Moore path space.

Question

Let $X$ a topological space and $MX$ the Moore path space of $X$ there is two maps from $\alpha,\omega: MX\rightarrow X$ (evaluation in 0 and evaluation at the total length). The classical path object $X^{I}$ is a subspace of $MX$. Is is true that $(\alpha,\omega): MX\rightarrow X\times X$ is a fibration and the inclusion $X^{I}\rightarrow MX$ a weak equivalence?

Answer 1

To show that $(\alpha,\omega)$ is a fibration, we must define a path-lifting function $L$ as follows. The arguments are a length $d\geq 0$, a Moore path $u:[0,d]\to X$, a path $v:[0,1]\to X$ starting at $u(0)$, and a path $w:[0,1]\to X$ starting at $u(d)$. The output $L(d,u,v,w)$ must be a path in $MX$ such that $L(d,u,v,w)(t)$ is a Moore path from $v(t)$ to $w(t)$. This is easy to arrange: we construct $L(d,u,v,w)(t)$ by gluing the reverse of $v|_{[0,t]}$ with $u$ and $w|_{[0,t]}$.

Also, there is an evident inclusion $f:X^I\to MX$, and a map $g:MX\to X^I$ given by $g(d,u)(t)=u(dt)$. Both of these are compatible with $\alpha$ and $\omega$. We have $gf=1$ on the nose, and there is a homotopy $h:1\simeq fg$ given by $$ h(s,(d,u)) = (1-s+sd, t \mapsto u(dt/(1-s+sd))))$$

To check that this is continuous we use the definition $$ MX = \{(d,u) : d \geq 0, u : [0,\infty) \to \mathbb{R}, u(x)=u(d) \text{ for } x\geq d \} $$ (and topologise this as a subspace of $[0,\infty)\times X^{[0,\infty)}$, using the standard compactly generated topology on the second factor). We then need to check that the map $$ (s,d,t) \mapsto \min(d,dt/(1-s+sd)) $$ is continuous on $[0,\infty)\times [0,1]\times [0,\infty)$. Of course one has to treat the point $(1,0,0)$ separately, but the argument is straightforward.

Answer 2

You can show the evaluation map is a weak fibration (I think this is the term: I mean a map homotopy equivalent in the category of spaces over $X\times X$ to a fibration -- namely $X^I\to X\times X$), which is good enough for many purposes, including the formation of homotopy pullbacks.