Bourbaki

Lemma 4

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

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

Let G be a connected topological group, X a Hausdorff topological space and f1,f2f_1, f_2 laws of left (resp. right) operation of G on X such that, for all x∈Xx \in X, the mappings s↦f1(s,x), s↦f2(s,x)s \mapsto f_1(s, x),\ s \mapsto f_2(s, x) of GG into XX are continuous. Suppose that there exists a neighbourhood VV of {e}×X\{e\} \times X in G×XG \times X such that f1f_1 and f2f_2 coincide on VV. Then f1=f2f_1 = f_2.

Let x∈Xx \in X and AA be the set of g∈Gg \in G such that f1(g,x)=f2(g,x)f_1(g, x) = f_2(g, x). Then AA is closed in GG. On the other hand, let g∈Ag \in A; we write y=f1(g,x)=f2(g,x)y = f_1(g, x) = f_2(g, x). There exists a neighbourhood UU of ee in GG such that f1(t,y)=f2(t,y)f_1(t, y) = f_2(t, y) for t∈Ut \in U, in other words such that f1(t′,x)=f2(t′,x)f_1(t', x) = f_2(t', x) for t′∈Ugt' \in Ug (resp. gUgU). Hence AA is open in GG and therefore A=GA = G.

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.