# A lemma on a sequence of three morphisms

Few months ago, I proved that when there is three morphisms of modules over a commutative ring with zero composition, i.e., a sequence $$A \xrightarrow{\alpha} B \xrightarrow{\beta} C \xrightarrow{\gamma} D$$ with $$\gamma\beta\alpha=0$$, then there is an induced exact sequence $$0 \to \text{ker }\alpha \xrightarrow{i_1} \text{ker }\beta\alpha \xrightarrow{\alpha'} \text{ker }\beta \xrightarrow{i_2} \text{ker }\gamma\beta/\text{im }\alpha \xrightarrow{\beta'} \text{ker }\gamma/\text{im }\beta\alpha \xrightarrow{i_3} \text{coker }\beta \xrightarrow{\gamma'} \text{coker }\gamma\beta \xrightarrow{i_4} \text{coker }\gamma \to 0.$$ in which all maps are canonical. Using this lemma, it was possible to prove for example the statements about exact couples in this article on my blog: https://fractalofideas.wordpress.com/2019/12/10/closed-complex-of-order-3/

The simple and elementary proof is given here, at the middle of the article: https://fractalofideas.wordpress.com/2019/09/23/how-to-prove-snake-lemma-without-diagrams/

As one may notice, this is nothing but the snake lemma, maybe in its simplest form possible. Is this worth to submit this result to some academic journal?

• The standard proof of the snake lemma seems pretty simple and elementary to me... Dec 9, 2019 at 18:05
• @lambda I agree with that. That’s my hesitation too. But I could say that this form of the lemma is most easy to apply in many situations.
– HyJu
Dec 9, 2019 at 18:11
• A modern point of view on such lemmas in abelian categories makes systematic use of the salamander lemma. See for example sbseminar.wordpress.com/2007/11/13/… (by MO founding member Anton Geraschenko). Dec 9, 2019 at 18:18
• @Todd Trimble Thank you for your comment and reference! I think the salamander lemma could also be proved using this lemma as well, which now I’m trying.
– HyJu
Dec 9, 2019 at 18:30
• Sorry, but I have to link to this: youtube.com/watch?v=etbcKWEKnvg Dec 9, 2019 at 18:33