Chapter IX in Stong's "Notes on Cobordism Theory".

Dwyer's paper is about homological stability. It does not contain a computation of a stable homology group.

What is your definition of ``equivalence onto its image''?

The map $BAut(BA)\rightarrow BAut(A)$ is given by the action on $\pi_1(BA)$. Note that $A$ is abelian, so $\pi_1(BA)$ is functorial in unpointed maps.

See mathoverflow.net/questions/457011/…

Now one can compute $BAut(K(A,1))\simeq K(A,2)_{hAut(A)}$ and under this equivalence those fibrations whose monodromy over the basepoint agrees with the given action correspond to those pointed maps $BG\rightarrow K(A,2)_{hAut(A)}$ that induce the given action on fundamental groups.

$H^2(G;A)$ is the group of pointed homotopy classes of maps $BG\rightarrow K(A,2)_{hAut(A)}$ that induce the given action $G\rightarrow Aut(A)$ on fundamental groups (here the h-subscript denotes homotopy orbits). Fibrations over $BG$ with fibre $BA\simeq K(A,1)$ and an identification of the fibre over a fixed basepoint are classified by pointed maps $BG\rightarrow BhAut(K(A,1))$ where $Aut(K(A,1))$ is the topological monoid of self-homotopy-equivalences of $K(A,1)$.

Yes, up to a sign. This is a special case of Theorem 4 on p. 133 of Hochschild—Serre‘s ‘Cohomology of group extensions‘.

There will be a new list soon: aimath.org/workshops/upcoming/kirbylist

The relation follows from the discussion in Section 2 of James' "On the iterated suspension''.

The fundamental group of the base of a fibre sequence acts on the fibre over the basepoint in the \emph{unpointed} homotopy category, so in general there is no induced action on the homotopy groups. The fundamental group of the total space however acts on the fibre in the \emph{pointed} homotopy category, and thus in particular on its homotopy groups. In your example, the action of $\pi_1(BG)=\pi_0(G)$ on $G/H$ is indeed given by left-translation, which is not pointed, but can be enhanced to a pointed action once one restricts it to $\pi_0(H)$ since $[1] \in G/H$ is now preserved.

That book has a second author.

Less fancy, classical Lefschetz duality gives $H^q(M)\cong H^q(\overline{M})\cong H_{n-q}(\overline{M},\partial \overline{M})\cong \widetilde{H}_{n-q}(\overline{M}/\partial \overline{M})\cong \widetilde{H}_{n-q}(M^+)$.

Makes sense. Thanks.