Let $X$ be a finite dimensional $K(\pi,1)$ manifold. Is it true that the space contractible loops of this manifold can be contracted to the space of constant loops on $X$? What if $X$ is a finite dimensional $CW$-complex?


I understand that the answer is positive if $X$ is a negatively curved manifold, since in this case there is a geometric argument.



The space of contractible loops of a finite dimensional $K(\pi,1)$
The statement is true for a $K(\pi,1)$ but not true for other $X$. Finite dimensionality is not relevant. 


Here's a sketch: let $X = K(\pi,1)$ and I will assume $X$ has the homotopy type of a CW complex. Let $\Omega\_0 X$ be the space of contractible based loops in $X$. It's easy to show that the homotopy groups of this space are trivial in every degree (in degree 0 by definition and in higher degrees since $X$ is a $K(\pi,1)$ space, using the fact that $\pi\_j(\Omega\_0X) = \pi\_{j+1}(X)$ for $j >0$). 


Let $L\_0 X$ be the space of contractible unbased loops. Then the sequence
$$
\Omega\_0 X \to L\_0 X \to X
$$
is a fibration, where the second map is evaluation at the basepoint of the circle. Since the fiber has trivial homotopy groups, it follows that the map $L\_0X \to X$ is a weak homotopy equivalence. It follows that the section $X\to L\_0X$ given by the constant loops is also a weak homotopy equivalence. It's therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.


