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Suppose we have two short exact sequences in an abelian category $$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$ $$0 \to A' \mathrel{\overset{f'}{\to}} B' \mathrel{\overset{g'}{\to}} C' \to 0 $$ and morphisms $a : A \to A', b : B \to B', c : C \to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence $$0 \to \ker a \to \ker b \to \ker c \to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c \to 0$$ where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.

It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $\ker a, \ker b, \operatorname{coker} b, \operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : \ker c \to \operatorname{coker} a$. However, I'm stuck showing exactness at $\ker c$ and $\operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?

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    $\begingroup$ In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8. $\endgroup$
    – t.b.
    Oct 22, 2011 at 17:09
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    $\begingroup$ You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof... $\endgroup$ Oct 22, 2011 at 19:05
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    $\begingroup$ @Arturo: It's on youtube of course. However, it's an "elementary" proof. $\endgroup$
    – t.b.
    Oct 22, 2011 at 19:18
  • $\begingroup$ Ask: if using elements is more natural and we have Freyd-Mitchell embedding, then why not just chase elements? $\endgroup$ Jan 23 at 1:24

4 Answers 4

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You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.

In any case, you can also construct the boundary map as follows:

We are given a map $b: B \to B'$. Let $B'' \hookrightarrow B$ denote the preimage in $B$ of $\ker c$. (If you want to desribe this in more categorical terms, it is the kernel of the composite $B \to C \to C'$.)

Then the map $B''\hookrightarrow B \rightarrow B'$ factors through the monomorphism $A' \hookrightarrow B'$ (using the fact that $A' =\ker(B' \to C')\, \, $). This then induces a map on quotients $ B''/A \to A'/\operatorname{im}A$, which is precisely the desired map $\ker c \to\operatorname{coker}a.$

Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.

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    $\begingroup$ Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though. $\endgroup$
    – Zhen Lin
    Oct 22, 2011 at 20:31
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    $\begingroup$ @ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me? $\endgroup$
    – user153312
    Jan 20, 2015 at 19:08
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    $\begingroup$ There are many ways to do it. Another way is to use the mapping cone construction. $\endgroup$
    – Zhen Lin
    Jan 20, 2015 at 21:17
  • $\begingroup$ How do you see that the universal map $B''\to\ker c$ is an isomorphism? $\endgroup$
    – Bubaya
    Mar 1, 2019 at 13:12
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    $\begingroup$ @Bubaya $B''\to\ker c$ is just an epimorphism, whose kernel is $A$. And proving this fact is a little bit cumbersome. You can refer to Stacks Project for details. $\endgroup$
    – Jerry.Li
    Jun 4, 2020 at 15:41
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There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).

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    $\begingroup$ Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book? $\endgroup$
    – robjohn
    Jan 14, 2013 at 7:37
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The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.

However the Secret Blogging Seminar says:

If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.

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  • $\begingroup$ By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $f\colon A\to X$ and $g\colon B\to X$? $\endgroup$
    – Yai0Phah
    Oct 7, 2016 at 16:10
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There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).

Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).

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    $\begingroup$ I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book. $\endgroup$
    – t.b.
    Oct 22, 2011 at 20:12

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