Realizing inner automorphisms on Eilenberg-MacLane spaces

Question

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.

Answer 1

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.