Bourbaki

Proposition 3

Tag 0B3G, permanent. Cite it as [Bourbaki, Tag 0B3G].

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

Let X, Y be two topological spaces, let f:X→Yf : X \to Y be a continuous mapping, let R be the equivalence relation f(x)=f(y)f(x) = f(y) on X, and let X→pX/R→hf(X)→iYX \xrightarrow{p} X/R \xrightarrow{h} f(X) \xrightarrow{i} Y be the canonical decomposition of f. Then the following three statements are equivalent:

a) f is an open mapping.

b) The three mappings p, h, i are open.

c) The equivalence relation R is open, h is a homeomorphism, and f(X)f(X) is an open subset of Y.

Also the preceding proposition remains true if "open" is replaced by "closed" throughout.

Since the injection i is continuous, it follows from Proposition 1 c) of no. 1 that if f is open then so is h∘ph \circ p; since p is surjective and continuous, Proposition 1 b) shows that h is open; h is in any case a continuous bijection; thus h is a homeomorphism, and therefore, from Proposition 1 a) of no. 1, p=h‾−1∘(h∘p)p = \overline{h}^{-1} \circ (h \circ p) is an open mapping. Also [no. 1, Proposition 1 b)] i∘hi \circ h is open; hence [no. 1, Proposition 1 a)] so is i=(i∘h)∘h‾−1i = (i \circ h) \circ \overline{h}^{-1}. This proves that a) implies b). Conversely, Proposition 1 a) of no. 1 shows that b) implies a). Finally, the equivalence of b) and c) follows immediately from the definitions.

The proof in the case of closed mappings is analogous, mutatis mutandis.

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

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