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I come now with a topological group question. Suppose a topological group $G$ acts on a topological space $X$. Suppose $G$ and $X/G$ are connected. Show $X$ is connected. Me and a few friends have been pondering this one for awhile to no avail.

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The map $q:X\to X/G,\ x\mapsto G\cdot x,$ is a quotient map because we equip $X/G$ with the quotient topology.

One can prove that if $q:X\to Y$ is a quotient map and $Y$, as well as all the fibers $q^{-1}(y),\ y\in Y$ are connected, then $X$ is connected, too, see my answer here.

In your case the fibers $q^{-1}(G\cdot x)$ are just the orbits $G\cdot x$. Since $m:G\times X\to X,\ (g,x)\mapsto g\cdot x$ is continuous, we see that $G\cdot x=m(G\times\{x\})$ is connected, being the continuous image of a space homeomorphic to $G$.

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  • $\begingroup$ I have proven the lemma about the pre-images of fibers already. I do not know how to show that the inverse image of a coset is connected, however. $\endgroup$ – Johnny Apple Aug 22 '13 at 2:33
  • $\begingroup$ @AnthonyVasaturo: $q^{-1}(Gx)=Gx$. Can you show that $Gx$ is connected using the fact that $G$ is connected? Use that $G\times X\to X$ is continuous. $\endgroup$ – Stefan Hamcke Aug 22 '13 at 2:34
  • $\begingroup$ I don't exactly see why this map is open nor do I see why G connected means coset Gx is connected. I tried by contradiction, and the only thing I could say was that if A and B formed a separation for Gx, then G was entirely contained in A or B. $\endgroup$ – Johnny Apple Aug 22 '13 at 2:42
  • $\begingroup$ @AnthonyVasaturo: Whoops, sorry, I realize that you do not need openness of that map, and openness is proven using the map is a quotient map, lol. $\endgroup$ – Stefan Hamcke Aug 22 '13 at 2:43
  • $\begingroup$ So the image of G x {x} must be connected since that is isomorphic to G, and X is a union of all such images, all of which intersect at a common point, making X connected? $\endgroup$ – Johnny Apple Aug 22 '13 at 2:48

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