always free, but the manifolds admitting such actions admit much more complicated than free-by-finite fundamental groups).
