# path space is contractible

let $(X,*)$ be a pointed topological space. I want to show that the path space $PX$ of paths $\gamma:[0,1]\to X;\; \gamma(0)=*$, is contractible. so i will show that the map $f:PX\to *; \gamma\mapsto \gamma(0)$ is a homotopy equivalence with homotopy inverse the map $g:*\to PX;\; *\mapsto \gamma_*$ where $\gamma_*$ is the constant path on $*$. First, $f\circ g=id_*$. It remains to show that $g\circ f$ is homotopic to $id_{PX}$. Define the map $H_t: PX\to PX; \gamma\mapsto \lambda$ where $\lambda$ is the path defined by $\lambda(t')=\gamma(t'(1-t))$ and this is clearly a homotopy. Is it correct? and can we say that the map $r:PX\to \{\gamma_*\}$ that sends all paths $\gamma\in PX$ to the constant path $\gamma_*$ is a deformation retraction of $PX$ onto $\{\gamma_*\}$?

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What topology are you placing on the path space? –  Chris Eagle Dec 6 '12 at 14:16
I think it is the compact open topology. –  palio Dec 6 '12 at 17:45
In the same context, we know that $\pi_*(\Omega X)=\pi_{*+1}(X)$. Thus can one conclude that the loop space of a contractible space is also contractible? –  MyIsmail Jan 1 at 8:58
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  Najib Idrissi Jan 1 at 10:02

Yes, I think all that is exactly right. In fact $H$ appears to give a strong deformation retraction to $\{\gamma_*\}$, since it's constant on $\gamma_*$.
The only slightly unclear part to me is why $H$ is continuous. Checking that requires going into the topology defined on $PX$. But that shouldn't be hard.
indeed the question is more general. given $(X,*)$ a based top space, when does the map $r:X\to \{*\};x\mapsto *$ defines a deformation retraction of $X$ onto $\{*\}$ isn't $r$ always continuous? do we have a continuity concern here? –  palio Sep 2 '12 at 0:08
$r$ is certainly continuous. It's also clearly a retraction. But a deformation retraction (described well in Wikipedia) is a homotopy between a retraction and the identity map of $X$. $r$ by itself will not define that, and in general such a deformation retraction will not exist unless $X$ is contractible. Does that help? –  Hew Wolff Sep 2 '12 at 2:18
i think we need to find a continuous choice of path that begins in $x\in X$ and ends in $*$ and this is not always possible. –  palio Sep 2 '12 at 7:00