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?

  • $\begingroup$ 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. $\endgroup$ – William Jan 26 '20 at 16:45
  • $\begingroup$ 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. $\endgroup$ – William Jan 26 '20 at 16:50
  • $\begingroup$ 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! $\endgroup$ – k-t Jan 27 '20 at 5:28
  • $\begingroup$ *helped to clear up $\endgroup$ – 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, $\iota(\Omega (X,x_0)) = (X,x_0)^{(S^1,*)} \subset X^{S^1}$.


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