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I'm attempting to prove that a space is connected and compact. I have a continuous function $f:X \rightarrow S^{1}$. $X$ is metrizable and locally connected. $f$ is non-constant, surjective and non-injective. Generally the fibers of $f$ are not connected. X is a one-dimensional CW complex, so a graph, which is of genus 2.

What additional properties of $X$ or $f$ are sufficient for such a proof? And how would I go about the proof?

Thanks!

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    Connected fibers should be enough, I think.2011-09-27
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    @MarianoSuárez-Alvarez Yes, I looked into this. Connected fibers + $f$ is a closed map, means that $f$ is a proper map and therefore the preimage of $S_{1}$ is compact? The only problem is that there are very few connected fibers.2011-09-27
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    @AliBaba : $f$ is closed, continuous and all fibres compact would imply that $X$ is compact as the inverse image of a compact set under a perfect map. I suppose $S_1$ is the unit circle, so $\mathbf{S}^1$ ? $f$ quotient (so closed and continuous would do) plus connected fibres means the inverse image of a connected set is connected.2011-09-27
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    @Ali: you should consider giving us more details about $X$ and the map.2011-09-27
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    @MarianoSuárez-Alvarez It's a one dimensional CW complex. Ummm, not really much more I can say.2011-09-27
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    Please do add that information to the question. If that is all you can say about $X$, then I cannot say anything else :)2011-09-27
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    @MarianoSuárez-Alvarez I really don't have any other restrictions on $X$ or $f$. I was wondering what restrictions would imply connectedness? Thanks a lot for the help so far!2011-09-27
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    At that level of generality, the problem is purely combinatorial.2011-09-27
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    @MarianoSuárez-Alvarez If $f$ is a closed map then the image's closedness must be equal to the preimage's closedness? Therefore if the only clopen subsets of $S^1$ are $S^1$ and $\Phi$, then the only clopen preimages must be $X$ and $\Phi$ therefore $X$ is connected?2011-09-27
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    @Ali: there are tons of continuous maps $X\to S^1$ from a disconnected compact graph $X$ of genus $2$ to $S^1$, all of which are closed. I don't know what you are asking in your last comment.2011-09-27
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    @MarianoSuárez-Alvarez Sorry, it was an attempt at a proof. I'll try to elaborate; Let $A \neq X \neq \Phi$. Suggest that $A$ is clopen. If $f$ was a open and closed map, then $f(A)$ would be clopen too. $f^{-1}(S^1)=X$ and $f^{-1}(\Phi)=\Phi$ therefore $A$ cannot be clopen, therefore $X$ is connected.2011-09-27
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    The first projection $X=S^1\times\{0,1\}\to S^1$ (with $\lbrace0,1\rbrace$ a discrete two-element set, so that $X$ is just two disjoint copies of $S^1$) is open and closed.2011-09-27
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    @MarianoSuárez-Alvarez Darn. What's wrong with my proof above then?2011-09-27
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    You seem to have overlooked the fact that $f[A]$ can be equal to $S^1$ even if $A\ne X$.2011-09-27
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    @AliBaba You're proof above doesn't quite make sense - is $\Phi$ supposed to be a subset of $X$ or of $S^1$? Either way, the statement $f^{-1}(\Phi) = \Phi$ doesn't make sense. If you meant something like $f^{-1}(f(\Phi)) = \Phi$, that's incorrect, as shown by the example of two circles mapping to one circle.2011-09-27
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    @MartianInvader: Ali Baba is apparently using $\Phi$ for the empty set.2011-09-27

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