---
tag: 410BCE4PAM
kind: definition
name: Mapping class group
msc: ["57K20", "20F38"]
defines: JV5NYG913V
uses: [0TD904MG06, 0QT00PB5G2, G5S1M9]
sources:
  - url: https://en.wikipedia.org/wiki/Mapping_class_group
---

## Definition

Let $S$ be a connected oriented [[surface|VGJJVBZ1MQ]], possibly with boundary $\partial S$.
Let $\mathrm{Homeo}^+(S, \partial S)$ be the group, under composition, of orientation-preserving [[homeomorphisms|0QT00PB5G2]] $f \colon S \to S$ with $f(x) = x$ for all $x \in \partial S$.
Two such homeomorphisms $f$ and $g$ are *isotopic* if there is a continuous map $H \colon S \times [0, 1] \to S$ such that:

1. each $H_t = H(\cdot, t)$ lies in $\mathrm{Homeo}^+(S, \partial S)$;
2. $H_0 = f$ and $H_1 = g$.

The *mapping class group* of $S$ is the quotient group
$$\mathrm{Mod}(S) = \mathrm{Homeo}^+(S, \partial S) / \mathrm{Homeo}_0(S, \partial S),$$
where $\mathrm{Homeo}_0(S, \partial S)$ is the normal subgroup of homeomorphisms isotopic to the identity.
Its elements, the isotopy classes, are *mapping classes*.

## Remarks

For a surface with marked points or punctures, the homeomorphisms may permute the marked points, and the isotopies must keep them fixed.
Some authors also allow orientation-reversing homeomorphisms; this gives the extended mapping class group, which contains $\mathrm{Mod}(S)$ with index $2$ when $S$ admits an orientation-reversing homeomorphism fixing $\partial S$.

Example: for the torus $T^2 = \R^2 / \Z^2$, the map $\mathrm{Mod}(T^2) \to SL(2, \Z)$ that records the action on the first homology group $H_1(T^2; \Z) \cong \Z^2$ is an isomorphism.
A matrix $A \in SL(2, \Z)$ maps $\Z^2$ onto itself, so it induces a homeomorphism of $T^2$, and since $\det A = 1$ it preserves orientation.

Non-example: let $A = S^1 \times [0, 1]$ be the annulus, with points written $(\theta, t)$, and let $T(\theta, t) = (\theta + 2\pi t, t)$, a [[Dehn twist|W890NRHGS4]].
The maps $H_s(\theta, t) = (\theta + 2\pi s t, t)$ for $s \in [0, 1]$ deform the identity into $T$, but for $0 < s < 1$ they rotate the boundary circle $S^1 \times \set{1}$, so they fail clause 1.
In fact $T$ is not isotopic to the identity in the sense above, and $\mathrm{Mod}(A) \cong \Z$, generated by the class of $T$.
For the closed disk, the Alexander lemma gives $\mathrm{Mod}(D^2) = 1$.