I don't understand a step of Borsuk-Ulam theorem, which i tagged with a star below.

$\underline{Borsuk-Ulam}$: If $f:S^2\rightarrow\mathbb R^2$ continuous, then $\exists x$, s.t. $f(x)=f(-x)$

according to the proof(by contradiction):

Assume, there is no such $x$, then define

$g:S^2\rightarrow\mathbb R^2$,$\quad$$g(x)=\frac{f(x)-f(-x)}{||f(x)-f(-x)||}$

$c:[0,1]\rightarrow S^2$,$\quad$$c(s)=(\cos(s), \sin(s),0)$

let $h:=g\circ c$ and $\bar h$ its lift. Then,

$\bar h(s+\frac{1}{2})=\bar h(s)+\frac{m}{2}$, $\quad$$m\in 2\mathbb Z+1$

$\star$$(\bar h(1)=\bar h(\frac{1}{2})+\frac{m}{2}=\bar h(0)+m\overset{\textbf{WHY}}=m)\overset{\textbf{WHY}}\Longrightarrow$ $h$ is not nullhomotopic.

The rest is clear, since $c$ is nullhomotopic, $h$ is also nullhomotopic and we got a contradiction.

  • $\begingroup$ How do you know $\tilde h(s + 1/2) = \tilde h(s) + m/2$ where $m$ is odd? $\endgroup$ – Robert Cardona Nov 30 '14 at 9:39
  • $\begingroup$ @Robert Cardona To be honest I cannot remember what I've done, maybe because of $h(s)=-h(s+1/2)$. If you understand a bit german and read professors writing, it is on page $11$ math.uzh.ch/index.php?file&key1=25869 $\endgroup$ – derivative Nov 30 '14 at 10:42
  • $\begingroup$ Thanks. I read through it, but it didn't explicitly explain the step. Thanks though! $\endgroup$ – Robert Cardona Nov 30 '14 at 17:54

Well, you've left out a bit of the logic here. We assume there is no such $x$ and then construct the odd function $g\colon S^2\to S^1$. It then restricts to an odd function on the equator, and oddness gives the line above the star. The first equality you have tagged with a "WHY" is not necessarily valid, as we do not know that $g(1,0,0)=(1,0)$. But, regardless, basic covering space theory tells us that $h\colon S^1\to S^1$ is nullhomotopic if and only if $\overline h(1) = \overline h(0)$. Indeed, if $\overline h(1) - \overline h(0)=m\in\Bbb Z$, this tells us that $[h] = m\in\pi_1(S^1)\cong\Bbb Z$.

  • $\begingroup$ i think one can select $\bar h(0)$ equal to 0. But i never heard of this basic covering space theory. $\endgroup$ – derivative Dec 23 '13 at 20:33
  • 1
    $\begingroup$ You can only "select" $\overline h(0)$ among the preimages of the point $h(0)$, and that point needn't be $c(0)$. It's probably best for you to read a little bit about covering spaces and homotopy lifting. I can't really give a whole course here. :) What book were you reading this special case of Borsuk-Ulam in? $\endgroup$ – Ted Shifrin Dec 23 '13 at 20:42
  • $\begingroup$ it is the proof of our professor $\endgroup$ – derivative Dec 23 '13 at 20:45
  • 1
    $\begingroup$ If you don't have a text and your professor really didn't teach you this, I do not understand. Munkres's book Topology has a very readable treatment of this. You can also look at Hatcher's Algebraic Topology (much more sophisticated and harder to read), which is available for free download. $\endgroup$ – Ted Shifrin Dec 23 '13 at 21:07
  • 1
    $\begingroup$ Yeah, I concur. Hatcher is quite challenging. Look at the discussion in Munkres. This is the whole point of his proof! $\endgroup$ – Ted Shifrin Dec 23 '13 at 21:17

We need the following theorem. It can be found in Munkres or deduced from Hatcher's proof that the fundamental group of $\pi_1(S^1) \cong \mathbb Z$.

Theorem 1: Lef $p : E \to B$ be a covering map; let $p(e_0) = b_0$. Let $f$ and $g$ be two paths in $B$ from $b_0$ to $b_1$; let $\tilde f$ and $\tilde g$ be their respective liftings to paths in $E$ beginning at $e_0$. If $f$ and $g$ are path homotopic, then $\tilde f$ and $\tilde g$ end at the same point of $E$ and are path homotopic.

Proof: See Munkres Theorem 54.3.

Lemma 2: Let $f$ be a loop in $X$. If $f$ is nullhomotopic, then there exists a nullhomotopy from $f$ which is a path homotopy.

Proof: Let $f : I \to X$ be a loop based at $x_0$. Let $a = h(1) = h(0)$. Suppose $f$ is nullhomotopic, then there exists a homotopy $H : I \times I \to X$ such that $H(s, 0) = f(s)$ and $H(s, 1) = e_c(s) = c$ where $c \in X$.

Define $\alpha_t : I \to X$ by $\alpha_t(s) = H(0, ts)$ and note that this is a path from $\alpha_t(0) = H(0, 0) = f(0) = a$ to $\alpha_t(1) = H(0, t)$.

Define $\beta : I \to X$ by $\beta_t(s) = H(1, t - ts)$ and note that this is a path from $\beta_t(0) = H(1, t)$ to $\beta_t(1) = H(1, 0) = f(1) = a$.

Define $h_t : I \to X$ by $h_t(s) = H(s, t)$ and note that this is a path from $h_t(0) = H(0, t)$ to $h_t(1) = H(1, t)$.

Hence, we've established that the following is well-defined: $\alpha_t * h_t * \beta_t$ is a loop based at $a$ for all $t \in I$. If we define $F : I \times I \to X$ by $F(s, t) = (\alpha_t * h_t * \beta_t)(s)$, then it is a path homotopy between $f$ and the constant map at $a$.

Lemma 3: Let $p : E \to B$ be a covering map. The lifting of a constant map is constant.

Proof: Let $f : I \to B$ be a constant path in $B$ and let $\tilde f$ be it's lifting from $B$ to $E$ via $p$. That is, $\tilde f : I \to E$ such that $p \circ \tilde f = f$.

Since $f$ is constant $\tilde f : I \to \pi^{-1}(c)$ where $f(s) = c$ for all $s \in I$. By property of covering maps, $\pi^{-1}(c)$ is discrete. Since $I$ is connected and $\pi^{-1}(c)$ is discrete, and a continuous map from a connected space to a discrete space must be constant, $\tilde f$ is constant.

Now we look at your $\star$ line:

Observe that $$\tilde h(1) = \tilde h \bigg( \frac12 + \frac12 \bigg) = \tilde h \bigg( \frac12 \bigg) + \frac{q}{2} = \tilde h \bigg(0 + \frac12 \bigg) + \frac{q}{2} = \tilde h(0) + q$$ and $$h(1) = h\bigg( \frac12 + \frac12 \bigg) = -h\bigg( \frac12\bigg) = -h\bigg(0 + \frac12\bigg) = h(0)$$ and so $h$ is a loop at basepoint $h(0) = h(1)$.

Suppose, by way of contradiction, that $h$ is nullhomotopic, then there exists a homotopy $H : I \times I \to S^1$ between $h$ and the constant map at $c$, $e_c$, for some $c \in S^1$.

By Lemma 1 we can suppose, without loss of generality, that $H$ is a path-homotopy in which case $c$ is the basepoint of $h$.

Since $e_c$ is constant, by Lemma 3 it's lifting, starting at $\tilde h(0)$, is constant.

Since $f$ and $e_c$ are path homotopic, by Theorem 1, $\tilde f$ and $\tilde e_c$ have the same end points which means that $\tilde h(0) = \tilde e_c(0) = \tilde e_c(1) = \tilde h(1)$ which implies that $q = 0$, a contradiction to $q$ odd. Conclude that $h$ is not nullhomotopic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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