Path space fibration

If ${\displaystyle f\colon X\to Y}$ is any map, then the mapping path space ${\displaystyle P_{f}}$ of ${\displaystyle f}$ is the pullback of the fibration ${\displaystyle Y^{I}\to Y,\,\chi \mapsto \chi (1)}$ along ${\displaystyle f}$. (A mapping path space satisfies the universal property that is dual to that of a mapping cylinder, which is a push-out. Because of this, a mapping path space is also called a mapping cocylinder.^[3])

Since a fibration pulls back to a fibration, if Y is based, one has the fibration

${\displaystyle F_{f}\hookrightarrow P_{f}{\overset {p}{\to }}Y}$

where ${\displaystyle p(x,\chi )=\chi (0)}$ and ${\displaystyle F_{f}}$ is the homotopy fiber, the pullback of the fibration ${\displaystyle PY{\overset {\chi \mapsto \chi (1)}{\longrightarrow }}Y}$ along ${\displaystyle f}$.

Note also ${\displaystyle f}$ is the composition

${\displaystyle X{\overset {\phi }{\to }}P_{f}{\overset {p}{\to }}Y}$

where the first map ${\displaystyle \phi }$ sends x to ${\displaystyle (x,c_{f(x)})}$; here ${\displaystyle c_{f(x)}}$ denotes the constant path with value ${\displaystyle f(x)}$. Clearly, ${\displaystyle \phi }$ is a homotopy equivalence; thus, the above decomposition says that any map is a fibration up to homotopy equivalence.

If ${\displaystyle f}$ is a fibration to begin with, then the map ${\displaystyle \phi \colon X\to P_{f}}$ is a fiber-homotopy equivalence and, consequently,^[4] the fibers of ${\displaystyle f}$ over the path-component of the base point are homotopy equivalent to the homotopy fiber ${\displaystyle F_{f}}$ of ${\displaystyle f}$.

By definition, a path in a space X is a map from the unit interval I to X. Again by definition, the product of two paths ${\displaystyle \alpha ,\beta }$ such that ${\displaystyle \alpha (1)=\beta (0)}$ is the path ${\displaystyle \beta \cdot \alpha \colon I\to X}$ given by:

${\displaystyle (\beta \cdot \alpha )(t)={\begin{cases}\alpha (2t)&{\text{if }}0\leq t\leq 1/2\\\beta (2t-1)&{\text{if }}1/2\leq t\leq 1\\\end{cases}}}$.

This product, in general, fails to be associative on the nose: ${\displaystyle (\gamma \cdot \beta )\cdot \alpha \neq \gamma \cdot (\beta \cdot \alpha )}$, as seen directly. One solution to this failure is to pass to homotopy classes: one has ${\displaystyle [(\gamma \cdot \beta )\cdot \alpha ]=[\gamma \cdot (\beta \cdot \alpha )]}$. Another solution is to work with paths of arbitrary lengths, leading to the notions of Moore's path space and Moore's path space fibration, described below.^[5] (A more sophisticated solution is to rethink composition: work with an arbitrary family of compositions; see the introduction of Lurie's paper,^[6] leading to the notion of an operad.)

Given a based space ${\displaystyle (X,*)}$, we let

${\displaystyle P'X=\{f\colon [0,r]\to X\mid r\geq 0,f(0)=*\}.}$

An element f of this set has a unique extension ${\displaystyle {\widetilde {f}}}$ to the interval ${\displaystyle [0,\infty )}$ such that ${\displaystyle {\widetilde {f}}(t)=f(r),\,t\geq r}$. Thus, the set can be identified as a subspace of ${\displaystyle \operatorname {Map} ([0,\infty ),X)}$. The resulting space is called the Moore path space of X, after John Coleman Moore, who introduced the concept. Then, just as before, there is a fibration, Moore's path space fibration:

${\displaystyle \Omega 'X\hookrightarrow P'X{\overset {p}{\to }}X}$

where p sends each ${\displaystyle f:[0,r]\to X}$ to ${\displaystyle f(r)}$ and ${\displaystyle \Omega 'X=p^{-1}(*)}$ is the fiber. It turns out that ${\displaystyle \Omega X}$ and ${\displaystyle \Omega 'X}$ are homotopy equivalent.

Now, we define the product map

${\displaystyle \mu :P'X\times \Omega 'X\to P'X}$

by: for ${\displaystyle f\colon [0,r]\to X}$ and ${\displaystyle g\colon [0,s]\to X}$,

${\displaystyle \mu (g,f)(t)={\begin{cases}f(t)&{\text{if }}0\leq t\leq r\\g(t-r)&{\text{if }}r\leq t\leq s+r\\\end{cases}}}$.

This product is manifestly associative. In particular, with μ restricted to Ω'X × Ω'X, we have that Ω'X is a topological monoid (in the category of all spaces). Moreover, this monoid Ω'X acts on P'X through the original μ. In fact, ${\displaystyle p:P'X\to X}$ is an Ω'X-fibration.^[7]