Why is the cohomological $\delta$-functor for $H^\ast(K(\pi,1),–)$ effaceable? Let $\pi$ be an abelian group and $\mathbf Z[\pi]$ its group ring. One obtains a cohomological $\delta$-functor from the short exact sequence of $\mathbf Z[\pi]$-modules
$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$
and the LES of cohomology
$$\cdots\rightarrow
H^q(K(\pi, 1),A)\rightarrow H^q(K(\pi, 1),B)\rightarrow H^q(K(\pi, 1),C)
\rightarrow H^{q+1}(A)\rightarrow\cdots$$
(I hope I have set this up right.) Then I would like to claim that the functor $\delta^n:H^q(K(\pi,1),C)\rightarrow H^{q+1}(K(\pi,1),A)$ is effaceable; i.e. there is a monomorphism of $\mathbf Z[\pi]$-modules $A\rightarrow P$ such that the induced map $H^q(K(\pi,1),A)\rightarrow H^q(K(\pi,1),P)$ is $0$, noting that
$$C^\ast(K(\pi,1);A)\cong\operatorname{Hom}_{\mathbf Z[\pi]}(C_\ast(\widetilde{K(\pi,1)}),A),$$
noting that $\widetilde{K(\pi,1)}$ is contractible.
 A: Let $P$ be the set of all functions $\pi\to A$.  Make $P$ an abelian group by pointwise addition, and a $\mathbb{Z}[\pi]$-module by $(g\varphi)(h)=\varphi(hg)$ for $\varphi\in P$ and $g,h\in\pi$.  Note that there is a canonical monomorphism $i:A\to P$ given by $i(a)(g)=ga$.  So it suffices to show that $H^q(K(\pi,1),P)=0$ for all $q>0$.
To show this, note that for any $\mathbb{Z}[\pi]$-module $B$, there is a natural isomorphism $\operatorname{Hom}_{\mathbb Z[\pi]}(B,P)\cong\operatorname{Hom}_{\mathbb{Z}}(B,A)$: given $\alpha:B\to P$, define $\beta:B\to A$ by $\beta(b)=\alpha(b)(1)$, and conversely, given $\beta:B\to A$, define $\alpha:B\to P$ by $\alpha(b)(g)=\beta(gb)$.  To see that these correspondences are inverse to each other, note that if $\alpha$ is $\mathbb{Z}[\pi]$-linear, it must satisfy $$\alpha(b)(g)=\alpha(g^{-1}(gb))(g)=(g^{-1}\alpha(gb))(g)=\alpha(gb)(1)=\beta(gb).$$
Thus we conclude that the cochain complex $\operatorname{Hom}_{\mathbb Z[\pi]}(C_\ast(\widetilde{K(\pi,1)}),P)$ is naturally isomorphic to $\operatorname{Hom}_{\mathbb Z}(C_\ast(\widetilde{K(\pi,1)}),A)$, which is just the ordinary cochain complex of $\widetilde{K(\pi,1)}$ with coefficients in (the underlying abelian group of) $A$.  Since $\widetilde{K(\pi,1)}$ is contractible, the cohomology of this complex vanishes in positive degrees.
(To motivate this construction, note that you could come up with it by working backwards: the most obvious way you can hope to show that $\operatorname{Hom}_{\mathbb Z[\pi]}(C_\ast(\widetilde{K(\pi,1)}),P)$ has no cohomology for some particular module $P$ is if in fact that cochain complex can be interpreted as cochains on $\widetilde{K(\pi,1)}$ with coefficients in some abelian group $A$.  If this is true, then you know the identity $\operatorname{Hom}_{\mathbb Z[\pi]}(B,P)=\operatorname{Hom}_{\mathbb{Z}}(B,A)$ must hold when $B=C_\ast(\widetilde{K(\pi,1)})$, and so you could guess that it holds for all modules $B$.  By Yoneda's lemma, that identity uniquely determines $P$ from $A$ up to isomorphism, so you can try to construct $P$ from $A$ so that it is true.  In particular, taking $B=\mathbb{Z}[\pi]$, you see that the underlying set of $P$ must be $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[\pi],A)$, or equivalently the set of functions from $\pi$ to $A$.  It then just takes a bit of fiddling to figure out the correct $\mathbb{Z}[\pi]$-module structure to put on $P$.  You then notice that if $A$ also has a $\mathbb{Z}[\pi]$-module structure, you can take $B=A$, and the identity map $A\to A$ corresponds to some $\mathbb{Z}[\pi]$-linear map $A\to P$.  You can then explicitly compute what this map is and see that it is injective.)
