# Connecting homomorphism and Baer sum in an abelian category

I would like to prove that the connecting homomorphism $$\delta \colon \mathrm{Hom}_{\mathcal{A}}(N,M_3) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$$ from part (2) of Lemma 12.6.4 of the Stacks Project is indeed a group homomorphism.

In the proof, I would like to argue as follows:

By the definition of $$\delta \colon \mathrm{Hom}_{\mathcal{A}}(N,M_3) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$$, we know that $$\delta(f)$$ for some $$f \in \mathrm{Hom}_{\mathcal{A}}(N,M_3)$$ is simply the image of $$[0 \to M_1 \xrightarrow{\alpha} M_2 \xrightarrow{\beta} M_3 \to 0] \in \mathrm{Ext}_{\mathcal{A}}(M_3,M_1)$$ under the morphism $$\mathrm{Ext}_{\mathcal{A}}(f,M_1) \colon \mathrm{Ext}_{\mathcal{A}}(M_3,M_1) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$$.

In particular, if we are given $$f,g \in \mathrm{Hom}_{\mathcal{A}}(N,M_3)$$, then $$\delta(f+g)$$ equals the image of $$[0 \to M_1 \xrightarrow{\alpha} M_2 \xrightarrow{\beta} M_3 \to 0] \in \mathrm{Ext}_{\mathcal{A}}(M_3,M_1)$$ under the morphism $$\mathrm{Ext}_{\mathcal{A}}(f+g,M_1) \colon \mathrm{Ext}_{\mathcal{A}}(M_3,M_1) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$$.

However, it is well-known that $$f+g \in \mathrm{Hom}_{\mathcal{A}}(N,M_3)$$ may be described as the composition $$f+g \colon N \xrightarrow{\Delta} N \times N \xrightarrow{f \times g} M_3 \times M_3 \xrightarrow{\Sigma} M_3.$$ Moreover, by the definition of the Baer sum (cf. also this answer by Roland), $$\delta(f) + \delta(g)$$ is given by the image of $$[0 \to M_1 \times M_1 \to (M_2 \times_f N) \times (M_2 \times_g N) \to N \times N \to 0] \in \mathrm{Ext}_{\mathcal{A}}(N \times N,M_1 \times M_1)$$ ((here $$M_2 \times_f N$$ denotes the fibre product of $$M_2 \xrightarrow{\beta} M_3 \xleftarrow{f} N$$, and $$M_2 \times_g N$$ is defined similarly)) under $$\mathrm{Ext}_{\mathcal{A}} \Big( (N \times N, M_1 \times M_1) \xrightarrow{(\Delta^{\mathrm{opp}}, \Sigma)} (N,M_1) \Big) \colon \mathrm{Ext}_{\mathcal{A}}(N \times N, M_1 \times M_1) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1).$$ Further, I am quite sure that $$[0 \to M_1 \times M_1 \to (M_2 \times_f N) \times (M_2 \times_g N) \to N \times N \to 0] \in \mathrm{Ext}_{\mathcal{A}}(N \times N,M_1 \times M_1)$$ is the image of $$[0 \to M_1 \times M_1 \to M_2 \times M_2 \to M_3 \times M_3 \to 0] \in \mathrm{Ext}_{\mathcal{A}}(M_3 \times M_3,M_1 \times M_1)$$ under $$\mathrm{Ext}_{\mathcal{A}}(f \times g, M_1 \times M_1) \colon \mathrm{Ext}_{\mathcal{A}}(M_3 \times M_3,M_1 \times M_1) \to \mathrm{Ext}_{\mathcal{A}}(N \times N,M_1 \times M_1)$$.

But now there is one final step missing in order to conclude my argument relating $$\delta(f+g)$$ with $$\delta(f)+\delta(g)$$ in terms of the Ext bifunctor $$\mathrm{Ext}_{\mathcal{A}} \colon \mathcal{A}^{\mathrm{opp}} \times \mathcal{A} \to \mathrm{Ab}$$. I think the main question here is: How can I obtain the short exact sequence $$0 \to M_1 \times M_1 \to M_2 \times M_2 \to M_3 \times M_3 \to 0$$ via pullback/pushout when starting with the given short exact sequence $$0 \to M_1 \to M_2 \to M_3 \to 0$$?

Any help is kindly appreciated! In particular, if you have a reference for (or are willing to insert here) a proof of the fact that $$\delta \colon \mathrm{Hom}_{\mathcal{A}}(N,M_3) \to \mathrm{Ext}_{\mathcal{A}}(N,M_1)$$ is a group homomorphism which does not follow the approach described above, then please feel free to let me know, too.

P.S.: I previously asked the same question on Mathematics StackExchange, but did not receive any comment/answer there.

• Isn't the connection morphism constructed using the snake lemma?
– user130903
Commented Jan 5, 2021 at 7:45
• @Zero: No, I don't think so. Commented Jan 5, 2021 at 7:56
• This just the usual long exact sequence you get for a left-exact functor like Hom(N,.). Look at Lang's book on Algebra.
– user130903
Commented Jan 5, 2021 at 8:40
• @Zero: Yes, I know, but I would like to prove this without using cohomology. Note also that $\mathrm{Ext}_\mathcal{A}(Y,X)$ is defined here to be the set of isomorphism classes of short exact sequences $0 \to X \to Z \to Y \to 0$ in $\mathcal{A}$. Commented Jan 5, 2021 at 10:12
• Then you can apply the Yoneda isomorphism.
– user130903
Commented Jan 5, 2021 at 10:49