$f:\mathbb{S}^1\rightarrow\mathbb{S}^1$ odd $\Rightarrow$ $\mathrm{deg}(f)$ odd (Borsuk-Ulam theorem)

Solution 1:

To keep more to formulas, consider the following: As you already pointed out, by the lifting property there is a unique continuous function $q: \mathbb R \to \mathbb R$ such that $$f(e^{2\pi i t}) = e^{2\pi i q(t)}$$

Since $f$ is odd we have $$e^{2\pi i q(t+1/2)} = f(e^{2\pi i (t+1/2)}) = f(-e^{2\pi it}) = - f(e^{2\pi it}) = e^{2\pi i (q(t)+1/2)}$$ so

$$q(t+1/2) \equiv q(t) + 1/2 \pmod {\mathbb Z}$$

Therefore the continuous function $t \mapsto q(t+1/2) - q(t) - 1/2$ only takes values in $\mathbb Z$ and thus has to be constant. So there is $n \in \mathbb Z$ such that $n = q(t+1/2) - q(t) - 1/2$ for all $t$.

Now

\begin{align*}\deg(f) &= q(1) - q(0)\\ &= [q(1) - q(1/2) - 1/2] + [q(1/2) - q(0) - 1/2] + 1 \\ &= 2n + 1\end{align*}

and we are done. Maybe you can find out what you've been missing by comparing this to your version of the proof (they are both more or less the same).

Solution 2:

Finally figured it out, using the comment of Miha.

It is enough to prove $\widetilde{f\circ q}(1/2+t)=\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(t)$, since $t:=1/2$ gives the desired formula.

In general, for a given $\alpha:I\rightarrow\mathbb{S}^1$, $\alpha:\partial I\mapsto1$, by the definition and uniqueness of liftings, if $\beta:I\rightarrow\mathbb{R}$, $\beta:0\mapsto0$, and $p\circ\beta=\alpha$, then $\beta=\widetilde{\alpha}$.

In our case, $\alpha:=f\circ q(1/2+t)$ and $\beta:=\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(t)$. We have $$\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(0)=(k+1/2)+0=\widetilde{f\circ q}(1/2+0)$$ and $$p\big(\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(t)\big)=e^{2\pi i ...}=p(\widetilde{f\circ q}(1/2))\cdot p(\widetilde{f\circ q}(t))$$ $$=-1\cdot f\circ q(t)=f(-q(t))=f\circ q(1/2+t),$$ which proves the claim.

Intuitively, when $t$ moves from $0$ to $1/2$, $q(t)$ moves in the upper half of the circle, $f\circ q(t)$ wraps around $\mathbb{S}^1$, and $\widetilde{f\circ q}(t)$ moves up and down on the spiral $\mathbb{R}$. But when $t$ moves from $1/2$ to $1$, $q(t)$ moves in the lower half of the circle, on which $f$ has the same values as on the upper half ($f(-x)=-f(x)$), so $f\circ q(t)$ is the same, and therefore the lift $\widetilde{f\circ q}(t)$ is also the same, just translated in $\mathbb{R}$ by $\widetilde{f\circ q}(1/2)$.