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let $X$ be a top space and $G$ a group acting on $X$. Consider $ F=\bigl\{(x,gx)\in X\times X\mid x\in X, g\in G\bigr\}\ $ i want to write an homeomorphic image of $F$.

for example take $G=\mathbb Z_2=\{1,-1\}$ acting on the sphere $S^d$ by multiplication. in This case $F=\{(x,y)\in X \times X \;| \; x=\pm y\}$ so we can write it as the disjoint union $F=F_1\sqcup F_2$ where $F_1=\{(x,x)\in S^d\times S^d\}\cong diagonal(S^d\times S^d)\cong S^d$ and $F_2=\{(x,-x)\in S^d\times S^d\}\cong S^d$ Hence $F$ is homeomorphic to a disjoint union of two copies of $S^d$,i.e. $F=S^d\sqcup S^d$

Is there a similar way to write $F$ for more general $X$ and $G$? thanks for help.

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    The two I wrote are the same; the one you wrote is not correct use of mathematical quantifiers. "For all $x$" etc are prefixes, not suffixes and not sentences in and of themselves. We (should) never write "$(a+b)+c = a+(b+c)\ \forall a,b,c$", we should write $\forall a,b,c \bigl( (a+b)+c = a+(b+c)$". What you wrote originally was the equivalent of an opening clause "Where for all $x\in X$ and all $g\in G$ the following holds..." and then you never said what "the following" was.2011-12-04

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In any group action, $g\colon X\to X$ induces a permutation, so for a fixed $g$, $F_g = \{(x,gx)\in F\times F\mid x\in F\}$ can be identified with $F$ in the obvious way.

The only question then is whether the sets $F_g$ form a partition. If $(x,y)\in F_g\cap F_h$, then $gx = hx$, hence $h^{-1}gx = x$. You want $F_g\cap F_h\neq\emptyset$ to imply $F_g=F_h$. That is, you want the following implication to hold: $\exists x(gx = hx)\Rightarrow \forall x (gx=hx).$

The action of $G$ on $X$ induces a group homomorphism $G\to S_X$, the permutation group of $X$. Let $N$ be the kernel. You want the induced action of $G/N$ on $X$ to be free (or fixed-point-free); that is: if $(gN)x = x$ for some $x$, then $gN= 1N$.

If the action of $G/N$ is free, then the $F_g$ partition $X$ into sets that are bijectable with $F$.

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    @palio: I don't know; I'm not very familiar with topological actions. You'll note that the above does not use, in any way whatsoever, the fact that you are acting on a topological set, just that you are acting on a set. Presumably, in order to get nice topological conditions, you are going to require that the actions be somehow related to the topology, e.g., that $g\colon X\to X$ (given by $g(x) = gx$) be a homeomorphism from each $g\in G$.2011-12-04