A non vanishing function has an nth root If $V$ is a simply connected domain, then will any  continuous function $f:V\to\mathbb{C}$ with no zeros in $V$ have a continuous $n^{th}$ root?
Now, if the function were holomorphic, then we could use the existence of $\frac{f'}{f}$ and the chain rule to get a logarithm of $f$, from which we can get the n-th root. But, will any non-vanishing fucntion do? And what if the domain is not simply connected? Will we have an easy counterexample? Any hints? Thanks beforehand.
 A: Yes, on a simply connected domain every (continuous) nonvanishing function has a logarithm and hence an $n^{th}$ root for every $n$. The point is that if $V$ is simply connected then, intuitively, a continuous function $f : V \to \mathbb{C}^{\times}$ can't "wrap around" $0$; formally, the induced map on $\pi_1$ vanishes which implies that $f$ lifts to the universal cover $\exp : \mathbb{C} \to \mathbb{C}^{\times}$, which says exactly that $f$ has a logarithm.
We can also get this result using the exponential sheaf sequence (which makes sense for continuous or smooth or holomorphic functions), which tells us that the obstruction to a map $f \in H^0(V, \mathbb{C}^{\times})$ lifting to a map $\tilde{f} \in H^0(V, \mathbb{C})$ is its image under the connecting homomorphism
$$\partial f \in H^1(V, 2 \pi i \mathbb{Z}) \cong \text{Hom}(\pi_1(V), \mathbb{Z}) \cong 0$$
which vanishes because $V$ is simply connected; this argument shows moreover that it suffices to assume only that $H^1(V) = 0$ (which is also enough in the previous argument, since this would also imply that the induced map on $\pi_1$ vanishes).
The same argument applied to the Kummer sequence
$$1 \to \mu_n \to \mathbb{C}^{\times} \xrightarrow{z \mapsto z^n} \mathbb{C}^{\times} \to 1$$
(where $\mu_n$ is the group of $n^{th}$ roots of unity) shows a sharper result: that the obstruction to a map $f \in H^0(V, \mathbb{C}^{\times})$ admitting an $n^{th}$ root $\tilde{f} \in H^0(V, \mathbb{C}^{\times})$ is its image under the connecting homomorphism
$$\partial f \in H^1(V, \mu_n) \cong \text{Hom}(\pi_1(V), \mathbb{Z}/n)$$
so for an $n^{th}$ root to exist it suffices that every map $\pi_1(V) \to \mathbb{Z}/n$ be zero (which happens e.g. if $\pi_1(V)$ is torsion and all its torsion has order relatively prime to $n$).
If $V$ isn't simply connected we have the universal counterexample given by $V = \mathbb{C}^{\times}$ and $f : V \to \mathbb{C}^{\times}$ the identity map.
