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Let $X$ and $Y$ be topological spaces and a surjective map $f:X\rightarrow Y $. Suppose that a group $G$ acts on $X$. and let $\pi:X\rightarrow X/G$ be the quotient map.

1) Under what conditions $f$ factors through $X/G$? I mean there exists $f':X/G\rightarrow Y$ such that $f=f'\circ \pi$

My guess is that if we have $f(gx)=f(x)$ for all $g\in G$ and $x\in X$ then this is the only required condition

2) Now suppose that another group $H$ acts also on $X$ and such that $f(gx)=gx$ for all $g\in G$ and $x\in X$ and $f(hx)=hx$ for all $h\in H$ and $x\in X$, then in this case how to factor $f$ in a way that takes into account the two actions of $G$ and $H$ and construct an injective induced map? for example consider a surjective map $f:S^1\times S^1\rightarrow Z$ where $Z$ is a given topological space. I want to construct an homeo to $Z$ by passing to the quotient: suppose we know that $f(y,x)=f(x,y)$ and $f(\bar x,\bar y)=f(x,y)$ this means that we have two actions:the action of the symmetric group $S_2$ permuting coordinates and the action of $\mathbb Z_2$ generated by "taking the conjugate" action how do we pass to the quotient by $S_2$ and $\mathbb Z_2$ to construct an injective map?

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Your guess in 1 is correct. Let f:X--->Y be a map of sets and suppose that G acts on X. Then f factors through X/G iff f(gx) = f(x) for all g in G and x in X. Concerning 2, I think you have an action of V_4 (the fourgroup of Klein) on S^1 times S^1. That is, you quotient out by the action of G x H. – Luffy Jun 21 '11 at 8:27
Please note that I have removed the "algebraic-topology" and "algebraic-geometry" tags since this question (as stated) has little to do with either subject. – Amitesh Datta Jun 21 '11 at 8:56
What exactly are the roles of the topologies in this question? – user92843 Jun 21 '11 at 20:09

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