Bourbaki

Proposition 9

Tag 0B2P, permanent. Cite it as [Bourbaki, Tag 0B2P].

General Topology, chapter I, § 4. Product of topological spaces

Let II be a directed set and JJ a cofinal subset of II. Let (Xα,fαβ)(X_\alpha, f_{\alpha\beta}) be an inverse system of topological spaces indexed by II; let X=lim←⁡XαX = \varprojlim X_\alpha and let fα:X→Xαf_\alpha : X \to X_\alpha be the canonical mapping. Then the family of sets fα−1(Uα)f^{-1}_\alpha(U_\alpha), where α\alpha runs through JJ and UαU_\alpha runs through a base Bα\mathfrak{B}_\alpha of the topology of XαX_\alpha for each α∈J\alpha \in J, is a base of the topology of XX.

From § 2, no. 3 we know that the finite intersections of sets of the form fα−1(Uα)f^{-1}_\alpha(U_\alpha) (α∈I,Uα\alpha \in I, U_\alpha open in XαX_\alpha) form a base of the topology of XX. If (αi)1≤i≤n(\alpha_i)_{1 \leq i \leq n} is a finite family of indices of II, then there exists γ∈J\gamma \in J such that αi≤γ\alpha_i \leq \gamma for 1≤i≤n1 \leq i \leq n; hence fαi=fαiγ∘fγf_{\alpha_i} = f_{\alpha_i\gamma} \circ f_\gamma; if we put

Vγ=⋂ifαiγ−1(Uαi), V_\gamma = \bigcap_i f^{-1}_{\alpha_i\gamma}(U_{\alpha_i}),

then we have

fγ−1(Vγ)=⋂ifαi−1(Uαi); f^{-1}_\gamma(V_\gamma) = \bigcap_i f^{-1}_{\alpha_i}(U_{\alpha_i});

but VγV_\gamma is open and is therefore a union of sets belonging to Bγ\mathfrak{B}_\gamma. Hence the result.

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 volume. Not translated.