Extension group and Baer sum

Question

Suppose $\mathcal{A}$ is an abelian category and $A$ and $B$ are two objects of $\mathcal{A}$, the extensions of $A$ by $B$ consist of isomorphism classes of short exact sequences of the form \begin{equation} 0 \rightarrow B \rightarrow E \rightarrow A \rightarrow 0 \end{equation} the set of which will be denoted by $\text{Ext}_{\mathcal{A}}(A,B)$. The Baer sum will induce a group structure on $\text{Ext}_{\mathcal{A}}(A,B)$, which seems a little ad hoc to me, and I don't understand well. Could anyone explain some intuitions behind the construction of the group structure of extensions and its natural connection to a derived functor of $\text{Hom}_{\mathcal{A}}$?

Answer 1

The Baer sum is not that surprising. Here is how I see it:

• The group structure on Hom-set may be recovered from the biproduct $\oplus$. Indeed, if $f:A\to B$ and $f':A\to B$ are two parallel arrows, then there is an obvious morphism $f\oplus f':A\oplus A\to B\oplus B$. Then $f+f'$ is the composition : $$ A\overset{\Delta}\longrightarrow A\oplus A\overset{f\oplus f'}\longrightarrow B\oplus B\overset{\nabla}\longrightarrow B$$ where $\Delta$ is the diagonal and $\nabla$ the codiagonal. Both these morphisms exist because $\oplus$ is both a product and a biproduct. In other words: $f+f'$ is given by the image of $f\oplus f'$ under the composition: $$\operatorname{Hom}(A\oplus A,B\oplus B)\overset{\nabla_*}\longrightarrow \operatorname{Hom}(A\oplus A,B)\overset{\Delta^*}\longrightarrow\operatorname{Hom}(A,B)$$

• If $e,e'$ are two extension of $B$ by $A$, then there is by obvious direct sum an extension $e\oplus e'\in\operatorname{Ext}^1(A\oplus A,B\oplus B)$. (Just take the direct sum of the two short exact sequence defined by $e,e'$).

• $\operatorname{Ext}^1$ is like a Hom-set, in particular it has the same functoriality.

So here you have the Baer sum: $e+e'$ is the extension which is the image of $e\oplus e'$ by: $$\operatorname{Ext}^1(A\oplus A,B\oplus B)\overset{\nabla_*}\longrightarrow\operatorname{Ext}^1(A\oplus A,B)\overset{\Delta^*}\longrightarrow\operatorname{Ext}^1(A,B)$$

Check that this is indeed the definition of the Baer sum. See here Baer Sum notation requires clearence. for more details.