Motion planning algorithm

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

• 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. – Ricardo Andrade Nov 9 '14 at 17:39
• @RicardoAndrade: Very clear, thanks – MyIsmail Nov 10 '14 at 9:13

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 $PX\simeq X\simeq *$.
• @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 $ev$. – Fernando Muro Oct 10 '14 at 21:00
• Sorry! $i:\partial I \hookrightarrow I$ is an unpointed cofibration so $i^*=ev$ is a fibration. – Jeff Strom Oct 10 '14 at 21:30