Bourbaki

Proposition 1

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

Algebra, chapter II, § 6. Inverse and direct limits of modules

Let (Eα,fαβ), (Eα′,fαβ′), (Eα′′,fαβ′′)(E_\alpha, f_{\alpha\beta}),\ (E'_\alpha, f'_{\alpha\beta}),\ (E''_\alpha, f''_{\alpha\beta}) be three inverse systems of AαA_\alpha-modules and (uα), (vα)(u_\alpha),\ (v_\alpha) two inverse systems of AαA_\alpha-linear mappings such that the sequences

0⟶Eα′→uαEα→vαEα′′ 0 \longrightarrow E'_\alpha \xrightarrow{u_\alpha} E_\alpha \xrightarrow{v_\alpha} E''_\alpha

are exact for all α\alpha. Then, writing u=lim⁡uα, v=lim⁡vαu = \lim u_\alpha,\ v = \lim v_\alpha, the sequence

0⟶lim⁡Eα′→ulim⁡Eα→vlim⁡Eα′′ 0 \longrightarrow \lim E'_\alpha \xrightarrow{u} \lim E_\alpha \xrightarrow{v} \lim E''_\alpha

is exact.

As uα−1(0)={0}u_\alpha^{-1}(0) = \{0\} for all α\alpha, it follows from Set Theory, III, § 7, no. 2, Proposition 2 that u−1(0)={0}u^{-1}(0) = \{0\}, hence uu is injective; further, the uα(Eα′)u_\alpha(E'_\alpha) form an inverse system of subsets of the EαE_\alpha and thus u(lim⁡Eα′)=lim⁡uα(Eα′)u(\lim E'_\alpha) = \lim u_\alpha(E'_\alpha). As uα(Eα′)=vα−1(0)u_\alpha(E'_\alpha) = v_\alpha^{-1}(0) by hypothesis, v−1(0)=lim⁡uα(Eα′)=u(lim⁡Eα′)v^{-1}(0) = \lim u_\alpha(E'_\alpha) = u(\lim E'_\alpha) (Set Theory, III, § 7, no. 2, Proposition 2), which completes the proof.

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.