Homotopy equivalence gives an outer automorphism of the fundamental group

Question

Let $X$ be a path-connected topological space. for $x \in X, G = \pi_1(X,x)$ Show that a homotopy equivalence $f \colon X \to X$ gives a well-defined element $g \in \operatorname{Out}(G)$.

How might one begin on this question?

Answer 1

Let $y = f(x)$. Then $f$ determines an isomorphism from $\pi_1(X, x)$ to $\pi_1(X, y)$. If we had $y = x$ then this would be an automorphism of $G$, but we don't. So what can we do instead?

We can fix a path between $x$ to $y$, which also induces an isomorphism from $\pi_1(X, x)$ to $\pi_1(X, y)$, and use this path to turn the isomorphism induced by $f$ into an isomorphism from $\pi_1(X, x)$ to $\pi_1(X, x)$. This gives an automorphism, but it depends on a choice of path. Now show that if you change the path then the class of this automorphism in $\text{Out}(G)$ doesn't change, so it's well-defined independent of a choice of path.