Bourbaki

Theorem 5

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

Algebra, chapter II, § 7. Vector spaces

Given two exact sequences of vector spaces (over the same field K*) and linear mappings*

0→E′→E→E′′→00→F′→F→F′′→0 \begin{align*} 0 &\to E' \to E \to E'' \to 0 \\ 0 &\to F' \to F \to F'' \to 0 \end{align*}
and two vector spaces G, H over K, the corresponding sequences
0→Hom(E′′,G)→Hom(E,G)→Hom(E′,G)→00→Hom(H,F′)→Hom(H,F)→Hom(H,F′′)→0 \begin{align*} 0 &\to \mathrm{Hom}(E'', G) \to \mathrm{Hom}(E, G) \to \mathrm{Hom}(E', G) \to 0 \\ 0 &\to \mathrm{Hom}(H, F') \to \mathrm{Hom}(H, F) \to \mathrm{Hom}(H, F'') \to 0 \end{align*}
are exact and split.

This follows from the fact that every vector subspace is a direct factor (no. 3, Proposition 4) and from § 2, no. 1, Propositions 1 and 2.

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.