Bourbaki

Lemma 1

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

Algebra, chapter X, § 3. Résolutions

Consider a diagram of A-modules and homomorphisms

M′→α′Mf′↓f↓N′→β′NM→αM′′f↓k′′↓N→βN′′ \begin{array}{ccc} M' & \xrightarrow{\alpha'} & M \\ f' \downarrow & & f \downarrow \\ N' & \xrightarrow{\beta'} & N \end{array} \quad \begin{array}{ccc} M & \xrightarrow{\alpha} & M'' \\ f \downarrow & & k'' \downarrow \\ N & \xrightarrow{\beta} & N'' \end{array}

such that f∘α′=β∘f′,α∘α′=0,Ker β=Im β′,and f=k′′∘α+β∘k and where M′ is projective. There exists an A-homomorphism k′:M′→N′ such that f′=k∘α′+β′∘k′.f \circ \alpha' = \beta \circ f', \alpha \circ \alpha' = 0, \mathrm{Ker}\ \beta = \mathrm{Im}\ \beta', \text{and } f = k'' \circ \alpha + \beta \circ k \text{ and where } M' \text{ is projective. There exists an A-homomorphism } k': M' \to N' \text{ such that } f' = k \circ \alpha' + \beta' \circ k'.

In fact, put g=f′−k∘α′g = f' - k \circ \alpha'; we have

β∘g=β∘f′−β∘k∘α′=f∘α′−β∘k∘α′=k′′∘α∘α′=0. \beta \circ g = \beta \circ f' - \beta \circ k \circ \alpha' = f \circ \alpha' - \beta \circ k \circ \alpha' = k'' \circ \alpha \circ \alpha' = 0.
This implies Im (g)⊂Ker (β)=Im (β′)\mathrm{Im}\ (g) \subset \mathrm{Ker}\ (\beta) = \mathrm{Im}\ (\beta'). Since M′M' is projective, there therefore exists an A-homomorphism k′:M′→N′k': M' \to N' such that β′∘k′=g\beta' \circ k' = g, whence the lemma.

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

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