Solution
= Solution
The <standard simplicial decomposition of a sphere> $F^n$ is the boundary of the <cross-polytope> with vertices $\pm e_1,\ldots,\pm e_{n+1}$, 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 labels
$$
+e_{i_0},-e_{i_1},\ldots,(-1)^je_{i_j},\qquad i_0<\cdots<i_j.
$$
The 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. Thus
$$
p_j\equiv p_{j-1}\pmod2.
$$
Induction 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 $k\geq1$. 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 $\operatorname{sgn}(h_i)e_i$, 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 map
$$
h(x)=\frac{g(x)-g(-x)}{\|g(x)-g(-x)\|}
$$
would 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 is
$$
\boxed{\chi(KG(N,r))=N-2r+2\qquad(N\geq2r\geq2)}.
$$
The <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> $U_0,\ldots,U_d$ of $S^d$ admits a continuous <partition of unity> $\phi_0,\ldots,\phi_d$ subordinate to it. Apply the <Borsuk-Ulam theorem> to $(\phi_1,\ldots,\phi_d)$. Equal values at $x,-x$ and $\sum_i\phi_i=1$ 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 $v_1,\ldots,v_N\in S^d$ such that every $d+1$ of their vectors are <linearly independent>. For example, normalize $(1,t_i,\ldots,t_i^d)$ 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 $K=S^d\setminus\bigcup_{i=1}^dU_i$ 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 $U_0,U_1,\ldots,U_d$ 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.