Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. :-)

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.