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The standard simplicial decomposition of a sphere $F^n$ is the boundary of the cross-polytope with vertices $\pm e_1,\dots ,\pm e_{n+1}$, identified radially with $S^n$. A face chooses distinct coordinate indices and one sign for each; it never contains an antipodal pair.

A regular antipodal triangulation of a sphere is invariant under the antipodal map and has the nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ as subcomplexes. In particular each equator separates its sphere into two triangulated hemispheres. These conditions are preserved by barycentric subdivision.

For a simplicial map $f:F\to F^n$, a positive alternating simplex of dimension $j$ has distinct image labelsThe opposite alternating pattern is negative; all other simplices, including collapsed images, are neutral.

To prove the antipodal alternating-simplex parity lemma, let $p_j$ count positive $j$-simplices in the equatorial $S^j$. There is exactly one positive vertex in $S^0$, so $p_0=1$. A $j$-simplex has an odd number of positive alternating facets precisely when it is positive or negative alternating. Indeed an alternating sequence has exactly one positive alternating deletion; a nonalternating sequence has either zero or two. For a collapsed image, a positive facet can occur only when exactly one label is repeated; deleting either copy gives two such facets.

Count incidences of positive $(j-1)$-faces with $j$-simplices in one closed hemisphere. Interior faces contribute twice and equatorial faces once. The count modulo two is therefore $p_{j-1}$. It is also the number of positive or negative $j$-simplices in that hemisphere modulo two. Because $f$ is antipodal, each alternating pair $\sigma ,-\sigma$ has one positive and one negative member, with one member in each hemisphere. ThusInduction gives odd $p_k$, proving the existence of a positive $k$-simplex. In particular, an antipodal simplicial map $F\to F^{k-1}$ is impossible: a positive $k$-simplex would require $k+1$ distinct coordinate labels but only $k$ are available.

The Borsuk-Ulam theorem states that every continuous map $g:S^k\to {\mathbb{R}}^k$ has some $x$ with $g(x)=g(-x)$. First rule out a continuous antipodal map $h:S^k\to S^{k-1}$ for $k\ge 1$. By uniform continuity, take a sufficiently fine regular barycentric subdivision so that endpoints of each edge have images less than $2/\sqrt{k}$ apart. Label a vertex by $\mathrm{sgn}(h_i)e_i$, where $i$ is the smallest index maximizing $|h_i|$. Since $h$ is unit length, this maximum is at least $1/\sqrt{k}$; two adjacent vertices cannot have opposite labels. The labels therefore define an antipodal simplicial map to $F^{k-1}$, contradicting the preceding result.

If the stated Borsuk-Ulam theorem failed, the mapwould be just such a continuous antipodal map. This proves the required version; $k=0$ is immediate.

For the Kneser conjecture, equivalently Lovász theorem on Kneser graphs, the claim isThe Kneser graph joins disjoint $r$-subsets of $[N]$. Let $d=N-2r+1$. Colour each $r$-set by its least element in $[d]$ when it has one, and otherwise by one final colour. A colour class indexed by $i$ is an intersecting family because all its members contain $i$; the final class lies in a $(2r-1)$-set and is also intersecting. This proves the upper bound $d+1$.

For the lower bound, first deduce the open-cover Lusternik-Schnirelmann-Borsuk theorem. An open cover $U_0,\dots ,U_d$ of $S^d$ admits a continuous partition of unity ${\varphi }_0,\dots ,{\varphi }_d$ subordinate to it. Apply the Borsuk-Ulam theorem to $({\varphi }_1,\dots ,{\varphi }_d)$. Equal values at $x,-x$ and ${\sum }_i{\varphi }_i=1$ make all values equal. At least one common value is positive, so some $U_i$ contains an antipodal pair.

Now suppose the $r$-sets could be coloured with only $d$ colours, each an intersecting family. Choose $N$ points $v_1,\dots ,v_N\in S^d$ such that every $d+1$ of their vectors are linearly independent. For example, normalize $(1,t_i,\dots ,t_i^d)$ for distinct real $t_i$; the Vandermonde determinant proves this property. The general-position hemisphere count says that at most $d$ points lie on any equator. Since $N-d=2r-1$, at least one of the opposite open hemispheres determined by $x,-x$ contains an $r$-set of the labelled points.

Let $U_i$ be the open set of directions whose positive open hemisphere contains an $r$-set of colour $i$. It contains no antipodal pair, as such a pair would give two disjoint sets of that colour. The compact set $K=S^d\setminus {\bigcup }_{i=1}^d{U}_i$ also contains no antipodal pair: otherwise each of the two opposite hemispheres would have at most $r-1$ points, contradicting the count $2r-1$.

If $K$ is nonempty, compactness gives a positive set distance between $K$ and $-K$; an open neighbourhood $U_0$ of radius less than half that set distance still has no antipodal pair. If $K$ is empty, take $U_0=\varnothing$. In either case $U_0,U_1,\dots ,U_d$ is an open cover with no member containing an antipodal pair, contradicting the Lusternik-Schnirelmann-Borsuk theorem. Hence at least $d+1$ colours are required.

Triangulate the sphere by the boundary complex of the cross-polytope with simplicial vertices $\pm e_0,\dots ,\pm e_n$: a face contains at most one simplicial vertex from each antipodal pair. Take its barycentric subdivision. Its simplicial vertices are the nonempty faces $F$ of the original complex, and its simplices are strictly nested chains $F_0\subset \cdots \subset F_k$.

Form a quotient complex whose simplicial vertices are pairs $\{F,-F\}$, and whose simplices are the images of these chains. No chain contains both $F$ and $-F$. Moreover a chain of face-orbits lifts uniquely up to simultaneous negation: after choosing a largest face, every smaller face has at most one representative contained in it, because an original face contains no antipodal simplicial vertices. Thus the quotient is an honest simplicial complex, without two different simplices with the same simplicial vertex set. Its realization is homeomorphic to the sphere modulo its antipodal action. This provides a triangulation of Real projective space ${\mathbb{RP}}^n$. The subdivision avoids the duplicate-simplex problem of naively identifying simplicial vertices in the unsubdivided cross-polytope boundary.

The boundary simplicial complex of the cross-polytope has vertices $\pm e_i$, and its faces choose at most one vertex from each opposite pair. Radial projection identifies its geometric realization with the round sphere $S^n$. It is a regular antipodal triangulation of a sphere and is useful as a target for signed vertex labels: a vertex map is a simplicial map precisely when no edge receives opposite labels.