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Consider a path connected topological space $X$, one can equip its path space $PX=\{ \gamma: [0,1] \longrightarrow X \; continuous\}$ with the compact open topology. We call a motion planning algorithm of $X$, any continuous section $s:X\times X \longrightarrow PX$ of the evaluation $$ev: PX \longrightarrow X\times X, \; \gamma\mapsto (\gamma(0), \gamma(1))$$ M. Farber has shown that such section exists if and only if $X$ is contractible. Hence, we denote $\mathcal{M}(X)$ the space of such sections.

Question: Is $\mathcal{M}(X)$ also contractible when $X$ is contractible?

Any remarks or comments are welcome, thank

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  • $\begingroup$ The map $ev : PX \to X\times X$ you give is always a fibration. If $X$ is contractible, $ev$ is also a homotopy equivalence. By a standard result in the theory of fibrations, it follows that $ev$ is a homotopy equivalence over $X\times X$, i.e. there is a section $s$ of $ev$ and a homotopy $H : PX \times I \to PX$ between the identity of $PX$ and $s\circ ev$ such that $ev\circ H$ is a constant homotopy. Finally, the homotopy $(f,t) \mapsto H_t \circ f$ provides a contraction of $\mathcal{M}(X)$. By the way, there is also a completely elementary construction of such a contraction. $\endgroup$
    – Ricardo Andrade
    Nov 9, 2014 at 17:39
  • $\begingroup$ @RicardoAndrade: Very clear, thanks $\endgroup$
    – MyIsmail
    Nov 10, 2014 at 9:13

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I think so, assuming that $X$ is a CW-complex, of course. The evaluation map is a fibration, and $\mathcal M(X)$ is the fiber of the following fibration between mapping spaces $$\operatorname{map}(X\times X,PX)\longrightarrow \operatorname{map}(X\times X,X\times X)$$ at the identity in $X\times X$. Since $X\times X$ is contractible, then so is the target mapping space, therefore the fiber is weakly equivalent to the source, which is also contractible since $PX\simeq X\simeq *$.

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    $\begingroup$ Then you can say noting, but what non-CW-example would be interesting for you? $\endgroup$
    – Fernando Muro
    Oct 10, 2014 at 10:35
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    $\begingroup$ Why do you need a CW complex? $\endgroup$
    – Jeff Strom
    Oct 10, 2014 at 13:17
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    $\begingroup$ @JeffStrom So that the claimed fibrations are indeed fibrations, maybe? I don't know if all this works in a bigger generality, but you're the right person to answer this! $\endgroup$
    – Fernando Muro
    Oct 10, 2014 at 15:02
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    $\begingroup$ @JeffStrom what do you mean by $*\rightarrow X$? This is an unbased context, I think. And if $X$ were based, I can't see the relation with $ev$. $\endgroup$
    – Fernando Muro
    Oct 10, 2014 at 21:00
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    $\begingroup$ Sorry! $i:\partial I \hookrightarrow I$ is an unpointed cofibration so $i^*=ev$ is a fibration. $\endgroup$
    – Jeff Strom
    Oct 10, 2014 at 21:30

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