Degree of an odd map If $\gamma : S^1 \rightarrow S^1$, $\gamma(t) = e^{2 \pi i t}$ and $\gamma_+$ denotes $\gamma$ restricted to $[0, 0.5]$ i.e. the upper half of $S^1$ and $f:S^1 \rightarrow S^1$ is an odd map, how do I see that the angle swept by $f \circ \gamma_+$ is $\pi + 2 \pi$? 
Thanks for your help!
 A: I consider the map $z \mapsto - z$ from $S^1$ to itself ($S^1$ is hereafter identified to the unit circle of $\mathbb C$). An odd map is a map $\gamma : S^1 \to S^1$ obeying the law $\gamma(- z) = -\gamma(z)$. The restriction $\gamma_+$ of $\gamma$ to the top half-circle is a map $C_+ \to S^1$. Its endpoints are $\gamma(1)$ and $\gamma(-1) = -\gamma(1)$. So this path goes from a point to its antipodal. As such, it makes an odd number of half-turns and it sweeps an angle of $\pi$ modulo $2\pi$.
As any map $C_+ \to S^1$ sending $1$ and $-1$ to antipodal points can be extended to an odd map $S^1 \to S^1$, this result cannot be sharpened: for any angle $\theta$ equal to an odd number of half-turns, there's an odd map $\gamma$ whose $\gamma_+$ sweeps an angle equal to $\theta$.
Please note that in the other canonical model of $S^1$ (the quotient $\mathbb R / \mathbb Z$) you have to replace $z \mapsto -z$ by the transformation $t \mapsto t + .5$ to define an odd map. The transformation $t \mapsto -t$ of $\mathbb R/\mathbb Z$ corresponds to $z \mapsto \overline z$, and it is not true that if $\gamma : S^1 \to S^1$ obeys $\gamma(\overline z) = \overline{\gamma(z)}$, its restriction $\gamma_+$ sweeps an odd number of half-turns (the constant map $\gamma(z) = 1$ is a counterexample).
