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

Question

I'm having trouble understanding the proof of Borsuk-Ulam theorem ($n=2$) that we did in our class. The only problematic part is the last sentence in the proof of lemma 1.

$\mathbb{S}^1\subseteq\mathbb{C}$. We know that $\mathrm{deg}:\pi_1(\mathbb{S}^1)\rightarrow \mathbb{Z}$, $\mathrm{deg}([\alpha]):=\tilde{\alpha}(1)$, is an isomorphism, where $\tilde{\alpha}:I\rightarrow\mathbb{R}$ is the unique lifting of the loop $\alpha:I\rightarrow\mathbb{S}^1$, i.e. $p\circ\tilde{\alpha}=\alpha$, where $p:\mathbb{R}\rightarrow\mathbb{S}^1$, $p(t)=e^{2\pi it}$.

Lemma 1: continuous $f:\mathbb{S}^1\rightarrow\mathbb{S}^1$ odd (i.e. $f(-x)=-f(x)$) $\Longrightarrow$ $\mathrm{deg}(f)\in\mathbb{Z}$ odd.

Proof: Without loss of generality: $f(1)=1$ (if not, then we compose $f$ with a rotation; the degree and odd-ness is preserved). If $q:I\rightarrow\mathbb{S}^1$, $q(t):=e^{2\pi i t}$, then $q(t+1/2)=-q(t)$. Now $f\circ q:I\rightarrow\mathbb{S}^1$ is a loop at $1$, $\widetilde{f\circ q}$ its lifting, and $\mathrm{deg}(f)=\widetilde{f\circ q}(1)$. Clearly $p\circ\widetilde{f\circ q}(1/2)=f\circ q(1/2)=-1$, so by the definition of $p$, we have $\widetilde{f\circ q}(1/2)=k+1/2$ for some $k\in\mathbb{Z}$. (So far, I understand everything; here is what troubles me.) Since $f$ is odd, we have $\widetilde{f\circ q}(1)=k+1/2+k+1/2=2k+1$. WHY?

Lemma 2: continuous $f:\mathbb{S}^2\rightarrow\mathbb{S^1}$ is not odd.

Proof: $\mathbb{S^1}\overset{i}{\hookrightarrow}\mathbb{S}^2\overset{f}{\rightarrow}\mathbb{S^1}$. If $f$ is odd, then so is $f\circ i$. By Lemma 1, $f\circ i$ has odd degree, so it isn't nullhomotopic. But since $i$ is not surjective, it is nullhomotopic, hence so is $f\circ i$, $\rightarrow\leftarrow$. $\blacksquare$

Theorem (Borsuk-Ulam, $n=2$): $\forall$ continuous $f:\mathbb{S}^2\rightarrow\mathbb{R}^2$ $\exists x\in\mathbb{S}^2$: $f(-x)=f(x)$.

Proof: If the theorem were not true, then $F(x):=\frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$, would be a well defined continuous odd map $\mathbb{S}^2\rightarrow\mathbb{S}^1$, $\rightarrow\leftarrow$ (Lemma 2). $\blacksquare$

Answer 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).

Answer 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 $\widetilde{f\circ q}(1/2)$.