During my studies, I always wanted to see a "purely category-theoretical" proof of the Snake Lemma, i.e. a proof that constructs all morphisms (including the snake) and proves exactness via universal properties. It was an interest little shared by my teachers and fellow students, but I have recently found the time to pursue it again.

There is a wonderful book on category theory containing such a proof: The Handbook of Categorical Algebra, Volume 2, by Francis Borceux. I have a question about the proof, however, which I can't seem to resolve.

The Snake Lemma is Lemma 1.10.9, and I have a problem with one of the preliminaries: Namely, the "restricted" Snake Lemma 1.10.8.

Edit: I scanned the diagrams in question from the book. The following is what we want, i.e. we want to construct $\omega$ from the rest of the diagram where all squares commute and all rows and columns are exact. The claim of the Lemma

The construction is then as follows: $\Delta$ and $\Gamma$ are obtained by pull-back and we define $\Sigma:=\mathrm{Ker}(\Delta)$. Dually with $\Lambda$, $\Xi$ and $\Upsilon$.

enter image description here

On page 46, he says that

By lemma 1.10.1 and its dual, there are morphisms $\Psi$ and $\Omega$ making the diagram commutative and the outer columns exact.

I can not verify this statement. For instance concerning $\Psi$, it seems to me that in order to apply lemma 1.10.1, one would require that the sequence $(\Gamma,\lambda)$ is exact, but I do not see how that would follow from the construction. What am I doing wrong?!

Edit: Lemma 1.10.1 is the statement that in the following diagram, with commutative squares (1) and (2) and exact rows $(\zeta,\eta)$ and $(0,\nu,\xi)$ with $\gamma=\mathrm{Ker}(\theta)$, $\delta=\mathrm{Ker}(\lambda)$ and $\varepsilon=\mathrm{Ker}(\mu)$, there exist unique morphisms $\alpha$ and $\beta$ making the diagram commutative. Additionally, $(\alpha,\beta)$ is exact.

enter image description here

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    $\begingroup$ Simply for your information: Lemma $ 5 $ of Section $ 4 $ of Chapter $ 8 $ of Categories for the Working Mathematician, Second Edition by Saunders Mac Lane also gives a purely categorical proof of the Snake Lemma. :) $\endgroup$ Feb 26, 2013 at 8:56
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    $\begingroup$ It would be helpful if you could post some pictures of the relevant diagrams, to make the question more self-contained. $\endgroup$
    – Zhen Lin
    Feb 26, 2013 at 13:13
  • $\begingroup$ @Haskell Curry: Thanks for the tip, but Saunders Mac Lane has never really worked that well for me. If I get no answer to this question I might check it out, but usually I prefer Borceux' writing style. $\endgroup$ Feb 26, 2013 at 16:53
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    $\begingroup$ @Zhen Lin: There you go! $\endgroup$ Feb 26, 2013 at 16:54
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    $\begingroup$ @Jesko: Jonathan Wise has written up a direct proof of the Snake Lemma. math.stanford.edu/~jonathan/papers/snake.pdf. You can also find a proof in the first paper on abelian categories. Not Grothendieck's Tohoku, but "Exact Categories and Duality" D. A. Buchsbaum, published in 1955. $\endgroup$ Mar 9, 2013 at 3:27

1 Answer 1


In any abelian category you can introduce the notion of element. An element $y$ of an object $Y$ of an abelian category $\mathcal{A}$ is an equivalence class of pairs $(X,h)$, $X \in Ob(\mathcal A)$, $h: X \to Y$ by the equivalence relation $$ (X,h) =(X',h') \iff \exists Z \in Ob(\mathcal A), u:Z \to X, u':Z \to X'\, s.t. \, hu=hu', $$ where $u$ and $u'$ must be epimorphisms. Using the notion of element you can prove the statement in the category of abelian groups. See Gelfand, Manin "Metheods of homological algebra for details.

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    $\begingroup$ I think the primary interest in the posting of the question was to patch/explain the proof in Borceux (see Jesko's comment on the question from March 23rd). $\endgroup$
    – Mark S.
    Dec 29, 2013 at 19:33
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    $\begingroup$ @MarkS. is correct. I primarily want to understand the reasoning in Borceux' book. I hope you don't take my downvote as a personal offense, but this is precisely what I did not want. $\endgroup$ Dec 29, 2013 at 21:08
  • $\begingroup$ @JeskoHüttenhain It does seem to be a good answer to your title, but not your question "What am I doing wrong?!" However, the convention on math.stackexchange is that a question in the body takes priority, over the title. $\endgroup$
    – Mark S.
    Dec 29, 2013 at 21:10
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    $\begingroup$ @user46336 I like your answer because this answers the question I am curious about for so long - why every result proved by diagram chasing in R-Mod is always true in abelian category. For times this seemed to me really mysterious but amazing, just like some principles that always work but cannot be precisely formulated. But now I find this principle could be "exact"(LOL). $\endgroup$ Jun 6, 2014 at 2:36
  • $\begingroup$ I tried but still have no idea how this works. Could you illustrate it using categories that really have elements, such as abelian groups? $\endgroup$ Jun 6, 2014 at 4:25

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