Showing that for continuous logarithms $g_1, g_2$ of a function on a connected set, the difference $g_1 −g_2$ is a constant

Question

If $S$ is connected, $\ f$ is continuous and has continuous logarithms $g_1$ and $g_2$ on $S$, and continuous arguments $\theta_1$ and $\theta_2$, then $g_1 −g_2$ and $\theta_1-\theta_2$ are constant on $S$

If $f$ has continuous logarithms $g_1,g_2$:

$f(s)=e^{g_1(s)}=e^{g_2(s)}\implies g_1(s)-g_2(s)=2\pi ik(s)\ $, for some $k(s)\in\mathbb Z$

now if $g_1$ and $g_2$ are continuous, so is $k$, hence it must be a constant , but where did we use the fact that, $S$ is connected, do you have an counterexample such that the result is false for a not connected set ?

Answer 1

The correct conclusion is that $$g_1(s) - g_2(s) = 2 \pi i k(s)$$ for some continuous function $k: S \mapsto \Bbb Z$, rather than for some constant we denote $k(s)$. Now, since $k$ is continuous and $\Bbb Z$ is discrete, $k$ is constant on each connected component. In particular, if $S$ is connected, it only has one component as $k$ is a bona fide constant as desired.

This also tells you how to construct a counterexample: Pick a set $S$ with two connected components, say, $S_1, S_2$, and a function $f: S \to \Bbb C$ that admits a continuous logarithm $g$. Then, by construction for any $C \in 2 \pi i \Bbb Z - \{ 0 \}$, the function $$\widetilde{g}(s) := \left\{ \begin{array}{cc} g(s), & s \in S_1 \\ g(s) + C, & s \in S_2 \end{array} \right.$$ is also a continuous logarithm of $g$ but $\widetilde{g} - g$ is not constant: It is $0$ on $S_1$ and $C \neq 0$ on $S_2$.

Answer 2

Notice that $s\to k(s)$ must take integer values only so if it's continous than it's constant which gives you $g_1-g_2=2i\pi k$ for some constant $k$.

Where we used here $S$ connected? we used in order to deduce the implication : $k(s)$ continuous and $k(s)\in \Bbb N$ implies that $k(s)$ is constant, but how? the genralisation of the intermediate value theorem says that if $S$ is connected than $k(S)$ is connected but the only connected subsets of $\Bbb N$ in $\Bbb C$ are those of the form $\{k\}$