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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general-topology
algebraic-topology
homotopy-theory
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0What topology are you placing on the path space? – 2012-12-06
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0I think it is the compact open topology. – 2012-12-06
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2In 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? – 2015-01-01
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