Another description for the map $\text{Ext}^1_\mathbb{Z}(A,G)\to H^2(G,A)$

Question

Group extensions of $G$ by $A$ $0\to A\to E\to G\to 0$ up to equivalence (where $G$ and $E$ may be nonabelian) are in bijection with the second group cohomology $H^2(G,A)=\text{Ext}^2_{\mathbb{Z}[G]}(\mathbb{Z},A).$

On the other hand, if $G$ is abelian, and we seek only abelian extensions, such extensions are (up to equivalence) in bijection with $\text{Ext}^1_{\mathbb{Z}}(G,A)$ (whence the $\text{Ext}$ functor gets its name).

Hence there is an inclusion $\text{Ext}^1_{\mathbb{Z}}(G,A)\hookrightarrow \text{Ext}^2_{\mathbb{Z}[G]}(\mathbb{Z},A).$ Perhaps we could say it’s induced by the inclusion functor of abelian groups into groups, acting on group extensions. Other than the description given, using the bijection to group extensions, is there another way to see this map?

Answer 1

$H^2(G, A)$ can also be written $[BG, B^2 A]$, where

• for spaces $X, Y$, $[X, Y]$ denotes the set of homotopy classes of maps $X \to Y$,
• $BG$ denotes the Eilenberg-MacLane space $K(G, 1)$, and
• $B^2 A$ denotes the Eilenberg-MacLane space $K(A, 2)$.

If $G$ is also abelian, then $BG$ and $B^2 A$ can both be modeled by simplicial abelian groups, and we can ask for homotopy classes of maps between them as simplicial abelian groups (more intrinsically this might be equivalent to asking for homotopy classes of infinite loop space maps; I'm not sure). We can then pass to chain complexes by Dold-Kan, and we get that the resulting set of homotopy classes of maps is $\text{Ext}^1(G, A)$.