TITLE: Homotopy type of set of self homotopy-equivalences of a surface
QUESTION [14 upvotes]: Let $\Sigma$ be an oriented topological surface.  For simplicity, assume that the genus of $\Sigma$ is at least $2$.  There are a number of classical results on the homotopy types of various groups of self-maps of $\Sigma$:
1) Earle and Eells proved that the components of $\text{Diff}(\Sigma)$ are contractible.
2) Hamstrom proved that the components of $\text{Homeo}(\Sigma)$ are contractible.
3) Peter Scott proved that the components of $\text{Homeo}^{\text{PL}}(\Sigma)$ are contractible, where by $\text{Homeo}^{\text{PL}}(\Sigma)$ we mean the group of PL self-homeomorphisms of $\Sigma$.
Of course, in 1 and 3 we are fixing a $C^{\infty}$ or $\text{PL}$ structure on $\Sigma$, respectively.
Another standard fact is that every self homotopy-equivalence of $\Sigma$ is homotopic to a homeomorphism.  This leads me to my question.  Denote by $\text{HE}(\Sigma)$ the set of self homotopy-equivalences of $\Sigma$.  Are the components of $\text{HE}(\Sigma)$ contractible?
Another related question is as follows.  There is a beautiful alternate proof of the above theorem of Earle and Eells due to Earle and McMullen (see their paper "Quasiconformal Isotopies"; their proof uses complex analysis).  The proofs of Hamstrom's theorem and Scott's theorem are very complicated -- are there any alternate approaches to them in the literature?

REPLY [15 votes]: A couple comments. For the result about diffeomorphism groups there is a very nice alternative proof due to A. Gramain in the Annales Scient. E.N.S. v.6 (1973), pp. 53-66, that uses no analysis, just basic differential topology. Another approach, which I'm not sure is written down anywhere in detail, is to take the proof of the corresponding result for Haken 3-manifolds (due independently to Ivanov and myself) and scale it back from 3-manifolds to surfaces. This too uses no analysis, just basic topology.  I don't recall Scott's method for the PL case, but it might be similar to this.
For the result about homotopy equivalences it's best to look first at the space of homotopy equivalences that fix a basepoint. This has contractible components for any $K(\pi,1)$ space, by an elementary obstruction theory argument. (Just deform families of maps to the identity map, cell by cell, which is possible since the obstructions lie in the higher homotopy groups of the space.) By evaluating arbitrary homotopy equivalences at the basepoint one gets a fibration where the total space is the space of all homotopy equivalences, the base space is the $K(\pi,1)$ space, and the fiber is the earlier space of basepoint-preserving homotopy equivalences.  Looking at the long exact sequence of homotopy groups for this fibration then gives the desired result.  Triviality of the center of $\pi$ comes in at this point to show that the boundary map from $\pi_1$ of the base to $\pi_0$ of the fiber is injective.