# Topology: Proving a space is connected

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. –  Mariano Suárez-Alvarez Sep 27 '11 at 17:18
@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. –  Ali Baba Sep 27 '11 at 17:23
@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. –  Henno Brandsma Sep 27 '11 at 17:30
@Ali: you should consider giving us more details about $X$ and the map. –  Mariano Suárez-Alvarez Sep 27 '11 at 17:49
@MarianoSuárez-Alvarez It's a one dimensional CW complex. Ummm, not really much more I can say. –  Ali Baba Sep 27 '11 at 17:53

Suppose you have a compact graph $X_0$ and a map $f_0: X_0 \to S^1$ satisfying your hypotheses. Let $v$ be a vertex in $X_0$, and $X = X_0 \cup_v {\mathbb R}^+$, where the ray ${\mathbb R}^+$ is attached by identifying $0$ with $v$, and is given a CW structure by setting its $0$-skeleton to be the naturals $\mathbb N$. Now extend $f_0$ to $f$ by mapping the "whisker" ${\mathbb R}^+$ to $f_0(v)$.
It seems that $X$ and $f:X \to S^1$ also satisfy your hypotheses: if $X_0$ is a graph of genus 2, so is $X$; if $f_0:X_0 \to S^1$ is continuous, surjective and non-injective, so is $f$. Without some further assumptions to rule out this sort of construction, you'll never get compactness for $X$.
But he is not trying to prove compactness of $X$: he is assuming $X$ is compact. –  Mariano Suárez-Alvarez Sep 27 '11 at 22:06