Lemma 1

Consider a commutative diagram of commutative groups:

Then there exists one and only one homomorphism $u_1 : \operatorname{Ker}(a) \to \operatorname{Ker}(b)$ and one and only one homomorphism $u_2 : \operatorname{Coker}(a) \to \operatorname{Coker}(b)$ such that the diagrams

and

are commutative, i and j denoting the canonical injections and p and q the canonical surjections.

If $x \in \mathrm{Ker}(a)$, then $a(x) = 0$ and $b(u(x)) = u'(a(x)) = 0$, hence $u(x) \in \mathrm{Ker}(b)$, and the existence and uniqueness of $u_1$ are then immediate. Similarly, we have $u'(a(A)) = b(u(A)) \subset b(B)$, then by taking quotients $u'$ gives a homomorphism $u_2 : \mathrm{Coker}(a) + \mathrm{Coker}(b)$, which is the only homomorphism for which (15) is commutative.

Let us now start with the commutative diagram (10) of commutative groups; there corresponds to it by Lemma 1 a diagram

where $i, j, k$ are the canonical injections, $p, q, r$ the canonical surjections and $u_1, u_2$ (resp. $v_1, v_2$) the homomorphisms canonically associated with $u, u'$ (resp. $v, u'$) by Lemma 1. It is immediately verified that this diagram is commutative.