Bourbaki

Proposition 17

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

Algebra, chapter II, § 1. Modules

For every A-module E and every mapping f:T→Ef : T \to E, there exists one and only one A-linear mapping g:As(T)→Eg : A_s^{(T)} \to E such that f=g∘ϕf = g \circ \phi.

The condition f=g∘ϕf = g \circ \phi means that g(et)=f(t)g(e_t) = f(t) for all t∈Tt \in T, which is equivalent to g(ξet)=ξf(t)g(\xi e_t) = \xi f(t) for all ξ∈A\xi \in A and all t∈Tt \in T and also means that g∘jtg \circ j_t is the linear mapping ξ↦ξf(t)\xi \mapsto \xi f(t) of AsA_s into EE for all t∈Tt \in T; the proposition is therefore a special case of no. 6, Proposition 6.

The linear mapping gg is said to be determined by the family (f(t))t∈T(f(t))_{t \in T} of elements of EE; by definition

g(∑t∈Tξtet)=∑t∈Tξtf(t). g\left( \sum_{t \in T} \xi_t e_t \right) = \sum_{t \in T} \xi_t f(t).

The kernel RR of gg is the set of (ξt)∈As(T)(\xi_t) \in A_s^{(T)} such that ∑tξtf(t)=0\sum_t \xi_t f(t) = 0; it is sometimes said that the module RR is the module of linear relations between the elements of the family (f(t))t∈T(f(t))_{t \in T}. The exact sequence

0⟶R⟶As(T)⟶gE 0 \longrightarrow R \longrightarrow A_s^{(T)} \overset{g}{\longrightarrow} E
is said to be determined by the family (f(t))t∈T(f(t))_{t \in T}.

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.