Bourbaki

Proposition 7

Tag 0B3K, permanent. Cite it as [Bourbaki, Tag 0B3K].

General Topology, chapter I, § 5. Open mappings and closed mappings

Let RR be an open equivalence relation on a topological space XX, and let φ:X→X/R\varphi : X \to X/R be the canonical mapping. If AA is any subset of XX which is saturated with respect to RR, then the closure (resp. the interior) of φ(A)\varphi(A) in X/RX/R is φ(A‾)\varphi(\overline{A}) (resp. φ(A∘)\varphi(\overset{\circ}{A})).

Each of the two assertions of the proposition can be deduced from the other by taking complements and using formula (2) of § 1, no. 6 and the fact that if BB is a saturated subset of XX then φ(∁B)=∁φ(B)\varphi(\complement B) = \complement_{\varphi}(B). By virtue of Proposition 6, A‾\overline{A} is saturated; hence φ(A‾)\varphi(\overline{A}) is closed in X/RX/R, and since A⊂A‾A \subset \overline{A} we have φ(A)⊂φ(A‾)\varphi(A) \subset \varphi(\overline{A}), so that φ(A)⊂φ(A‾)\varphi(A) \subset \varphi(\overline{A}). But since φ\varphi is continuous, φ(A‾)⊂φ(A)\varphi(\overline{A}) \subset \varphi(A) (§ 2, no. 1, Theorem 1), and the result follows.

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

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Transcribed from the printed volume. Not translated.