# Realizing inner automorphisms on Eilenberg-MacLane spaces

Let $$G$$ be a discrete group and let $$(X,x_0)$$ be a based Eilenberg-MacLane space for $$G$$, so there is a fixed isomorphism $$\pi_1(X,x_0) = G$$ and the universal cover $$\widetilde{X}$$ is contractible. The homology of $$X$$ is thus the same as the homology of $$G$$.

Choose some $$\gamma \in G$$, and let $$c_{\gamma} \in \text{Aut}(G)$$ be the inner automorphism that conjugates elements of $$G$$ by $$\gamma$$. It is standard that $$c_{\gamma}$$ acts as the identity on $$H_k(G)$$ for all $$k$$.

Since $$(X,x_0)$$ is an Eilenberg-MacLane space for $$G$$, there is a based map $$\psi_{\gamma}\colon (X,x_0) \rightarrow (X,x_0)$$ that induces $$c_{\gamma}$$ on $$\pi_1(X,x_0)$$, and in fact $$\psi_{\gamma}$$ is unique up to based homotopy.

Question: Is it possible to write down $$\psi_{\gamma}$$ in some natural way where in particular it is obvious that it acts as the identity on $$H_k(X)$$? All the proofs I know that inner automorphisms act trivially on homology are either very algebraic or are based on very specific models of Eilenberg--MacLane spaces that mimic the algebraic constructions, and I'd like to see this fact topologically/geometrically.

Let's assume that the space $$X$$ is reasonable in the sense that the inclusion $$x_0 \hookrightarrow X$$ is a cofibration, i.e. has the homotopy extension property. This will hold, for instance, if $$X$$ is a CW complex and $$x_0$$ is a vertex.
Represent $$\gamma$$ by a path $$\rho\colon [0,1] \rightarrow X$$ with $$\rho(0)=\rho(1)=x_0$$. We can then use the homotopy extension property to extend $$\rho$$ to a homotopy $$\phi_t\colon X \rightarrow X$$ such that $$\phi_0 = \text{id}$$ and $$\phi_t(x_0) = \rho(t)$$ for all $$t \in [0,1]$$. Since $$\phi_{1}$$ is homotopic to the identity (but through a homotopy where the basepoint moves!), it clearly induces the identity on homology.
So it is enough to prove that $$\phi_1\colon (X,x_0) \rightarrow (X,x_0)$$ induces $$c_\gamma$$ on $$\pi_1$$. Let $$(\widetilde{X},\widetilde{x}_0) \rightarrow (X,x_0)$$ be the based universal cover of $$(X,x_0)$$. There is a unique lift $$\Phi\colon (\widetilde{X},\widetilde{x}_0) \rightarrow (\widetilde{X},\widetilde{x}_0)$$ of $$\phi_1$$. Identify the orbit of $$\widetilde{x}_0$$ under the deck group with $$G$$, so $$1 = \widetilde{x}_0$$. It is enough to prove that $$\Phi(g) = \gamma^{-1} g \gamma$$ for all $$g \in G$$.
The path $$\rho$$ lifts to a collection of paths that connect $$g \in G$$ to $$g \gamma$$ for all $$g \in G$$. Lifting the homotopy $$\phi_t$$, we get a homotopy $$\widetilde{\phi}_t\colon \widetilde{X} \rightarrow \widetilde{X}$$ starting at the identity that slides each $$g \in G$$ along these paths. You might think that $$\widetilde{\phi}_1$$ is $$\Phi$$, but while it is a lift of $$\phi_1$$, it is not $$\Phi$$ since it takes the basepoint $$1 = \widetilde{x}_0$$ to $$\gamma$$. We have to correct this by multiplying $$\widetilde{\phi}_1$$ by the element $$\gamma^{-1}$$ of the deck group, so $$\Phi(x) = \gamma^{-1} \cdot \widetilde{\phi}_1(x)$$ for all $$x \in \widetilde{X}$$. In particular, for $$g \in G$$ we have $$\Phi(g) = \gamma^{-1} g \gamma$$ as desired.
• @IJL: It depends upon which source you consult. I was going by what the OP wrote (fundamental group $G$, universal cover contractible). While I personally insist that Eilenberg—MacLane spaces be CW complexes (even better than having a nondegenerate basepoint!), not everyone is so fastidious. – Andy Putman Mar 18 at 14:11