Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 109 / 4 / a / 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, thenwould 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.