Continuous function can be written as an exponential

Question

Suppose a function $f: \mathbb{R} \to \mathbb{C}$ is continuous, never becomes $0$ and $f(0) = 1$. I want to proof that there exists a unique continuous function $g:\mathbb{R} \to \mathbb{C}$ such that $g(0) = 0$ and $$ \forall x \in \mathbb{R}: f(x) = e^{g(x)}.$$ I managed to proof the uniqueness of this $g$, but can't prove its existence. Is it the complex logarithm of $f$ and, if so, why is it continuous?

Answer 1

Since $\exp : \mathbb{C} \to \mathbb{C}^*$ is a covering space, the homotopy lifting property for this covering space implies the desired result. In particular, from the linked page: if $(Z, z)$ is a pointed space which is simply connected, and we have a continuous map $f : Z \to \mathbb{C}^*$ with $f(z) = 1$, then there is a unique lifting $g : Z \to \mathbb{C}$ such that $g(z) = 0$ and $f = \exp \mathop{\circ} g$. The desired result is the special case where $(Z, z) = (\mathbb{R}, 0)$.

(If you are not familiar with covering spaces and path lifting, let me know and I should be able to reproduce the proof of the homotopy lifting property for this particular case.)