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

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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, $f\circ i$ isn't surjective, so it is nullhomotopic, $\rightarrow\leftarrow$. $\blacksquare$

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). $\blacksquare$

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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 Lampret 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
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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
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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

2 Answers 2

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

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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 Lampret May 30 '11 at 13:22

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).

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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 Lampret 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

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