Understanding Hatcher's proof of Borsuk-Ulam theorem for $n=2$

Question

I am trying to understand the proof of the Borsuk-Ulam theorem for $S^2$ given in Hatcher's "Algebraic Topology" (Th. 1.10), as another person does here, but we are stuck at different places, so I hope this question is not considered a duplicate.

Borsuk-Ulam: If $f:S^2\rightarrow\mathbb R^2$ continuous, then there exists $x\in S^2$, such that $f(x)=f(-x)$.

Arguing by contradiction, assume there is no such $x$, then we can define

$g:S^2\rightarrow S^1$,$\quad$$g(x)=\frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$. Plugging in $-x$, we see that $g(-x)=-g(x)$.

Define a loop circling the equator $\eta:[0,1]\rightarrow S^2,\quad\eta(s)=(\cos(2\pi s), \sin(2\pi s),0)$ and let $h:=g\circ \eta:[0,1]\to S^1$ be the composed loop. Direct check gives:

$h(s+\frac12)=g(\eta(s+\frac12))=g((\cos(2\pi s+\pi),\sin(2\pi s+\pi),0))=g(-\eta(s))=-h(s)$

for all $s\in[0,\frac12]$. Let $\tilde h:[0,1]\to \mathbb R$ be a lifting of $h$.

Then there goes the phrase: 'the equation $h(s+\frac12)=-h(s)$ implies that $\tilde h(s+\frac12)=\tilde h(s)+\frac q2$ for $q$ some odd integer...'

Did anyone try to understand why this cryptic statement may be true?

Answer 1

OK: The universal cover of the circle that Hatcher uses is $$ p: t\mapsto e^{2\pi i t}, p: {\mathbb R}\to S^1\subset {\mathbb C}, $$ where we think of the circle $S^1$ as the unit circle in the complex plane. Now, it is just a direct calculation to show that for any two points $z, -z\in S^1$, any two elements $t_1\in p^{-1}(z), t_2\in p^{-1}(-z)$, differ by a "half-integer", i.e. a number of the form $n + \frac{1}{2}$, where $n$ is an integer.

Answer 2

This is a more detailed explanation of studiosus answer.

Lemma 1: Let $p : \mathbb R \to S^1$ be the standard covering map defined by $p(s) = e^{2\pi i s}$ and let $z \in \mathbb S^1$. If $t \in p^{-1}(z)$ and $s \in p^{-1}(-z)$, then $t - s = q/2$ where $q$ is an odd integer.

Proof: Let $p : \mathbb R \to S^1$ be the standard covering map defined by $p(s) = e^{2\pi i s}$ and let $z \in \mathbb S^1$. Let $t \in p^{-1}(z)$ and $s \in p^{-1}(-z)$. Define $x = t - s$ and observe that $x = t - s$ if and only if $2 \pi i x = 2 \pi i (t - s)$ if and only if $$ e^{2 \pi i x} = e^{2 \pi i (t - s)} = \frac{e^{2 \pi i s}}{e^{2 \pi i t}} = \frac{p(t)}{p(s)} = \frac{z}{-z} = -1.$$

Taking the natural log of both sides we get $$\begin{align} 2 \pi i x = \ln (-1) (\star) \end{align}$$ Recall by Euler's Identity, that $e^{i \pi} = e^{\pi i (2n + 1)} = -1$ for $n \in \mathbb Z$, so $$\begin{align} \ln(-1) = \pi i (2n + 1) (\star \star) \end{align}$$ for some $n \in \mathbb Z$. Putting both Equations $(\star)$ and $(\star \star)$, we get $\pi i (2n + 1) = 2 \pi i x$ for some $n \in \mathbb Z$. Hence $x = \frac{q}{2}$ for some $n \in \mathbb Z$ where $q = 2n + 1$ is odd.

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Let $p : \mathbb R \to S^1$ be the standard covering map defined by $p(t) = e^{2\pi i t}$.

Let $\tilde h$ be the lifting of $h$ from $S^1$ to $\mathbb R$ via $p$, so we get the following $p \circ \tilde h = h$.

Note that we have $\tilde h(s + 1/2) \in p^{-1}\big(h(s + 1/2) \big)$ and $\tilde h(s) \in p^{-1}\big(h(s) \big)$.

But since $h(s + 1/2) = -h(s)$, they are antipodal points, which means Lemma 1 tells us $\tilde h(s + 1/2) = \tilde h(s) + q/2$ where $q$ is an odd integer.