$S^1$-quotient of the space of unbased contractible loops of a finite dimensional $K(\pi,1)$ Let $X$ be a finite dimensional $K(\pi,1)$ manifold. Recall that the space of unbased null-homotopic loops $L_0(X)$ is the space of contractible continuous maps $S^{1}\to X$. There is an obvious $S^1$-action on $L_0(X)$. 
Is it true that the quotient $L_0(X)/S^1$ is homotopy equivalent to $X$? Moreover, is there is a deformation retraction from $L_0(X)/S^1$ to the space of constant loops in it? What if $X$ is a finite dimensional $CW$-complex?
I think I can prove this statement in the case when $X$ is a smooth compact negatively curved manifold and we consider smooth maps from $S^1$ it. (I believe smoothness should not be essential.)
This is the follow up of The space of contractible loops of a finite dimensional $K(\pi,1)$
 A: THIS ANSWER IS WRONG (in that it claims the result in excess generality)
Suppose that $X$ is a $K(\pi,1)$ and a CW complex. Let $E$ be a universal covering space of $X$, so that $E$ is contractible and there is a free $\pi$-action on $E$ with orbit space $E/\pi\cong X$. (I really mean free and properly discontinuous -- what Hatcher's book calls a "covering space action".)
This gives $L(E)$ a $\pi$-action, again free, with orbit space $L(E)/\pi\cong L_0X$. And $L(E)$ has also $S^1$ acting, commuting with the $\pi$-action, in such a way that 
$$
(L(E)/S^1)/\pi=L(E)/(S^1\times\pi)=(L(E)/\pi)/S^1\cong L_0(X)/S^1.
$$
On the other hand, a deformation retraction of $E$ to a point $p$ yields an $S^1$-equivariant deformation retraction of $L(E)$ to $L(p)$ (i.e. one that is compatible with the $S^1$-action), and this in turn yields a deformation retraction of $L(E)/S^1$ to $L(p)/S^1$. 
So $L(E)/S^1$ is contractible and has a free action of $\pi$ with orbit space $L_0(X)/S^1$. This implies that $L_0(X)/S^1$ is a $K(\pi,1)$.
THE ERROR in the answer as written is in the assertion that the action of $\pi$ on $L(E)/S^1$ is free. But if we add the hypothesis that the group $\pi$ is torsionfree (which is true in the kinds of cases that the question is asking about) then this is all right: if $g\in\pi$ fixes the $S^1$-orbit of the loop $\gamma:S^1\to X$, then $g\circ \pi=\pi\circ R$ for some rotation $R:S^1\to S^1$; since the action of $\pi$ on $E$ is free (in the strong sense), this cannot happen unless $g$ and $R$ have (the same) finite order. 
