$W_2(f,0) = \frac{1}{2} \# f^{-1}(l)$?

Question

There must be some rather straight forward reason for $W_2(f,0) = \frac{1}{2} \# f^{-1}(l)$ but I really get stuck with why. Could someone help me out?

Answer 1

This follows from the proof of the Borsuk-Ulam theorem preceding this theorem. (Starting on page 91 of Guillemin and Pollack's Differential Topology)

Guillemin and Pollack show that $$W_2(f,0) \equiv \tfrac{1}{2}\# f^{-1}(l) \pmod 2$$ in their proof of the Borsuk-Ulam theorem. But the conclusion of the Borsuk-Ulam theorem says that for a function $$f: S^k \longrightarrow \Bbb R^{k+1} \setminus \{0\}$$ such that $$f(-x) = -f(x) \text{ for all } x \in S^k,$$ we have that $$W_2(f,0) = 1.$$ Hence $f^{-1}(l)$ cannot be empty for any line $l$ through the origin of $\Bbb R^{k+1}$.