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$$g(x)=\frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$. Plugging in $-x$, we see that $g(-x)=-g(x)$.
Define a loop circling the equator $\eta:[0,1]\rightarrow S^2,\quad\eta(s)=(\cos(2\pi s), \sin(2\pi s),0)$ and let $h:=g\circ \eta:[0,1]\to S^1$ be the composed loop. Direct check gives:
$h(s+\frac12)=g(\eta(s+\frac12))=g((\cos(2\pi s+\pi),\sin(2\pi s+\pi),0))=g(-\eta(s))=-h(s)$
for all $s\in[0,\frac12]$. Let $\tilde h:[0,1]\to \mathbb R$ be a lifting of $h$.
Then there goes the phrase: 'the equation $h(s+\frac12)=-h(s)$ implies that $\tilde h(s+\frac12)=\tilde h(s)+\frac q2$ for $q$ some odd integer…'
Did anyone try to understand why this cryptic statement may be true?
Best Answer
OK: The universal cover of the circle that Hatcher uses is $$ 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.