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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?

Silvia Ghinassi
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victor
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  • You have a homotopy equivalence so you also have a homotopy of curves, then you have a map between $G$ And since it is an equivalence, it is an automorphism too. Now, you just have to prove that it induces an outer automorphism, but since it is path connected you also have inner automorphism between fundamental groups with different base points, so it is in some sense module these inner automorphisms. – user40276 Apr 12 '14 at 03:20

1 Answers1

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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.

Qiaochu Yuan
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