$\newcommand{\Aut}{\operatorname{Aut}}$Let $G$ be an abelian group.


It seems to be a well-known fact (for example [here](https://mathoverflow.net/questions/349880/classifying-space-for-fibrations-with-eilenberg-maclane-space-fibers-and-nontriv?noredirect=1&lq=1)) that $B\Aut(K(G,1))$, the classifying space of the topological monoid of (unbased) self homotopy equivalences of the Eilenberg-Maclane space $K(G,1)$, has nontrivial homotopy groups $\pi\_1 = \Aut(G),\pi\_2 = G$.


There is a fibration of topological monoids
$$\Aut\_\*(K(G,1)) \to \Aut(K(G,1)) \to K(G,1)$$


where the first map is the inclusion of the fiber, and the second map is evaluating at the basepoint $e$ ($\Aut\_\*$ is the topological monoid of *based* homotopy equivalences).


One can readily show that $\Aut\_\*(K(G,1))$ is homotopy equivalent to $\Aut(G)$ endowed with the discrete topology. Taking classifying spaces we get a fibration sequence


$$K(\Aut(G),1) \to B\Aut(K(G,1))\to K(G,2)$$


which induces the LES in homotopy


$$\cdots \to 0 \to \pi\_2(X) \to G \overset{\phi}{\to} \Aut(G) \to \pi\_1(X) \to 0 \to \cdots$$


where $X$ is $B\Aut(K(G,1))$.


To get the stated result, it suffices to show that the map $\phi: G \to \Aut(G)$ sends $g$ to conjugation by $g$. Why is this true? I appreciate any references.



Computing the homotopy type of $B\operatorname{Aut}(K(G,1))$ using a fibration sequence: why is $G \to \text{Aut(G)}$ given by conjugation?
This argument is rather elementary. Maybe we should later move this to MathStackExchange. Anyway:


As mentioned above in comments, $K(G,1)$ is not a topological monoid and $K(G,2)$ doesn't exist, so the initial setting is the fibration


$Aut\_\*(K(G,1))\hookrightarrow Aut(K(G,1))\twoheadrightarrow K(G,1)$


where $Aut(K(G,1))$ is the topological monoid of self-homotopy equivalences and $Aut\_\*(K(G,1))$ is the submonoid of pointed ones, for $K(G,1)$ has a base-point $x\_0$. This base-point can be taken as the unique vertex of a $CW$-model in order to apply the HEP (homotopy extension property). We obviously identify $\pi\_1(K(G,1),x\_0)=G$.


We have an isomorphism $\pi\_0(Aut\_\*(K(G,1)))\cong Aut(G)$ which sends a pointed self-equivalence $f\colon K(G,1)\to K(G,1)$ to $\pi\_1(f,x\_0)\colon G\to G$.


We now want to compute the morphism


$$\partial\colon G=\pi\_1(K(G,1),x\_0)\longrightarrow \pi\_0(Aut\_\*(K(G,1)))\cong Aut(G)$$


which is part of the long exact sequence of homotopy groups induced by the previous fibration.


Looking at the definition of this morphism in any basic text dealing with homotopy groups, we can proceed as follows:


1. Pick $g\in G$.
2. Choose a representing closed path at $x\_0$ $$\gamma\_g\colon [0,1]\longrightarrow K(G,1).$$
3. Lift it to a non-closed path starting at the base-point of $Aut(K(G,1))$, which is the identity in $K(G,1)$, $$\tilde{\gamma}\_g\colon [0,1]\longrightarrow Aut(K(G,1)).$$
4. $\partial(g)$ is the connected component of $\tilde{\gamma}\_g(1)$.


Now let's proceed wisely in our example:


If we apply the HEP to the pair given by $\gamma\_g$ and the identity in $K(G,1)$, we obtain an unbased homotopy


$$H\colon [0,1]\times K(G,1)\longrightarrow K(G,1)$$


such that $H(0,-)$ is the identity in $K(G,1)$ and $H(-,x\_0)=\gamma\_g$.


All maps $H(t,-)$ are homotopic to the identity in $K(G,1)$, hence they are self-homotopy equivalences, therefore we can choose the lifting in 3 as $$\tilde{\gamma}\_g(t)=H(t,-).$$


Then $\partial(g)$, regarded as an automorphism of $G$, is precisely the morphism induced in $\pi\_1(-,x\_0)$ by the based map $H(1,-)\colon K(G,1)\to K(G,1)$. This map is based because $\tilde{\gamma}\_g$ lifts a closed path.


Let us compute $\pi\_1(H(1,-),x\_0)$. Let $\mu\colon [0,1]\to K(G,1)$ be a closed path based at $x\_0$ representing an element $h\in G$. The homotopy $H(-,\mu)$ is an unbased homotopy between $H(0,\mu)=\mu$ and $H(1,\mu)$. Since $H(-,x\_0)$ is the path $\gamma\_g$ we have $H(1,\mu)\simeq \gamma\_g\*\mu\*\gamma\_g^{-1}$. Therefore $\partial(g)$ maps $h$ to $ghg^{-1}$.


