Proposition 15

Let X be a topological space on which a topological group G operates continuously and transitively. For X to be a topological homogeneous space (relative to G) it is sufficient that for some point x_0 ∈ X the mapping s → s.x_0 transforms each neighbourhood of e in G into a neighbourhood of x_0 in X.

Every x ∈ X can be written as x = t.x_0 for some t ∈ G. If V is a neighbourhood of e, then V.x = (Vt).x_0 is a neighbourhood of x, for we can write (Vt).x_0 = t((t^{-1}Vt).x_0), and the assertion follows from the facts that t^{-1}Vt is a neighbourhood of e in G and that y → t.y is a homeomorphism of X onto itself (no. 4, Lemma 1). It follows that if U is any open subset of G and if x is any point of X, then U.x is open in X; for if t ∈ U, then t^{-1}U is a neighbourhood of e, hence (t^{-1}U).x is a neighbourhood of x, and t.((t^{-1}U).x) = U.x is a neighbourhood of t.x. Hence U.x is open in X, so that the mapping s → s.x of G into X is open. This completes the proof.