Nonuniform guard fibres

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Here $P_7$ means that every subfamily of at most seven edges has a transversal of size at most two.

The safe projection in Prime guard families assumes that omitted guard fibers have small transversal number, or that every seven allowed fibers have three disjoint edges in their common intersection. Neither condition is needed for the following $P_7$-specific projection. Its hypothesis concerns each fiber separately. It does not provide an extraction theorem for arbitrary families, and does not settle Problem 644.

A graph on the seven row indices

Lemma. Let $\mathcal M$ be a nonempty intersecting family of subsets of $[7]$, each of size at least two. There exists a nonbipartite, 2-degenerate graph $G$ on $[7]$ such that every member of $\mathcal M$ contains an edge of $G$.

A graph is 2-degenerate if its vertices can be ordered so that each vertex has at most two earlier neighbors.

Proof. If $\{a,b\}\in\mathcal M$, take the edge $ab$ and join both $a$ and $b$ to all five other vertices. Every member of $\mathcal M$ meets $\{a,b\}$ and has at least two points, so it contains an edge. The graph contains a triangle, and deleting the five degree-two vertices leaves just $ab$.

If every member has at least four points, take the disjoint union $K_3\sqcup K_2\sqcup K_2$. Its independence number is three.

In the remaining case choose $A\in\mathcal M$ of size three and put $B=[7]\setminus A$. There exist $a\in A$ and a pair $P\subseteq B$ for which $\{a\}\cup P\notin\mathcal M$. Otherwise, choosing different $a,b\in A$ and complementary pairs $P,Q\subseteq B$ would give two disjoint members $aP,bQ$.

Make $A$ a triangle. Give the two vertices of $P$ label $a$, and give the other two vertices of $B$ the two different remaining labels in $A$. Join each vertex of $B$ to the two vertices of $A$ different from its label. This graph is nonbipartite and 2-degenerate: delete the four degree-two vertices in $B$, followed by the triangle.

If a member $M\in\mathcal M$ were independent, it would meet $A$ in exactly one point: it must meet $A$ by intersection, and cannot meet it twice because $A$ is a triangle. Every other point of $M$ would have the same label as that point. Since $|M|\ge3$, the only possibility is $M=\{a\}\cup P$, which was excluded. $\square$

Projection of arbitrary guard fibers

Let $U,W$ be disjoint finite sets. Let $\mathcal A$ be a family of nonempty subsets of $U$, and for each $A\in\mathcal A$ let $\mathcal G_A$ be a nonempty family of subsets of $W$, all of size at most $s$. Define

$$\mathcal Q=\{A\cup B:A\in\mathcal A,\ B\in\mathcal G_A\}. \tag{1}$$

Assume the following avoidance condition in every fiber:

$$\text{for each }X\subseteq W\text{ with }|X|\le2s, \quad \text{some }B\in\mathcal G_A\text{ is disjoint from }X. \tag{2}$$

For fibers of nonempty sets, this is exactly $\tau(\mathcal G_A)>2s$. Condition (2) also permits an empty guard edge, without needing a convention for its transversal number.

Theorem. Under (2),

$$\boxed{\mathcal Q\text{ has }P_7 \quad\Longleftrightarrow\quad \mathcal A\text{ has }P_7,\qquad \tau(\mathcal A)\ge\tau(\mathcal Q).} \tag{3}$$

Proof. A two-cover of any core subfamily covers its chosen lifts, and a global core transversal covers $\mathcal Q$. Thus only reflection of $P_7$ needs proof.

Suppose $\mathcal A$ fails $P_7$. Repeat edges, if necessary, to obtain a bad tuple $A_1,\ldots,A_7$. For each point $x$ in their union, let $M_x=\{i:x\notin A_i\}$. These missing sets are pairwise intersecting, since two disjoint ones would give a two-cover. Each has at least two elements: if it had just one, a point of that remaining nonempty row would complete a two-cover; an empty missing set would give a one-cover.

Apply the graph lemma to the $M_x$, and order the rows so that every vertex of $G$ has at most two earlier neighbors. Choose $B_i\in\mathcal G_{A_i}$ successively, avoiding the union of the guards at those earlier neighbors. This union has size at most $2s$, so (2) allows the choice. Thus

$$ij\in E(G)\quad\Longrightarrow\quad B_i\cap B_j=\varnothing.$$

The row spectrum of each guard point is an independent set of $G$.

The lifted tuple $A_i\cup B_i$ has no two-cover. Two core points already fail on the original tuple. A core point $x$ and a guard point fail because $M_x$ contains a graph edge, while the guard spectrum is independent. Two guard points cannot cover all rows, since that would cover $V(G)$ by two independent sets and make $G$ bipartite. Points outside the tuple's union cannot help. Thus $\mathcal Q$ fails $P_7$, proving (3). $\square$

The coefficient two cannot be uniformly replaced by a smaller coefficient for all $s$. At $s=1$, take three disjoint singleton core edges, and let every fiber consist of the singleton guards $\{u\},\{v\}$. Each fiber has transversal number two. The lifted family has a global two-cover, whereas the core fails $P_3$. This observation does not assert asymptotic sharpness as $s\to\infty$.

Limitation

The fibers need not coincide or have any common allowed edge, and their ground supports may overlap arbitrarily. However, the theorem only applies when every fiber satisfies (2). It does not show that a high-transversal family has a useful core--guard partition, or that the fibers failing (2) can be discarded with $o(k)$ loss. Controlling the aggregate cost of such low-transversal fibers is still missing. Thus (3) is a safe projection criterion, not a rank-sensitive normalization or an unrestricted amplifier.