at.algebraic topology - Homotopy type of set of self homotopy-equivalences of a surface

Hi Andy,

Here is a proof for the case with marked points (see below for some ideas for the case of closed surfaces).

Proof: straight-line homotopy.

Less tersely: let $HE_0(Sigma,ast)$ be the identity component of the monoid of self-homotopy equivalences of $Sigma$ fixing the basepoint; in particular, each $fin HE_0(Sigma,ast)$ is homotopic rel $ast$ to the identity. Fix a hyperbolic metric on $Sigma$ and thus an identification of the universal cover $widetilde{Sigma}$ with the hyperbolic plane $mathbb{H}^2$, and a basepoint $ast$ in $mathbb{H}^2$.

Each $fcolon Sigmato Sigma$ has a unique lift to $mathbb{H}^2$ fixing the basepoint. because $f$ acts trivially on $pi_1(Sigma)$, $f$ commutes with the deck transformations. Thus we may take the straight-line homotopy $f_t(x)=tx-(1-t)f(x)$, where by this convex combination I mean to move with unit speed along the geodesic from $f(x)$ to $x$. Since the deck transformations act by isometries on $mathbb{H}^2$, this homotopy descends to $Sigma$; each $f_t$ is still a homotopy equivalence. We can perform this straight-line homotopy for all $f$ simultaneously; since the lifts of $f$ are uniformly continuous, this homotopy is continuous on $HE_0(Sigma,ast)$ and gives a contraction to the identity.

There must be some work needed to get from this to the case for closed surfaces, because this proof works for a genus 1 surface with marked point, and of course $HE_0(T^2)$ is homotopy equivalent to $T^2$ itself. But it seems to me like a reduction should be possible; I think the important thing is that $pi_1(Sigma)$ is centerless.

Acknowledgement: I learned this idea from Rita Jimenez Rolland, based on conversations she had with Mladen Bestvina about the related case of $text{Aut}(F_n)$.

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