Bourbaki

Proposition 21

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

Algebra, chapter II, § 1. Modules

Every exact sequence of A-modules

0⟶G→gE→fF⟶0 0 \longrightarrow G \xrightarrow{g} E \xrightarrow{f} F \longrightarrow 0

in which F is a free A-module, splits (no. 9). To be precise, if (bλ)λ∈L(b_\lambda)_{\lambda \in L} is a basis of F and, for each λ∈L\lambda \in L, aλa_\lambda is an element of E such that f(aλ)=bλf(a_\lambda) = b_\lambda, the family (aλ)λ∈L(a_\lambda)_{\lambda \in L} is free and generates a supplementary submodule to g(G).

There exists one and only one linear mapping h:F→Eh : F \to E such that h(bλ)=aλh(b_\lambda) = a_\lambda for all λ∈L\lambda \in L (Corollary 3 to Proposition 17). As h is a linear section associated with f, the proposition follows from I, § 4, no. 9, Proposition 15.

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.