Bourbaki

Proposition 1

Tag 03OC, permanent. Cite it as [Bourbaki, Tag 03OC].

Theory of Sets, chapter III, § 7. Inverse limits and direct limits

Let I be an ordered set, let (Eα,fαβ)(E_\alpha, f_{\alpha\beta}) be an inverse system of sets relative to I, let E=lim←⁡EαE = \varprojlim E_\alpha be its inverse limit, and for each α∈I\alpha \in I let

fα:E→Eαf_\alpha : E \to E_\alpha

be the canonical mapping. For each α∈I\alpha \in I, let uαu_\alpha be a mapping of a set F into EαE_\alpha such that

fαβ∘uβ=uαwhenever α⩽β.(5)f_{\alpha\beta} \circ u_\beta = u_\alpha \qquad \textit{whenever } \alpha \leqslant \beta. \tag{5}

Then :

(a) there exists a unique mapping uu of F into E such that

uα=fα∘ufor all α∈I;(6)u_\alpha = f_\alpha \circ u \qquad \textit{for all } \alpha \in I;\tag{6}

(b) the mapping uu is injective if and only if, for each pair of distinct elements yy, zz of F, there exists α∈I\alpha \in I such that uα(y)≠uα(z)u_\alpha(y) \neq u_\alpha(z).

For the relation uα=fα∘uu_\alpha = f_\alpha \circ u means that for each y∈Fy \in F we have

prα(u(y))=uα(y);\mathrm{pr}_\alpha(u(y)) = u_\alpha(y);

the element u(y)∈∏α∈IEαu(y) \in \prod_{\alpha \in I} E_\alpha is uniquely determined by u(y)=(uα(y))α∈Iu(y) = (u_\alpha(y))_{\alpha \in I}. It remains to be shown that u(y)∈Eu(y) \in E for all y∈Fy \in F, in other words, that

prα(u(y))=fαβ(prβ(u(y)))\mathrm{pr}_\alpha(u(y)) = f_{\alpha\beta}(\mathrm{pr}_\beta(u(y)))

whenever α⩽β\alpha \leqslant \beta. But this can be written in the form

uα(y)=fαβ(uβ(y))u_\alpha(y) = f_{\alpha\beta}(u_\beta(y))

and therefore follows from (5). The second part of the Proposition follows immediately from the definitions.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2004, Springer. Not translated.