Corollary 2

Tag 0427, permanent. Cite it as [Bourbaki, Tag 0427].

Commutative Algebra, chapter I, § 1. Diagrams and exact sequences

Suppose that the diagram (10) is commutative and the two rows are exact. Under these conditions:

(i) If b is injective and if a and v are surjective, then c is injective. (ii) If b is surjective and if c and $u'$ are injective, then a is surjective.

To prove (i) consider the diagram

\begin{array}{ccccc} u(A) & \xrightarrow{w} & B & \xrightarrow{v} & C \\ a' \downarrow & & b \downarrow & & c \downarrow \\ u'(A') & \xrightarrow{w'} & B' & \xrightarrow{v'} & C' \end{array}

where $a'$ is the mapping with the same graph as the restriction of b to $u(A)$ and w and $w'$ are the canonical injections; clearly this diagram is commutative and its rows exact. Moreover $w'$ is injective and by hypothesis v is surjective; then by Proposition 2, (iii) we have an exact sequence

0 = \mathrm{Ker}(b) \longrightarrow \mathrm{Ker}(c) \xrightarrow{d} \mathrm{Coker}(a') = 0

since b is injective and $a'$ is surjective; whence $\mathrm{Ker}(c) = 0$ .

To prove (ii) consider the diagram

\begin{array}{ccccc} A & \xrightarrow{u} & B & \xrightarrow{w} & v(B) \\ a' \downarrow & & b \downarrow & & c' \downarrow \\ A' & \xrightarrow{w'} & B' & \xrightarrow{\sigma'} & v'(B') \end{array}

where this time $c'$ is the mapping with the same graph as the restriction of c to v(B) and w and w' have respectively the same graphs as v and v'; this diagram is commutative and its rows exact. Moreover w is surjective and by hypothesis u' is injective; then by Proposition 2, (iii) we have an exact sequence

0 = \operatorname{Ker}(c') \xrightarrow{d} \operatorname{Coker}(a) \longrightarrow \operatorname{Coker}(b) = 0

since b is surjective and c' is injective; whence Coker (a) = 0.

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

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Cites

Proposition 2