How to show that any continuous map $f:\mathbb{C}^*\to \mathbb{C}^*$ is of the form $f(z)=z^me^{g(z)}?$

Question

Given a nowhere vanishing continuous map $f:\mathbb{C}^*\to \mathbb{C}^*$, how do I show that there exist an integer $m\in \mathbb{Z}$ and a continuous map $g:\mathbb{C}^*\to \mathbb{C}$ such that $f$ is of the form $$f(z)=z^me^{g(z)}, ~~~~~ \forall z\in \mathbb{C}^*.$$ Where $\mathbb{C}^*= \mathbb{C}\setminus\{0\}.$

I come across this problem while trying to show that the cokernel of the exponential map in the following sequence $$0\to \mathbb{Z}\stackrel{2\pi i}{\rightarrow} \mathcal{O}(\mathbb{C}^*)\stackrel{exp} {\rightarrow}\mathcal{O}^*(\mathbb{C^*})\to 1$$ is isomorphic to $\mathbb{Z}=H^1(\mathbb{C}^*, \mathbb{Z}).$ Where $\mathcal{O}(\mathbb{C^*})$ denotes the group of all continuous maps from $\mathbb{C}^*$ to $\mathbb{C}$ and $\mathcal{O}^*(\mathbb{C^*})$ denotes the group of all continuous maps from $\mathbb{C}^*$ to $\mathbb{C}^*$.

I have proved that for all integers $m\in \mathbb{Z}$, the function sending $z\mapsto z^m$ belongs to distinct classes of the quotient group $\frac{\mathcal{O}^*(\mathbb{C^*})}{exp\left(\mathcal{O}(\mathbb{C^*})\right)}$ and I want to show that these are the only classes in the quotient. For that I have to solve the above problem.

Is there an elementary proof for this problem where I dont have to use cohomology of the punctured complex plane. Thanks in advance.

Answer 1

We hide everything fancy in

Lemma. Let $U\subseteq \Bbb C$ be simply connected and $h\colon U\to\Bbb C^\times$ continuous. Then there exists continuous $\ell_h\colon U\to \Bbb C$ such that $h=\exp\circ \ell_h$. Moreover, such $\ell_h$ is determined up to an additive constant $\in2\pi i\Bbb Z$.

Proof. Easy for sufficiently small open sets. These can be glued together. $\square$

With $U_1:=\Bbb C\setminus (-\infty,0]$, $U_2:=-U_1=\Bbb C\setminus [0,+\infty)$, $f_i:=f|_{U_i}$ we thus find continuous functions $\ell_{f_j}\colon U_j\to \Bbb C$ such that $f(z)=e^{\ell_{f_j}(z)}$ for $z\in U_j$. Moreover, $\ell_{f_1}-\ell_{f_2}$ is a locally constant multiple of $2\pi i$, i.e., there exist $m_+,m_-\in\Bbb Z$ such that such that $\ell_{f_1}(z)-\ell_{f_2}(z)=2\pi i m_+$ for positive and $\ell_{f_1}(z)-\ell_{f_2}(z)=2\pi im_-$ for negative imaginary part of $z$. In fact, we may assume wlog that $\ell_{f_1}(z)=\ell_{f_2}(z)$ on the upper half plane and $\ell_{f_2}(z)=\ell_{f_1}(z)+2\pi i m$ on the lower half plane (with $m=m_+-m_-$).

Using the identity instead of $f$ on $U_1$, we likewise find $\ell_{\operatorname{id}_j}\colon U_j\to\Bbb C$ (in other words, branches of $\ln$) such that $z=e^{\ell_{\operatorname{id}_j}(z)}$ for $z\in U_j$. Again, we may assume $\ell_{\operatorname{id}_1}(z)=\ell_{\operatorname{id}_2}(z)$ on the upper half plane and then can verify that $\ell_{\operatorname{id}_2}(z)=\ell_{\operatorname{id}_1}(z)+2\pi i$ on the lower half plane.

It follows that $$ g(z):=\begin{cases}\ell_{f_1}(z)-m\ell_{\operatorname{id}_1}(z)&z\in U_1\\\ell_{f_2}(z)-m\ell_{\operatorname{id}_2}(z)&z\in U_2\end{cases}$$is a well-defined continuos map $\Bbb C^\times\to\Bbb C$ with $$ e^{g(z)}=\frac{f(z)}{z^m}$$ for all $z\in\Bbb C^\times$.