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The following result is well-known:

Let $U\subset \mathbb{C}$ be simply connected and let $f\colon U \to \mathbb{C}^\ast$ be holomorphic, then there exists $g\colon U \to \mathbb{C}$ holomorphic such that $\operatorname{exp}(g(z))=f(z)$ for all $z\in U$.

I am wondering if (and if so, why) this stays true if we only assume that $f$ is smooth. Feel free to restrict your attention to open balls. This answer suggests that this is true. It would also help, if someone could give the details for why $$ (log(f))(x)=\int_{\gamma} \frac{df}{f} $$ is well-defined for $\gamma$ a differentiable path in a manifold $M$ with vanishing first cohomology group connecting a fixed point $x_0$ and $x$, and why it is smooth with $exp(log(f)(x))=f(x)$ (see this answer).

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user363120
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  • $g(z) = \int_1^z \frac{1}{s} ds$ isn't analytic on $\mathbb{C}^*$ because $\int_{|z|=1}\frac{1}{s} ds \ne 0$. So your last sentence should mean really that $\int_\gamma \frac{df}{f} = 0$ for any closed curve $\gamma$ – reuns Mar 12 '17 at 07:14

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Sure. The point is that $\exp : \mathbb{C} \to \mathbb{C}^*$ is a covering map. Hence, by the lifting theorem, if $U \subset \mathbb{C}$ is simply connected and $f : U \to \mathbb{C}^*$ is continuous, it can be lifted to $\mathbb{C}$, i.e. there exists $g : U \to \mathbb{C}$ (continuous) such that $f(z) = \exp(g(z))$. Since $\exp$ is analytic, this $g$ will be analytic if $f$ is so, and smooth if $f$ is so.

Pedro
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  • Thanks for your answer. So, the existence of a continuous map $g$ with the desired (topological) properties can be proved using topological methods only? That's interesting, the constructions I have seen all used analytical methods. – user363120 Mar 12 '17 at 01:13
  • Probably, yeah. – Pedro Mar 12 '17 at 01:31