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 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.

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.

**Answer**

In any abelian category you can intriduce 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.

**Attribution***Source : Link , Question Author : Jesko Hüttenhain , Answer Author : user46336*