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Suppose $X$ is a normal space (compact, if this would help here), and $f : X \to \mathbf{C}$ is continuous. Set $K = f^{-1}(\{0\})$ and let $E \supset K$ be some open neighborhood of $K$ such that $f$ is bounded away from $0$ on $X \backslash E$.

My question is the following. Is it possible to construct (or prove the mere existence of) a continuous map $g: X \to \mathbf{C}$ such that $g$ vanishes nowhere and it coincides with $f$ on $X \backslash E$? My intuition tells me yes. However, I have difficulties constructing such a $g$.

If $u : X \to [0,1]$ is an Urysohn function which vanishes outside $E$ and equals $1$ on $K$, setting $g := f + u$ will work when $f$ is real-valued and positive on $E$, but otherwise it might brake down. This can be avoided by writing $f(x) = |f(x)| e^{ i \arg f(x) }$, and then setting $g(x) := ( |f(x)| + u(x) ) e^{i \arg f(x) }$, however this is ill-defined whenever $f(x) = 0$, so then the question would be if $x \mapsto e^{i \arg f(x) }$ can be continuously extended from $X \backslash K$ to $X$, in such a way that this extension does not vanish. If we think of this as a map from $\overline{X \backslash K}$ (continuously extended to $\partial K$) into $\mathbf{C}$, then Tietze's extension theorem gives us an extension, but this might vanish on $K$. This seems like a dead end to me.

There are probably some other constructions that I'm overlooking. Or maybe there exists some cunning counter-example?

Any advice is welcome. :-)

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up vote 2 down vote accepted

Here is a counterexample: Let $X = \mathbb{C}$, $f(z) = z$, and $E= \{ z \in \mathbb{C} : |z| <1\}$. If such a $g$ would exist, then it would be a map of the closed unit disk into the punctured plane with $g(z)=z$ for $|z|=1$. Then $h(z) = \frac{g(z)}{|g(z)|}$ would be a continuous retraction of the closed unit disk onto its boundary, and we know that such a map does not exist.

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Indeed, nice example. I guess I was hoping too hard for this to be true, that I became blind for the simple examples. – ZulfiqarIII Nov 16 '12 at 18:13

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