In one textbook I found an exercise about the characterization of all logarithmic elements (i.e. such $a$ that can be written as $a=\exp(b)$ for some $b$) and all invertible elements in the $C^\ast$-algebra $C(S^1)$ of complex-valued continuous functions on a circle.
There were suggested the following characterizations
a) $f$ is logarithmic in $C(S^1)$ $\Leftrightarrow$ $\deg(f)=0$ and $f(z)\neq 0$ for all $z\in S^1$.
b) $f$ is invertible in $C(S^1)^{\times}$ $\Leftrightarrow$ there exists a logarithmic element $g$ in $C(S^1)$ and $f(z)=z^{\deg(f)}g(z)$,
where $\deg(f)$ is the Brouwer degree of the map $\frac{f}{|f|}:S^1\rightarrow S^1 \subset \mathbb{C}$.
What is the simplest method to show that ? I tried to start with (b) and then (a) is just a special case, but probably (a) will be necessary as a lemma to prove (b). I do not see how to start without using too sophisticated machinery (e.g. Čech cohomology), therefore I am looking for some elementary proof.
This is immediate from basic covering space theory. The map $\exp:\mathbb{C}\to\mathbb{C}\setminus\{0\}$ is a covering map. Any loop $f:S^1\to\mathbb{C}\setminus\{0\}$ lifts to a path $\tilde{f}:[0,1]\to\mathbb{C}$ (unique once you choose $\tilde{f}(0)$), and this lift is a loop iff $[f]\in\pi_1(\mathbb{C}\setminus\{0\})$ is in the image of the induced map $\exp_*:\pi_1(\mathbb{C})\to\pi_1(\mathbb{C}\setminus\{0\})$. Since $\pi_1(\mathbb{C})$ is trivial, this just means $[f]$ must be trivial, i.e. $\deg(f)=0$. This proves (a).
For (b), note that if you define $g(z)=f(z)z^{-\deg(f)}$, then $\deg(g)=0$. It then follows from (a) that $g$ has a logarithm.