Consider the following commutative diagram in an abelian category:$$\require{AMScd} \begin{CD} @. A @>{f}>> B @>{g}>> C @>>> 0\\ @. @VV{a}V @VV{b}V @VV{c}V \\ 0@>>> A' @>>{f'}> B' @>>{g'}> C' \end{CD}$$

The snake lemma gives a morphism $\delta:\ker c\to \operatorname{coker} a$ making the sequence $$\ker a \to \ker b\to \ker c\to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c$$ exact. I wonder if the morphism $\delta$ is unique such that the induced sequence above is exact.

The proof of the snake lemma using elements makes me believe that the answer is yes, but I'm not really sure.

(This is basically the same as Uniqueness of the connecting homomorphism in theory of abelian categories but my question adds a condition, so I think that it was worth making a new question. If someone is against this, I can delete my question.)

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    $\begingroup$ Don't think so. Say $A=C'=0$, $B=C\neq 0$, $A'=B'=B\oplus B$, with $g$ and $f'$ the identity functions, and $b\colon B\to B\oplus B$ is the inclusion into the first component. The sequence you get is $0\to 0\to B\to B\oplus B\to B\to 0$, and all you need is for $\delta$ to map $B$ to $B\oplus 0$ isomorphically. If $B$ has nontrivial automorphisms, you can have lots of $\delta$ that work. $\endgroup$ Feb 26, 2021 at 20:04

1 Answer 1


In order for the sequence to be exact, all you need is an isomorphism between $\def\coker{\operatorname{coker}}\ker c/\operatorname{im}(\ker b\to \ker c)$ and $\ker(\coker a\to\coker b)$. By the snake lemma, these objects are isomorphic, but they may well have non-trivial automorpshims that you can sneak into $\delta$. For abelian groups, you may for example always replace $\delta$ with $-\delta$ (which is the same as $\delta$ only when the quotients in question are of exponent 2). Incidentally, $-\delta$ is also natural.


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