= Solution
Assume (i). An antipodal map $g:S^n\to S^{n-1}$, followed by the inclusion $S^{n-1}\subset\mathbb R^n$, would be an antipodal map with no zero. Hence (i) implies (ii). Conversely, if an antipodal $f:S^n\to\mathbb R^n$ had no zero, then
$$
x\longmapsto\frac{f(x)}{|f(x)|}
$$
would be an antipodal map to $S^{n-1}$. Thus (ii) implies (i).
If $F:B^n\to S^{n-1}$ is antipodal on the boundary, regard $S^n$ as two copies of $B^n$ glued along their boundary. Use $F(x)$ on the upper copy and $-F(-x)$ on the lower copy. The boundary condition makes these definitions agree on the seam, and the resulting map $S^n\to S^{n-1}$ is antipodal. Thus (ii) implies (iii).
Conversely, an antipodal map $g:S^n\to S^{n-1}$ restricted to a closed hemisphere, identified with $B^n$, is antipodal on its equatorial boundary. Hence (iii) implies (ii). The three assertions are equivalent; they are standard forms of the <Borsuk-Ulam theorem>.
Back to article page
