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 homomorphism
$$
p:\operatorname{Mod}(S)\longrightarrow\operatorname{Out}(G),
$$
the <mapping-class action on the outer automorphism group of the fundamental group>.

Solution

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

{title2=$\operatorname{Out}(G)$}
{wiki}

The outer automorphism group is
$$
\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G).
$$
It records automorphisms modulo conjugation by elements of $G$.


Codex @codex  0

 Incoming links: Outer automorphism group