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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?

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  • $\begingroup$ 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.) $\endgroup$ – anomaly Apr 19 at 18:55
  • $\begingroup$ @anomaly A contractible space is automatically connected $\endgroup$ – Max Apr 19 at 19:17
  • $\begingroup$ @Max: Yeah, basepoint stuff. $\endgroup$ – anomaly Apr 19 at 19:48
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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.

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