Proposition 1

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

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

Consider a commutative diagram of A-modules

\tag{4} \begin{array}{ccccc} M & \xrightarrow{u} & N & \xrightarrow{v} & P\\ \downarrow f & & \downarrow g & & \downarrow h\\ M' & \xrightarrow{u'} & N' & \xrightarrow{v'} & P' \end{array}

Suppose that the two lines of (4) are exact. Then:

(i) If $h$ is injective, one has

\tag{5} \operatorname{Im}(g)\cap\operatorname{Im}(u')=\operatorname{Im}(u'\circ f)=\operatorname{Im}(g\circ u).

(ii) If $f$ is surjective, one has

\tag{6} \operatorname{Ker}(g)+\operatorname{Im}(u)=\operatorname{Ker}(v'\circ g)=\operatorname{Ker}(h\circ v).

Let us prove (i). It is clear that one has

\operatorname{Im}(u'\circ f)=\operatorname{Im}(g\circ u)\subset\operatorname{Im}(g)\cap\operatorname{Im}(u').

Inverse support, let $y'\in\operatorname{Im}(g)\cap\operatorname{Im}(u')$ . There exists $y\in N$ such that $y'=g(y)$ . Since $v'\circ u'=0$ , we have $0=v'(y')=v'(g(y))=h(v(y))$ , whence $v(y)=0$ since $h$ is injective. Since $(u,v)$ is an exact sequence, there exists $x\in M$ such that $y=u(x)$ , whence $y'=g(u(x))$ .

Let us prove (ii). Since $v\circ u=0$ and $v'\circ u'=0$ , it is clear that

\operatorname{Ker}(g)+\operatorname{Im}(u)\subset\operatorname{Ker}(v'\circ g)=\operatorname{Ker}(h\circ v).

Conversely, let $y\in\operatorname{Ker}(v'\circ g)$ . Then $g(y)\in\operatorname{Ker}(v')$ , and there exists $x'\in M'$ such that $u'(x')=g(y)$ since the sequence $(u',v')$ is exact. Since $f$ is surjective, there exists $x\in M$ such that $f(x)=x'$ , whence $g(y)=u'(f(x))=g(u(x))$ ; we conclude that $y-u(x)\in\operatorname{Ker}(g)$ , which completes the proof.

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

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