Proof of the Borsuk-Ulam Theorem The Borsuk-Ulam Theorem says the following:
For any continuous map $g: S^n \rightarrow \mathbb{R}^n$ there exists $x \in S^n$ such that $g(x)=g(-x)$.
I'm trying to work through the proof given in Allen Hatchers "Algebraic Topology" but I don't understand the very last step. His proof goes like this:
Let $f(x)=g(x)-g(-x)$ with $g$ as above. So $f(-x)=-f(x)$. We need to show that $f(x)=0$ for some $x$. If this is not the case we can replace $f(x)$ by $\frac{f(x)}{|f(x)|}$ to obtain a new map $f: S^n \rightarrow S^{n-1}$. Now, if we can show that the restriction of this $f$ to the equator $S^{n-1}$ is nullhomotopic, then we're done by previous propositions. But this is exactly the step that I don't understand:
Why is $f|_{S^{n-1}}$ nullhomotopic?
Hatcher simply says that it is nullhomotopic via the restriction of $f$ to one of the hemispheres bounded by $S^{n-1}$. What does he mean by that and why is it true?
 A: It may help to visualize the case $n=2$ first. The equator is the boundary of (say) the south hemisphere of $S^n$, which is a disk $D^n$. Let $h : S^{n-1} \to S^{n-1}$ be the restriction of $f$ to the equator $S^{n-1}$; then there is a continuous extension $D^n \to S^{n-1}$ that agrees with $h$ on the boundary (namely the restriction of $f$ to the hemisphere). It's then a known fact that if there is such an extension, then $h$ is nullhomotopic.
A: He is saying that $f$ is homotopic to the constant map since by looking at one of the hemispheres which has the equatorial sphere as its boundary we can connect $x\in S^{n-1}\subset S^{n}$ by a subarc of a great circle to the north (south) pole. The image of this arc under $f$ (unrestricted) is a path connecting $f(x)$ to $f(N)$ where $N$ denotes the north pole. We can do this with any $x\in S^{n-1}$. By parametrizing the arcs $p_{x}$ by arclength we have that $p_{x}(0)=x$ and $p_{x}(1)=N$ then we can define the homotopy by:
$$h(x,t)=f(p_{x}(t))$$
Then $h(x,0)=f\vert_{S^{n-1}}$ and $h(x,1)=f(N)$. So $f\vert_{S^{n-1}}$ is nullhomotopic.
