What is the path space mod the loop space? Suppose $(X, \ast)$ is a pointed, connected topological space. Is $PX/ \Omega X$ (weakly) homotopy equivalent to anything nice?
Here
$$
PX = \{f: [0, 1]\to X \mbox{  s.t. } f(0)= \ast\}
$$
and
$$
\Omega X = \{f: [0, 1]\to X \mbox{  s.t. } f(0), f(1)= \ast\}
$$
In the example of the unit circle, I think that $P S^1/ \Omega S^1 \cong S^1.$ Based on that I'm led to the hypothesis that $P X/ \Omega X \cong X.$
More generally, we have a fibration of pairs
$$
(PX, \Omega X) \to (X, \ast)
$$
with fiber $(\Omega X, \Omega X).$ If we take the long exact sequence associated to a fibration we would appear to get the desired result, but this can't be right because no similar results holds for trivial fibrations.
 A: Usually, the result will be $\Sigma \Omega X$ (the answer for $S^1$ should actually be an infinite wedge of circles), this is not particularly special to path and loop spaces, but rather to pairs of spaces where the total space is contractible.
If $X$ is a based CW complex or manifold $\Omega X \hookrightarrow PX$ should be a cofibration; I am not sure where to find a reference for this fact, but maybe in some of Milnor's work on the homotopy type of function spaces between CW complexes.
Anyways, this means that $PX/\Omega X \simeq PX \cup \operatorname{cone}(\Omega X)$. The latter maps to $\Sigma (\Omega X)$ which we model as $\operatorname{cone}(\Omega X)/(\Omega X) \times \{0\}$ by collapsing $PX$ to a point. Again, in reasonable circumstances like $X$ a CW complex or manifold, this map will be a homotopy equivalence by work of Milnor. I am not sure if it is true in general, it is really a question about if one needs to take CW approximations before gluing in cones to calculate homotopy pushouts.
