I am doing this to try to figure out The action of $\pi_1(BK) \curvearrowright H_*(BG)$ for the fibration $BG \hookrightarrow BH \to BK$ .
Let the fibration $F \hookrightarrow E \xrightarrow{p} B$ be given. It is well known that there is a well defined map $\pi_1(B) \to Aut(h_*(F))$:
Fact:
For this fibration and $[\alpha] \in \pi_1(B)$, we have an $\{A_t\}_{t \in [0,1]}: F_b \to E$, s.t.
- $A_0=Id_{F_b}$,
- $p \circ A_t(b)=\alpha(t)$
- $A_1$ is a homotopy equivalence map $F_b \to F_b$. Since the (free, basepointed) homotopy class of $\alpha$ does not affect the (free) homotopy class of $A:[0,1]\times F_b \to E$, $A_1^*: h_*(F_b) \circlearrowleft$ is well defined.
Fact #2 Since $F$ is path connected, there is basepointed homotopy in the same free homotopy class as $A_1$. Unfortunately, there are several different maps, in different basepointed homotopy classes $A_1: F_b \to F_b$ that come from the same basepointed homotopy class of $\alpha$. These result from the choices of paths from $A_1(b)$ to $B$ - thus the choices are the same up to inner automorphisms of $\pi_1(F_b)$
remark: Fact #2 shows that we don't have a well defined map $\pi_(B) \to Aut(\pi_1(F_b,b))$ in general. When $\pi_1(F_b)$ is abelian, $Inn\pi_1(F_b)=0$ so there is a well defined map.
Question: Is the following assesment correct?
Specialize to the fibration $BG \xrightarrow{\phi} BH \xrightarrow{psi} BK$ associated to the exact sequence of discrete groups $1 \to G \to H \to K \to 1$. Choose a $k=[\alpha] \in \pi_1(B)$, and choose a basepointed homotopy class $[A_1]$ to get $[A_1]^*:G=\pi_1(F_b) \to G=\pi_1(F_b)$. We get a map $BG \to BG$.
As in the remark above, if $G$ is abelian (e.g. when the above extension is a central extension), then the automorphism on $\pi_1(BG)$ does not depend on the choice of $[A]$ and there is a well defined map $\pi_1(F_b)$ to itself.
Note that this happens exactly when there is an algebraically defined action of $K$ on $G$: "$kgk^{-1}$" $:= \phi^{-1}(h\phi(g) h^{-1})$ for a choice of a lift $h$ of $k$, is well defined iff the inner automorphisms of $G$ are trivial.