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?



Fundamental group of a manifold with an $S^1$-action
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](http://www.ams.org/journals/tran/1968-131-01/S0002-9947-1968-0219086-9/S0002-9947-1968-0219086-9.pdf).


(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).


