= Solution
Write $T_a=\Theta(a)$ for the total of $H'$ assigned to a point $a\in H$. The <duad-syntheme duality on six points> is constructed entirely from incidence, as follows.
For a <duad> $d=\{a,b\}$, define $\Theta(d)$ to be the unique <syntheme> common to $T_a$ and $T_b$. This is a bijection between the fifteen <duads> and the fifteen <synthemes> of $H'$, by the last incidence count in part (a).
For a <duad> $d'$ of $H'$, the three <synthemes> containing $d'$ each belong to two totals. Each total contains exactly one of these <synthemes>, since its five matchings cover every <duad> exactly once. Their three pairs of totals therefore partition all six totals. Pulling these pairs back to $H$ gives a <syntheme> $s$. Different $d'$ give different $s$, since two distinct <synthemes> of $H'$ containing $d'$ have intersection exactly that <duad>. There are fifteen of each, so this construction is bijective. Define $\Theta(s)=d'$ by its inverse. Equivalently, the three <synthemes> $\Theta(d)$ for $d\in s$ have common <duad> $\Theta(s)$. In particular,
$$
\boxed{d\in s\quad\Longleftrightarrow\quad \Theta(s)\in\Theta(d).}
$$
For a total $T$ of $H$, map its five <synthemes> to five <duads> of $H'$. Any two of the original <synthemes> are disjoint. Their image <duads> must intersect: if two image <duads> were disjoint, the unique <syntheme> containing both in $H'$ would give a common <duad> in the original two <synthemes> through the pairs-of-totals construction. Conversely intersecting <duads> cannot lie together in a <syntheme> and give disjoint original <synthemes>. Five distinct pairwise-intersecting edges must form the full star at one point. Indeed two edges meeting at a point either force every other edge through that point or leave only the three edges of a triangle, which cannot contain five edges. Define $\Theta(T)$ to be the star's center. Distinct totals give distinct stars; since there are six of each, this is a bijection to the points of $H'$.
It remains to extend to unordered three-versus-three partitions. Start with a partition $B\mid B^c$ of $H'$. Its six cross <synthemes> are the <perfect matchings> between the two triples, parametrized by permutations in $S_3$. Two of these are disjoint exactly when the quotient of their permutations is a three-cycle. Hence the six cross <synthemes> split into two classes of three: within a class any two are disjoint, and between classes any pair shares a <duad>. This unordered division into two classes is independent of the chosen orderings of the triples.
Every total has exactly two cross <synthemes>. To see this, any <syntheme> has either one or three cross <duads>. If a total has $h$ all-cross <synthemes>, it covers $3h+(5-h)$ cross <duads>; the whole complete graph has nine, so $h=2$. Its two cross <synthemes> belong to the same parity class. Conversely any pair in one class extends to a unique total. Thus the six totals split into two triples, the three totals arising from pairs in each parity class. Pulling them back through the original point-total bijection defines a partition $A\mid A^c$ of $H$.
Under the <duad> mapping, its six internal <duads> become exactly the six cross <synthemes> of $B\mid B^c$: a <syntheme> in a parity class belongs to the two totals formed by pairing it with the other two members. These incidences give the three edges of a triangle on each triple of totals. Therefore the partition is characterized by
$$
\boxed{d\text{ is internal to }A\mid A^c\quad\Longleftrightarrow\quad
\Theta(d)\text{ is a cross syntheme for }B\mid B^c.}
$$
This correspondence is injective: the six cross <synthemes> determine all nine cross <duads> of $K_{3,3}$, whose bipartition is unique up to interchange. There are $\binom63/2=10$ partitions on each side, so it is bijective. Define $\Theta(A\mid A^c)=B\mid B^c$ by the inverse of the construction above.
The inverse incidence rule is also useful. If a <duad> $d'$ is internal to $B\mid B^c$, none of the cross <synthemes> contains it, so its inverse <syntheme> has no internal <duad> of $A\mid A^c$ and is entirely cross. If $d'$ is cross, exactly two cross <synthemes> contain it, so the inverse <syntheme> has two internal <duads> and one cross <duad>. Thus internal <duads> and cross <synthemes> exchange roles in both directions. All the extensions are natural: they use intersections and incidence, with no auxiliary ordering left in the answer.
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