Bourbaki

Example

Tag 0BL6, permanent. Cite it as [Bourbaki, Tag 0BL6].

General Topology, chapter III, § 4. Groups operating properly on a topological space; compact ness in topological groups and spaces with operators

Let GG be a group and let HH be a subgroup of GG. Then HH operates freely by (left or right) translations on GG.

Let GG be a group operating freely on a set XX, let RR be the equivalence relation defined by GG on XX, and let C⊂X×XC \subset X \times X be the graph of RR. If (x,y)∈C(x, y) \in C, then there exists s∈Gs \in G such that s.x=ys.x = y; and ss is unique, since s.x=s′.xs.x = s'.x implies s′−1s.x=x{s'}^{-1}s.x = x, and therefore s′−1s=e{s'}^{-1}s = e (as GG operates freely). If we make correspond to (x,y)∈C(x, y) \in C the unique s∈Gs \in G such that s.x=ys.x = y, we define a mapping φ:C→G\varphi : C \to G, which we shall call the canonical mapping of CC into GG. With this notation:

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

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Cites

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Transcribed from the printed volume. Not translated.