# Extension group and Baer sum

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 $$0 \rightarrow B \rightarrow E \rightarrow A \rightarrow 0$$ 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 $\text{Hom}_{\mathcal{A}}$?

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.

• From its definition, two extensions $0 \rightarrow B \rightarrow E \rightarrow A \rightarrow 0$ and $0 \rightarrow B \rightarrow E' \rightarrow A \rightarrow 0$ are isomorphic if there exists an isomorphism $f: E \rightarrow E'$ which together with the identify morphisms of $A$ and $B$ form a commutative diagram. I am wondering why we don't use the definition that there exists automorphisms $f_A$ of $A$ and $f_B$ of $B$, which together with $f$ form a commutative diagram? Commented Apr 6, 2018 at 17:37
• @Wenzhe Because this is not a well behaved notion (it does not define an additive functor, worse the set of such isomorphism class of extension does not have a group structure, look for example to the case of $0\to\mathbb{Z/3Z}\to E\to\mathbb{Z/3Z}\to 0$ : compute all the extensions there). And also because this notion is not useful. Commented Apr 6, 2018 at 19:20