# Fundamental group of group of homeomorphism of a compact surface

I'm reading "A Primer on Mapping Class Group", and there is something I don't understand in the proof of Theorem 4.6.

Define $$\mathrm{Homeo}^+(S)$$ to be the group of orientation-preserving homeomorphisms of $$S$$ and define $$\mathrm{Homeo}_0(S)$$ the connected component of the identity in $$\mathrm{Homeo}^+(S)$$. The group $$\mathrm{Homeo}^+(S)$$ is endowed with the compact-open topology.

Theorem 1.14: Let $$S$$ be a compact surface, possibly minus a finite number of points from the interior. Assume that $$S$$ is not homeomorphic to $$S^2$$ , $$\mathbb{R}^2$$, $$D^2$$ , $$T^2$$ , the closed annulus, the once punctured disk, or the once punctured plane. Then the space $$\mathrm{Homeo}_0 (S)$$ is contractible.

Theorem 4.6: Let $$S$$ be a surface with $$\chi(S) < 0$$, possibly with punctures and/or boundary. Let $$(S,x)$$ be the surface obtained from $$S$$ by marking a point $$x$$ in the interior of $$S$$. Then the following sequence is exact:$$1 \to \pi_1(S,x)\to \mathrm{Mod}(S,x) \to \mathrm{Mod}(S) \to 1.$$

In the proof of Theorem 4.6, it is said : By Theorem 1.14 the group $$\pi_1 (\mathrm{Homeo}^+ (S))$$ is trivial.

I don't see how to deduce this fact from Theorem 1.14. Is it obvious that one can always find an homotopy from one homeomorphism to another?

• The group $\pi_1 X$ is trivial if $X$ is contractible, almost by definition. (And $X$ is contractible iff every connected component of it is.) – anomaly Apr 19 at 18:55
• @anomaly A contractible space is automatically connected – Max Apr 19 at 19:17
• @Max: Yeah, basepoint stuff. – anomaly Apr 19 at 19:48

By Theorem 1.14 $$\text{Homeo}_0(S)$$ is contractible, hence every connected component of $$\text{Homeo}^+(S)$$ is contractible (consider group structure of $$\text{Homeo}^+(S)$$). As each connected component is contractible the whole space must be simply connected.