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? $\endgroup$ Nov 30, 2014 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, 2014 at 10:42
  • $\begingroup$ Thanks. I read through it, but it didn't explicitly explain the step. Thanks though! $\endgroup$ Nov 30, 2014 at 17:54

2 Answers 2


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, 2013 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? $\endgroup$ Dec 23, 2013 at 20:42
  • $\begingroup$ it is the proof of our professor $\endgroup$
    – derivative
    Dec 23, 2013 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. $\endgroup$ Dec 23, 2013 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! $\endgroup$ Dec 23, 2013 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.


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