For a space \(X\) let \(\operatorname{Aut}(X)\) denote the group of homotopy classes of homotopy equivalences \(X \to X\). Show that for a CW complex \(K(\pi, 1)\), \(\operatorname{Aut}(K(\pi, 1))\) is isomorphic to the group of outer automorphisms of \(\pi\), that is, automorphisms modulo inner automorphisms.
Solution
Let \(K=K(\pi,1)\). A self-homotopy equivalence \(f:K\to K\) induces an automorphism \[ f_*:\pi_1(K,x_0)\longrightarrow \pi_1(K,f(x_0)). \] After choosing a path from \(f(x_0)\) back to \(x_0\), this becomes an automorphism of \(\pi\). Changing the path changes the resulting automorphism by an inner automorphism. Hence there is a well-defined homomorphism \[ \Phi:\operatorname{Aut}(K)\longrightarrow\operatorname{Out}(\pi). \]
Conversely, Hatcher’s Proposition 4A.2 identifies free homotopy classes of maps into a \(K(\pi,1)\) with homomorphisms on fundamental groups modulo inner automorphisms: \[ [K,K]\cong \operatorname{Hom}(\pi,\pi)/\operatorname{Inn}(\pi). \] Under this bijection, a map is a homotopy equivalence exactly when its induced endomorphism of \(\pi\) is an automorphism. Indeed, if \(f_*\) is an automorphism, choose a map \(g:K\to K\) inducing \(f_*^{-1}\); then \(gf\) and \(fg\) induce the identity outer automorphism, hence are freely homotopic to the identity. Thus \(f\) is a homotopy equivalence.
Therefore the preceding bijection restricts to a bijection \[ \operatorname{Aut}(K)\xrightarrow{\cong}\operatorname{Aut}(\pi)/\operatorname{Inn}(\pi), \] and composition corresponds to composition of outer automorphisms. Hence \[ \boxed{\operatorname{Aut}(K(\pi,1))\cong\operatorname{Out}(\pi).} \]