Suppose $M$ is a *connected* smooth manifold with a smooth $S^1$-action that fixes a point in $M$. Let $X$ be a $K(\pi,1)$-space and let $\varphi: M\to X$ be a continuous map. 


**Question.** How to prove that $\varphi$ can be homotoped to a map $\varphi': M\to X$ that sends each orbit of the $S^1$ action to a point in $X$? 


I would be grateful for a short proof or a reference. 


**Remark.** I can prove the statement when $X$ is a negatively curved manifold, but I am confident that the statement holds as well when $X$ is $K(\pi,1)$.



Maps to $K(\pi,1)$ spaces from manifolds with $S^1$-action
I would like to explain why the answer to the question is negative in general, following the suggestion of Tom Godwillie (however see **PS** for a positive answer if $dimX<\infty$)


Let us note first that there is an $S^1$-action on $\mathbb RP^2=M$ such that the quotient is a the segment.


Now, apply the idea of Tom Goodwille. Namely take $\Gamma=\mathbb Z\_2$, and consider the inclusion $\mathbb RP^2\to \mathbb RP^{\infty}=K(\mathbb Z\_2,1)$ that induces an isomorphism on $\pi\_1$. Clearly, this map is not contractible, so it gives a counterexample to the claim in the question. 


**PS**. However, using the remark of Misha it is also possible to prove that the answer to the question is positive in the case when the group $\pi=\Gamma=\pi\_1(X)$ has no torsion elements. In particular, it holds when $X$ is finite-dimensional.


**Sketch proof.** Denote $M/S^1$ by $Y$ and let $p$ be the projection map. I claim that *the kernel of the homomorphism $\phi^M\_Y:\pi\_1(M)\to \pi\_1(Y)$ induced by $p$ is generated by torsion elements*. Since $\Gamma$ has no torsion elements, it follows that any homomorphism $\phi^M\_X:\pi\_1(M)\to \Gamma$ factors through a homomorphism $\phi^M\_Y$, i.e. there exists $\phi^Y\_X:\pi\_1(Y)\to \pi\_1(X)$ such that $$\phi^M\_X=\phi^Y\_X\circ\phi^M\_Y.$$


Now, since $X$ is aspherical, the homomorphism $\phi^Y\_X$ is induced by a map $\psi: Y\to X$. Finally note that the original map $\varphi: M\to X$ and the composed map $\psi\circ p$ are homotopic, since they induce the same homomorphism on $\pi\_1(M)$.


