Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 156 / 2

Fix a basepoint $x_0$. A homeomorphism $f:S\to S$ induces an isomorphism from ${\pi }_1(S,x_0)$ to ${\pi }_1(S,f(x_0))$. Choosing a path from $x_0$ to $f(x_0)$ identifies the latter group with $G={\pi }_1(S,x_0)$. Changing the path conjugates the resulting automorphism by an element of $G$, while an isotopy changes nothing in the outer automorphism group. Hence there is a natural homomorphismthe mapping-class action on the outer automorphism group of the fundamental group.

The commutator subgroup $[G,G]$ is characteristic: every automorphism sends commutators to commutators and therefore preserves the subgroup they generate. Thus an automorphism $A$ inducesThis is independent of the coset representative $g$. If $A^{\prime }=c_h\circ A$ differs from $A$ by an inner automorphism, thenbecause the abelianization is abelian. Hence $q([A])=\overline{A}$ is independent of the representative of the outer class. Finally $\overline{AB}=\overline{A}\overline{B}$, sois a well-defined group homomorphism.

For the punctured torus, $G=⟨\alpha ,\beta ⟩\cong F_2$ and $G^{\mathrm{ab}}\cong {\mathbb{Z}}^2$. Dehn twists about the two standard curves act on homology, in suitable oriented bases, byThe supplied result says that these matrices generate $S{L}_2(\mathbb{Z})$. Thus the image ofcontains the infinite group $S{L}_2(\mathbb{Z})$. Its intermediate group $\mathrm{Out}(F_2)$ must therefore be infinite.

Take the three-punctured sphere $S_{0,3}$. Its full mapping class group is finite, isomorphic to the permutation group $S_3$, whereasand $\mathrm{Out}(F_2)$ is infinite by part c. Hence the finite subgroup $p(\mathrm{Mod}(S_{0,3}))$ has infinite index in $\mathrm{Out}({\pi }_1(S_{0,3}))$.