# What is the fiber of the path space fibration?

Schematically, I understand the path space fibration $$PX$$ over some path-connected, pointed topological space $$X$$ with base point $$x_o$$ as: $$\Omega X \hookrightarrow PX \twoheadrightarrow X,$$ where the first arrow is the inclusion and the second is the evaluation map.

If we understand $$PX$$ as the space of paths in $$X$$ (i.e., continuous maps $$p(t)$$ from the unit interval to $$X$$, with $$p(0) = x_o$$), then it seems that the natural definition of $$\Omega X$$ is the space of paths $$p(t)$$ with $$p(0) = p(1) = x_o$$. However, I have found conflicting statements from various sources:

• This MSE question says that $$\Omega X$$ is the space I have just described.
• The Wikipedia article for path space fibration says that $$\Omega X$$ is the loop space of $$X$$; the Wikipedia article for loop space defines that space as the space of continuous pointed maps from $$S^1$$ (with base point) to $$X$$.
• This MSE question points out that the space of continuous pointed maps from $$S^1$$ to $$X$$ is not the same as the space of paths in $$X$$ beginning and ending at the same point, which I think is correct due to the fact that elements of the latter may be discontinuous at $$x_o$$ between $$t=1$$ and $$t=0$$.

Having become thoroughly confused by everything I found on the internet, I turned to Hatcher, where I discovered a lengthy and fairly dense discussion on loop spaces which requires the use of "James reduced products," a concept I have not encountered before in my coursework. This leads me to believe that perhaps the whole situation is more complicated than I initially thought.

My question is, what is $$\Omega X$$, and what subtlety am I missing here that is leading to the confusion described above?

• I think your third point is wrong, you may be misinterpreting part of the linked question. Since $S^1 \cong [0,1] / \{0,1\}$ any pointed map $S^1 \to X$ gives you a path $[0,1] \to X$ that starts and ends at the basepoint, and vice versa; this a homeomorphism between $Map_\bullet(S^1, X)$ and $L(x,x)$. The question asserts that if $X$ has the structure of a smooth manifold then the space of smooth pointed maps $S^1 \to X$ is not the same as smooth loops $[0,1]\to X$, since the latter might not induce a smooth function from the circle. – William Jan 26 '20 at 16:45
• The smooth category is tricky to work in for this context, because the composition of two smooth loops/paths may not even be smooth. You would have to add an extra assumption that they loops/paths behave a certain way in some $\epsilon$ neighbourhood of the basepoint in order for the composition to again be smooth. However in the topological category there is no issue. – William Jan 26 '20 at 16:50
• Paul Frost answered explicitly what $\Omega X$ is, and @William helped to up my confusion - I had not been careful enough to distinguish "smooth" and "continuous." Thanks! – k-t Jan 27 '20 at 5:28
• *helped to clear up – k-t Jan 27 '20 at 17:45

The definition in your question is that for pointed spaces $$(X,x_0)$$. More precisely we should write $$P(X,x_0) = (X,x_0)^{(I,0)}$$ = set of all basepoint-preserving maps $$(I,0) \to (X,x_0)$$ with compact-open topology for the pointed path space, $$p : P(X,x_0) \to (X,x_0), p(u) = u(1)$$, and $$\Omega(X,x_0) = p^{-1}(x_0)$$ fot the pointed loop space. Both have as basepoint the constant path at $$x_0$$. Then $$\Omega(X,x_0)$$ is the fiber over the basepoint $$x_0 \in X$$.

You can do an analogous construction for unbased spaces $$X$$:

$$PX = X^I$$ = set of all maps $$I \to X$$ with compact-open topology is the free path space. The evaluation map $$p : PX \to X, p(u) = u(1)$$, is a fibration. Its fibers are the sets $$p^{-1}(x) = \{u \in X^I \mid p(u) = u(1) = x \} =(X,x)^{(I,1)}$$. The latter is homeomorphic to $$P(X,x)$$.

A third construction is the free loop space of a space $$X$$:

$$\mathcal L X = X^{S^1} .$$

It can be viewed as the unpointed version of $$\Omega (X,x_0)$$. There is a canonical embedding $$\iota : \Omega (X,x_0) \to \mathcal L X$$: Each $$u \in \Omega (X,x_0)$$ is a closed path $$u : I \to X$$ such that $$p(0) = p(1) = x_0$$ which determines a unique continuous $$\hat u : I/\{0, 1\} \to X$$ and via the identification $$I/\{0, 1\} = S^1$$ this gives us $$\iota(u) \in \mathcal L X$$.

Note that this construction also allows to identify $$\Omega (X,x_0)$$ with $$(X,x_0)^{(S^1,*)}$$. In fact, $$\iota(\Omega (X,x_0)) = (X,x_0)^{(S^1,*)} \subset X^{S^1}$$.