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 $\{\gamma_*\}$?
path space is contractible
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1 Answers
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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.
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0i 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. – 2012-09-02