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I'd like to know if an even group action on a topological space is necessarily a homeomorphism. In particular, we say an action $G \times X \to X$ is even if, for any $x \in X$, there is an open neighborhood $x \in U \subset X$, such that $gU \cap U = \emptyset, \forall g \in G$ not the identity.

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Are there any even group actions? For all $g$, $gX\cap X\neq\emptyset$. –  Neal Nov 18 '12 at 3:09
Moreover, how can an action be a homeomorphism? Your question does not make sense as stated... –  Mariano Suárez-Alvarez Nov 18 '12 at 3:10
@Neal: Fixed. I forgot to exclude the identity element. Consider R as a topological space, Z as the group with the action zr=z+r. This is an even action. It is clearly a bijection and bicontinuous, and so a homeomorphism of R. In general, one defines an action to be continuous which ensures that it is a homeomorphism. My question is basically, are even actions necessarily continuous? My instructor implied this is true, but I'm at loss to show it (probably missing something obvious). –  Kannaguchi O. Nov 18 '12 at 5:55
@Kannaguchi: echoing Mariano's question, what do you mean by the statement that an action is a homeomorphism? Are you actually asking whether the map $G \times X \to X$ is a homeomorphism? This is almost never true; if $g \neq 1$ then $(g, x)$ is sent to the same point as $(1, gx)$, so if $G$ has non-identity elements then this map can't even be injective. –  Qiaochu Yuan Nov 18 '12 at 6:58
If you assume that $g:X \rightarrow X$ is continuous, which is part of the definition of a group acting on a topological space, then it's automatically a homeomorphism, with inverse $g^{-1}$ –  uncookedfalcon Nov 18 '12 at 8:14

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