Let $f,g : \mathbb{S}^1 \to \mathbb{S}^1$ be two continuous maps where $\mathbb{S}^1$ is the unit circle. Prove that $\deg (f \circ g) = \deg f \deg g$.
I don't know at all how to do it : I first set a continuous function $\overline{f \circ g} : \mathbb{R} \to \mathbb{R}$ such that $f \circ g (e^{it})=e^{i \overline{f \circ g} (t)}$ for all $t \in [0,2\pi]$.
I've done the same for $f$ and $g$. Then $2 \pi \deg (f \circ g) = \overline{f \circ g} (2\pi)- \overline{f \circ g} (0)$ but I don't see any link with $\overline{f}$ nor $\overline{g}$.
I think all the thing is to prove that $\overline{f \circ g} = \bar f \circ \bar g $
Any help, or link ?
Thanks.