Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p:\mathbb{R}\rightarrow\mathbb{S}^1$, $p(t)=e^{2\pi it}$.

enter image description here

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). $\blacksquare$

share|cite|improve this question
When you say "$\widetilde{f \circ q}$ its lifting", what exactly do you know about it? Well, $\widetilde{f \circ q}(x+1) = \widetilde{f \circ q}(x) + \operatorname{deg}(f)$ for all $x$ and $\widetilde{f \circ q}(0) = 0$. Now, can you say what $\widetilde{f \circ q}(-1/2)$ is? – t.b. May 25 '11 at 22:41
$\widetilde{f\circ q}$ is defined on $I$ and $-1/2\notin I$, so I'm a little confused. Additionally, where did you get $\widetilde{f\circ q}(x+1)=\widetilde{f\circ q}(x)+\widetilde{f\circ q}(1)$? – Leon May 25 '11 at 22:52
Okay, let me try to explain it differently I misunderstood you somewhat. Forget what I said before. You haven't really used "oddness" yet: The lifting from $0$ to $1/2$ is "essentially the same" as the lifting from $1/2$ to $1$, just rotated halfways. That is: you start at $k +1/2$ and circle up or down along the spiral exactly the same way as before and that's because of oddness. – t.b. May 25 '11 at 23:11
Yes, that's true. But the first thing you say (oddness) means that you're doing the same thing twice. You're moving along a certain trajectory on the spiral, until you get to the parameter $1/2$ and land at $k + 1/2$ on the spiral. Then you're just making exactly the same movement once again, just rotated halfway, starting from $k + 1/2$ in the picture, until you get to the parameter $1$ and land at $k+1/2+k+1/2$. Therefore $\widetilde{f\circ q}(1)=\widetilde{f\circ q}(1/2)+\widetilde{f\circ q}(1/2)$. I wish I could just show you by pointing in the picture. – t.b. May 25 '11 at 23:32
Here's what Theo is saying in a bit more formulaic way: you can prove (using crucially this particular covering projection) that $\tilde{f}\circ q(1/2+t)$ and $\tilde{f}\circ q(1/2)+\tilde{f}\circ q(t)$ are lifts of the same map with the same starting point, so they must be equal. – Miha Habič May 29 '11 at 21:54

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 $t$.


\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).

share|cite|improve this answer
I don't understand your very first sentence. The map $q$, as I defined it, maps $q:I\rightarrow\mathbb{S}^1$. Moreover, the lifting property in our case states: for $f\circ q:I\rightarrow\mathbb{S}^1$ there exists unique $\widetilde{f\circ q}:I\rightarrow\mathbb{R}$ such that $p\circ\widetilde{f\circ q}=f\circ q$. I don't understand your proof. – Leon May 26 '11 at 12:04
@Leon: My $q$ is not your $q$ (should probalby have called it differently). It is rather the lifting of $f(e^{2\pi it})$. To your second question: When you know that a unique lifting exists for any closed interval $I$, it is not difficult to then also show that a unique lifting exists when we take the whole real line as our domain - think of it as a generalized lifting property. – Sam May 26 '11 at 12:09
up vote 4 down vote accepted

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

share|cite|improve this answer
That's the argument I had in mind but I was too lazy to spell it out :) Sorry that my comments seemed to have confused you more than they have helped. – t.b. May 30 '11 at 5:41
@Theo: No problem, as long as it's solved in the end. – Leon May 30 '11 at 13:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.