# Actions of G on corresponding orbits are equivalent for stable maps

Given actions of G on X and on Y, these actions are equivalent if and only if there is a bijection from $X$ \ $G \rightarrow Y$ \ $G$ so that actions of G on corresponding orbits are equivalent.

Def. An action of G on X and Y is equivalent if there is a stable bijection $f: X \rightarrow Y$.

My Attempt at a proof

$\Rightarrow$

Suppose $f:X\rightarrow Y$ is stable, then choose an $x\in X$ and let the corresponding orbit be $Gx$. Then we have a map from the set of orbits of X to the set of orbits of Y,

$h: X \rightarrow Y$

and

$h(Gx)=f(gx) \forall g\in G=g.f(x) \forall g\in G = h(Gy)$

thus each orbit of X corresponds one to one with an orbit of Y.

$\Leftarrow$

Well we know each orbit of X is mapped to a corresponding orbit of Y. So we take the map $h$ from orbit to orbit and then we show that for each x in that orbit (since orbits are disjoint or equal) we can move the g's in G to outside without affecting the y's.

-
I assume stable means the same thing as $G$-equivariant. (I also assume your question is for feedback on your proofs; you don't explicitly state any question.) In your $\Rightarrow$ direction you seem a bit trigger-happy with your '$\forall g\in G$'s. I can't follow what you're saying in the $\Leftarrow$ direction. When in doubt, use plain English over symbols. –  anon Feb 6 '13 at 6:29
Stable means $f(g.x)=g.f(x)$ for all g. I am asking for help on the proof, yes. The $\Leftarrow$ is underdeveloped. I had a few ideas going at once and wasn't careful with the editing. I'll try to make it more concise. –  rckrd Feb 6 '13 at 6:36