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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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    @MartianInvader: Ali Baba is apparently using $\Phi$ for the empty set.2011-09-27

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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 $X$.

Of course, I could be missing something essential here, and would welcome any corrections.

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    @MarianoSuárez-Alvarez: I see your point, perhaps connectivity is the real issue. After all, it was in the title (compactness wasn't).2011-09-27