Show that under the map \(\langle X, Y \rangle \to \operatorname{Hom}(\pi_n(X, x_0), \pi_n(Y, y_0))\), \([f] \mapsto f_*\), the action of \(\pi_1(Y, y_0)\) on \(\langle X, Y \rangle\) corresponds to composing with the action on \(\pi_n(Y, y_0)\), that is, \((\gamma f)_* = \beta_\gamma f_*\). Deduce a bijection of \([X, K(\pi, 1)]\) with the set of orbits of \(\operatorname{Hom}(\pi_1(X), \pi)\) under composition with inner automorphisms of \(\pi\).
Solution
1. The action of \(\gamma \in \pi_1(Y, y_0)\) on \([f] \in \langle X, Y \rangle\) is given by \(\gamma f\), the map obtained by “dragging” \(f\) along \(\gamma\).
definition of the \(\pi_1(Y)\)-action on homotopy classes.
2. On \(\pi_n(Y, y_0)\), the action of \(\gamma\) is the automorphism \(\beta_\gamma\) (conjugation by \(\gamma\)).
the standard \(\pi_1\)-action on higher homotopy groups.
3. The induced map \((\gamma f)_* : \pi_n(X) \to \pi_n(Y)\) equals \(\beta_\gamma \circ f_*\).
the action of \(\gamma\) on the map \(f\) corresponds, on the induced homomorphism, to post-composing with the action \(\beta_\gamma\) on \(\pi_n(Y)\).
4. Hence \((\gamma f)_* = \beta_\gamma f_*\).
3.
5. For \(Y = K(\pi, 1)\), \(\langle X, K(\pi,1) \rangle \cong \operatorname{Hom}(\pi_1(X), \pi)\) (since \(K(\pi,1)\) is an Eilenberg–MacLane space, maps are classified by \(\pi_1\)).
the classification of maps into a \(K(\pi,1)\).
6. The \(\pi_1(K(\pi,1)) = \pi\)-action on \(\langle X, K(\pi,1) \rangle\) corresponds, under 5, to composition with inner automorphisms of \(\pi\).
4 (the action \(\beta_\gamma\) on \(\pi_1(K(\pi,1)) = \pi\) is conjugation by \(\gamma\), an inner automorphism).
7. Hence \([X, K(\pi,1)]\) (the set of orbits) is in bijection with the orbits of \(\operatorname{Hom}(\pi_1(X), \pi)\) under composition with inner automorphisms of \(\pi\).
5 and 6.
8. Q.E.D.
4 and 7.