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.
Best Answer
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 $[h] = m\in\pi_1(S^1)\cong\Bbb Z$.