Let $G$ be a group and let $\phi: G \to G$ be the inner automorphism given by conjugation by an element $g' \in G$, i.e., $\phi(g) = g'^{-1} g g'$. I want to show that the induced map on classifying spaces $B\phi: BG \to BG$ is homotopic to the identity.
One way to do this is to recall that $B\phi$ is the map classifying the principal $G$-bundle $EG \times_{G,\phi} G \to BG$ (perhaps assuming that $G$ is discrete - although I would be interested to see a proof to extend this result to all groups if it's true), and in this case we can explicitly write down an isomorphism $EG \times_{G,\phi} G \xrightarrow{\cong} EG$, which proves that the classifying map is homotopic to the identity.
But I also want to prove this statement "hands-on" using an explicit model of $EG$ and $BG$, namely the Milnor infinite join construction. Recall that the model for $EG$ in this case is: \begin{equation*} G * G * \cdots = \{(s_i g_i)_{i = 1}^\infty : s_i \in [0,1], g_i \in G, \sum s_i = 1\} / {\sim} \end{equation*}
This comes equipped with a topology that makes the projections onto $G$ and $[0,1]$ continuous. It also comes equipped with a free action of $G$ on the right given by right multiplication on each coordinate. Let $BG = EG/G$ be the quotient space.
If I'm not mistaken, the induced map is given by \begin{equation*} (B\phi)[(s_i g_i)_{i=1}^\infty] = [(s_i \phi(g_i))_{i=1}^\infty] = [(s_i g'^{-1} g_i g')_{i=1}^\infty] = [(s_i g'^{-1} g_i)_{i=1}^\infty] \end{equation*}
I have three questions.
- This type of reasoning seems to show that right multiplication by any element in $G$ induces the identity map on $BG$. This doesn't look right. Moreover, if this is the case, then together with the result that conjugation induces a map homotopic to the identity, this would imply that left multiplication by any element in $G$ induces a map homotopic to the identity as well. However, I remember reading that under some conditions that \begin{equation*} \operatorname{Out}(G) := \operatorname{Aut}(G)/\operatorname{Inn}(G) \cong [BG,BG], \end{equation*} and so the only maps that are homotopic to the identity are those induced by inner automorphisms. So what am I messing up here?
- Assuming I'm on the right track, how do I finish the proof, i.e., what is the explicit homotopy between $B\phi$ and the identity? I'm thinking that if $G$ was connected as a topological group, then I can perhaps fix a path from the identity element $e \in G$ to $g'^{-1}$ and use it to construct the homotopy. However, $G$ need not be connected and anyway if this argument worked it would also show that many other induced maps are homotopic to the identity, which doesn't seem to be the case (see 1). But I also don't see why it wouldn't work for connected groups.
- If I'm not on the right track, how do I fix this argument?
Thanks!