# Faithful actions of finite groups on topological spaces

Suppose that $G$ is a finite group acting faithfully on a topological space $X$. In the smooth setting, one can deduce that for each $x$ in $M$, the induced map $$G_x \to Diff_x\left(M\right)$$ from the isotropy group of $x$ to the group of germs of locally defined diffeomorphisms is a monomorphism. Does this result hold true in the topological setting? (Replacing diffeomorphisms with homeomorphisms)

In the smooth setting, the result follows from the following standard lemma:

Lemma: *Let $M$ be a manifold and $G$ a finite subgroup of $\mathit{Diff}\left(M\right)$. Then for any smooth map $f:V \to M$ defined on a non-empty open connected submanifold of $M,$ such that $f\left(x\right) \in G \cdot x$ for all $x,$ there exists a unique element $g \in G$ such that $f=g|_V.$ *

I'm guessing the answer is NO for general topological spaces. However, a counter-example would be nice.

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The answer is no. Consider your space to be a wedge of a bunch of circles -- and don't use common wedge points. Then an involution can fix a neighbourhood of a point and still flip some of the circle wedge summands. – Ryan Budney May 10 '11 at 17:47
I suspect the answer is yes for connected topological manifolds but I don't have an argument off the top of my head. These kinds of questions were investigated in the 60's as far as I know. – Ryan Budney May 10 '11 at 18:57
@Ryan: Thanks. I'm glad to see that there is such a simple counter-example. – David Carchedi May 10 '11 at 23:03
maybe Ryan or David can repost this as an answer and close the question ? – HenrikRüping May 11 '11 at 12:23