I mean, I agree with you that it’s a good idea to have some lemmas first on morphisms of ses (short exact sequences), such as are implicit in your #3, before launching into the text of #1. This is the bread and butter of homological algebra. I am not sure about more minimal hypotheses to push these lemmas through, such as weakening to enrichment in commutative monoids as opposed to abelian groups. I’d think one would have to work throughout with kernel pairs instead of mere kernels.

I think adding words like “pullback functor” and “pushforward functor” could help some readers parse the notation $\Delta^*$ and $+_*$ more readily.

]]>I have slightly updated 3, with much additional information. P.S. I am getting now the point of the original notation, as in 1. It looked to me that both stars are attached to $\Delta$ what was quite an obstacle to understanding.

]]>In fact, my interest is if we have a proper notion of short exact sequence and we have finite biproducts but not an additive category. Then one can still define the associative addition, however one may lack inverses. I am curious about such cases of Baer sum (giving additive monoids, if I am right).

]]>My edits may be improperly reflected in the entry, as the $n$Lab stalled when I was sending the biggest change to the entry. I wanted to have the content of entry 3 there.

]]>2 Notation $+_* \Delta^*(\hat G_1 \oplus \hat G_2)$ does not reveal the additive nor abelian duality, nor the fact that the pullback and pushout steps can be interchanged. It s quite obscure to me; if I had not read it before I would not understand the treatment. I think the treatment a la MacLane with diagonal and codiagonal on the same footing, plus usage of operations of pullback and pushout as general operation on extensions (not of the same groups) is giving broad but clean context.

On the other hand, the notation $\nabla_M (E_1\oplus E_2) \Delta_P$, is symmetric and a special case of reconstruction of additive structure from the categorical structure in additive categories, a trick also used for stable infinity categories.

]]>This is my treatment (the old one is still there). I think it is more clear, though some more diagrams could be useful.

In any category with products, for any object $C$ there is a diagonal morphism $\Delta_C:C\to C\times C$; in a category with coproducts there is a codiagonal morphism $\nabla_C: C\coprod C\to C$ (addition in the case of modules). Every additive category is, in particular, a category with finite biproducts, so both morphisms are there. Short exact sequences in the category of $R$-modules, or in arbitrary abelian category $\mathcal{A}$, form an additive category (morphisms are commutative ladders of arrows) in which the biproduct $0 \to A_i \to H_{i} \to G_i \to 0$ for $i = 1,2$ is $0\to A_1\oplus A_2 \to H_1\oplus H_2\to G_1\oplus G_2\to 0$.

Now if $0\to M\to N\to P\to 0$ is any extension, call it $E$, and $\gamma:P_1\to P$ a morphism, then there is a morphism $\Gamma' = (id_M,\beta_1,\gamma)$ from an extension $E_1$ of the form $0\to M\to N_1\to P_1\to 0$ to $E$, where the pair $(E_1,\Gamma_1)$ s unique up to isomorphism of extensions, and it is called $E\gamma$. In fact, the diagram

$\array{ N_1&\to &P_1\\ \downarrow\beta_1 && \downarrow\gamma\\ N&\to &P }$is a pullback diagram. Every morphism of abelian extensions $(\alpha,\beta,\gamma):E\to E'$ in a unique way decomposes as

$E\stackrel{(\alpha,\beta_a,id)}\longrightarrow E\gamma \stackrel{(id,\beta_1,\gamma)}\longrightarrow E'$for some $\beta_a$, with $\beta_1$ as above. In short, the morphism of extensions factorizes through $E\gamma$.

Dually, for any morphism $\alpha:M\to M_2$, there is a morphism $\Gamma_2 = (\alpha,\beta_2,id_P)$ to an extension $E_2$ of the form $0\to M_2\to N_2\to P$; the pair $(E_2,\Gamma_2)$ is unique up to isomorphism of extensions and it is called $\alpha E$.

In fact, the diagram

$\array{ M&\to &N\\ \downarrow\alpha && \downarrow\beta_2\\ M_2&\to &N_2 }$is a pushout diagram. Every morphism of abelian extensions $(\alpha,\beta,\gamma):E\to E''$ in a unique way decomposes as

$E\stackrel{(\alpha,\beta_2,id)}\longrightarrow \alpha E \stackrel{(id,\beta_ b,\gamma)}\longrightarrow E''$for some $\beta_b$, with $\beta_2$ as above. In short, the morphism of extensions factorizes through $\alpha E$.

There are the following isomorphisms of extensions: $(\alpha E)\gamma\cong \alpha (E\gamma)$, $id_M E \cong E$, $E id_P \cong P$, $(\alpha'\alpha)E\cong\alpha' (\alpha E)$, $(E\gamma)\gamma' \cong E(\gamma\gamma')$.

The Baer’s sum of two extensions $E_1,E_2$ of the form $0\to M\to N_i\to P\to 0$ (i.e. with the same $M$ and $P$) is given by $E_1+E_2 = \nabla_M (E_1\oplus E_2) \Delta_P$; this gives the structure of the abelian group on $Ext(P,M)$ and $Ext:\mathcal{A}^{op}\times\mathcal{A}\to Ab$ is a biadditive (bi)functor. This is also related to the isomorphisms of extensions $\alpha (E_1+E_2)\cong \alpha E_1+\alpha E_2$, $(\alpha_1+\alpha_2) E \cong \alpha_1 E+ \alpha_2 E$, $(E_1+E_2)\gamma \cong E_1\gamma + E_2\gamma$, $E(\gamma_1+\gamma_2)\cong E\gamma_1 + E\gamma_2$.

]]>What you copied in #1 looks notationally just fine to me – is there a problem with it?

]]>I am going to rewrite a part of the Baer sum, the section “On short exact sequences”, partly following S. MacLane, *Homology*, 1963 (he does the version for extensions of $R$-modules). I am not fully understanding and would like to discuss the issue, but I think the current notation is a bit hiding. So here is the version of the section before my update, so it can be reversed if somebody not happy:

For $0 \to A \to \hat G_{i} \to G \to 0$ for $i = 1,2$ two short exact sequences of abelian groups, their **Baer sum** is

The first step forms the pullback of the short exact sequence along rhe diagonal on $G$:

$\array{ A \oplus A &\to& A \oplus A \\ \downarrow && \downarrow \\ \Delta^* (\hat G_1 \oplus \hat G_2) &\to& \hat G_1 \oplus \hat G_2 \\ \downarrow && \downarrow \\ G &\stackrel{\Delta_G}{\to}& G\oplus G }$The second forms the pushout along the addition map on $A$:

$\array{ A \oplus A &\stackrel{+}{\to}& A \\ \downarrow && \downarrow \\ \Delta^* (\hat G_1 \oplus \hat G_2) &\to& +_* \Delta^*(\hat G_1 \oplus \hat G_2) \\ \downarrow && \downarrow \\ G &\to& G }$ ]]>