Lemma 1

Tag 0DOV, permanent. Cite it as [Bourbaki, Tag 0DOV].

Algebra, chapter X, § 1. Compléments d’algèbre linéaire

Consider a commutative diagram of A-modules

\tag{7} \begin{array}{ccc} M & \xrightarrow{u} & N\\ \downarrow f & & \downarrow g\\ M' & \xrightarrow{u'} & N' \end{array}

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

\tag{8} \begin{array}{ccc} \operatorname{Ker}(f) & \xrightarrow{u_1} & \operatorname{Ker}(g)\\ \downarrow i & & \downarrow j\\ M & \xrightarrow{u} & N \end{array}

and

\begin{array}{ccc} M' & \xrightarrow{u'} & N' \\ p \downarrow & & q \downarrow \\ \text{Coker } (f) & \xrightarrow{u_2} & \text{Coker } (g) \end{array}

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

Indeed, if $x \in \text{Ker } (f)$ , we have $f(\bar{x}) = 0$ and $g(u(x)) = u'(f(x)) = 0$ , hence $u(x) \in \text{Ker } (g)$ , and the existence and uniqueness of $u_1$ are then immediate. Similarly, we have

u'(f(M)) = g(u(M)) \subset g(N),

hence

u'

gives by passing to the quotients a homomorphism

u_2 : \text{Coker } (f) \to \text{Coker } (g),

which is the only homomorphism for which (9) is commutative.

Let us now start from a commutative diagram (4) of A-modules; by virtue of Lemma 1, there corresponds to it a commutative diagram

\begin{array}{ccccccccc} \text{Ker } (f) & \xrightarrow{u_1} & \text{Ker } (g) & \xrightarrow{v_1} & \text{Ker } (h) \\ i \downarrow & & j \downarrow & & k \downarrow \\ M & \xrightarrow{u} & N & \xrightarrow{v} & P \\ f \downarrow & & g \downarrow & & h \downarrow \\ M' & \xrightarrow{u'} & N' & \xrightarrow{v'} & P' \\ p \downarrow & & q \downarrow & & r \downarrow \\ \text{Coker } (f) & \xrightarrow{u_2} & \text{Coker } (g) & \xrightarrow{v_2} & \text{Coker } (h) \end{array}

where $i, j, k$ are the canonical injections, $p, q, r$ the canonical surjections, $u_1, u_2$ (resp. $v_1, v_2$ ) the homomorphisms deduced from $u, u'$ (resp. $v, v'$ ) by Lemma 1.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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