Bourbaki

Proposition 2

Tag 05N1, permanent. Cite it as [Bourbaki, Tag 05N1].

Algebra, chapter II, § 2. Modules of linear mappings. Duality

If the exact sequence

(7)

0⟶F′→uF→vF′′⟶0 0 \longrightarrow F' \xrightarrow{u} F \xrightarrow{v} F'' \longrightarrow 0

splits (in other words, if u(F′)u(F') is a direct factor of FF), the sequence

(8)

0⟶Hom⁡(E,F′)→uˉHom⁡(E,F)→vˉHom⁡(E,F′′)⟶0 0 \longrightarrow \operatorname{Hom}(E, F') \xrightarrow{\bar{u}} \operatorname{Hom}(E, F) \xrightarrow{\bar{v}} \operatorname{Hom}(E, F'') \longrightarrow 0

is exact and splits. Conversely, if sequence (8) is exact for every A-module E, the exact sequence (7) splits.

The first assertion follows from the fact that

Hom⁡(E,F′)⊕Hom⁡(E,F′′) \operatorname{Hom}(E, F') \oplus \operatorname{Hom}(E, F'')

is canonically identified with Hom⁡(E,F′⊕F′′)\operatorname{Hom}(E, F' \oplus F'') by means of the Z\mathbf{Z}-linear mapping Hom⁡(1E,j′)+Hom⁡(1E,j′′),j′:F′→F′⊕F′′\operatorname{Hom}(1_E, j') + \operatorname{Hom}(1_E, j''), j' : F' \to F' \oplus F'' and j′′:F′′→F′⊕F′′j'' : F'' \to F' \oplus F'' being the canonical injections (§ 1, no. 6, Corollary 1 to Proposition 6). Conversely, if sequence (8) is exact for E=F′′E = F'', there is an element g∈Hom⁡(F′′,F)g \in \operatorname{Hom}(F'', F) such that v∘g=1F′′v \circ g = 1_{F''} and the conclusion follows from § 1, no. 9, Proposition 15.

Remark (2). The results of this no. are valid without modification for all commutative groups with operators.

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

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Cites

Transcribed from the printed edition, 1998, Springer. Not translated.