Continuity of the projection into the complex sphere

Question

Suppose that we are given a continuous function $f\colon [0,1]\to \mathbb{C}$. Let us define

$$h(x) = \frac{f(x)}{|f(x)|} $$

whenever $f(x)\neq 0$. Can we always extend $h$ to a continuous function $h\colon [0,1]\to \mathbb{C}$?

It seems so, however I am not sure how to proceed if $f$ has infinitely many zeros (for example when the zero set of $f$ looks like the Cantor set).

Answer 1

The continuous, complex-valued function $f(x) = x - \frac{1}{2}$ on $[0, 1]$ gives rise to an $h$ having no continuous extension.

Answer 2

No. Let $f(x):=x-\frac{1}{2}$.

Then: $h(x)=1$ for $x>\frac{1}{2}$ and $h(x)=-1$ for $x<\frac{1}{2}$.

No definition of $h(\frac{1}{2})$ makes $h$ continuous.