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It is well known that two connecting homomomorphisms each belonging to a diagram are combined via naturality (if the two diagrams are commutatively related).

But what about the uniqueness of the connecting homomorphism itself? - i.e only one diagram and at start fixed chosen kernels and cokernels of all morphisms in the diagram

My answer: It is strict unique. My proof: Take two times this one diagram and relate them together commutatively via the identical morphisms. Then given two connecting homomorphisms d and d' of the two same diagramms, they must be equal by naturality.

My question: I want to prove the naturality of the connecting morphism via arrow categories (its very elegant). But therefore I need the strict uniqueness which I can only show using naturality (see above).

So: How can one see directly the strict uniqueness without using naturality? All should live in an abelian category.

Thank you very much for an answer.

user282600
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  • Is your question why is the connection morphism in the snake lemma unique? – Nex Oct 22 '15 at 11:25
  • Sorry, I forgot to mention the words "snake lemma". Yes, that is my question. I'm looking for a uniqueness-proof of the connecting homomorphism in the snake lemma in abelian categories, a proof without using naturality of the connecting homomorphism. – user282600 Oct 22 '15 at 20:24
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    I think one needs to carefully state this question, in particular what qualifies as a connecting morphism. For example if $A$ is an object in an abelian category with non-trivial automorphism. There is an obvious morphism of exact sequences from $0\to A\to A$ to $A\to A\to 0$. Any automorphism of $A$ will connect the two sequences $0\to 0\to A$ and $A\to 0\to 0$, obtained by taking kernels and cokernels respectively, forming a longer one. – Nex Oct 22 '15 at 20:54
  • Maybe there exists a unique connecting morphism making the long sequence exact? – Gabriel Feb 26 '21 at 18:38

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