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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 $\text{Hom}_{\mathcal{A}}$?

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Wenzhe
    Commented Apr 6, 2018 at 17:37
  • $\begingroup$ @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. $\endgroup$
    – Roland
    Commented Apr 6, 2018 at 19:20

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