Bourbaki

Example

Tag 0B1L, permanent. Cite it as [Bourbaki, Tag 0B1L].

General Topology, chapter I, § 3. Subspaces, quotient spaces

Let XX be a topological space, (Xi)i∈I(X_i)_{i \in I} a covering of XX, YY the sum of the subspaces XiX_i of XX; then there is a partition (Yi)i∈I(Y_i)_{i \in I} of YY into subspaces which are both open and closed, and for each i∈Ii \in I there is a homeomorphism fi:Yi→Xif_i : Y_i \to X_i. Let f:Y→Xf : Y \to X be the continuous mapping which agrees with fif_i on YiY_i for each i∈Ii \in I, and let RR be the equivalence relation f(x)=f(y)f(x) = f(y); the quotient space Y/RY/R is thus obtained by "pasting together" the YiY_i (§ 2, no. 5). Consider the bijection g:Y/R→Xg : Y/R \to X associated with ff; in general gg is not a homeomorphism, as is shown by the example in which each XiX_i consists of a single point and XX is not discrete. However, if the interiors of the XiX_i cover XX, or if (Xi)(X_i) is a locally finite closed covering of XX, then gg is a homeomorphism: for if UU is any open subset of YY which is saturated with respect to RR, then for each i∈Ii \in I the set

f(U)∩Xi=fi(U∩Yi) f(U) \cap X_i = f_i(U \cap Y_i)
is open in XiX_i, and the assertion follows from Proposition 3 of no. 1.

The following proposition gives a simple sufficient condition for gg to be a homeomorphism:

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.