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Orienting equal-order non-isomorphic targets

PDF p. 2: Section 3, ordinary classes, equal-order case

Statement

Let H1 and H2 be non-isomorphic finite simple graphs of the same order, each with at least one vertex. Their labels can be chosen so that H2 contains no copy of H1.

The order of a graph is the number of its vertices. A copy of H1 in H2 is a subgraph of H2 isomorphic to H1; it need not be induced. A graph isomorphism is a bijection of vertex sets preserving and reflecting adjacency.

Facts & Assumptions

[F1]

A copy of a finite simple graph J in a finite simple graph L exists exactly when an ordinary embedding J→L exists. An ordinary embedding is an injective vertex map that preserves edges (Ordinary embeddings as injections preserving edges).

[F2]

If two finite simple graphs admit ordinary embeddings in both directions, then they are isomorphic (Rigidity of mutual ordinary embeddings).

[F3]

The order of a finite graph is the cardinality of its vertex set (Order, size, complement and independence number).

Proof

Given: Non-isomorphic finite simple graphs H1,H2 of equal order, each with at least one vertex.

Goal: Choose their labels so that H2 has no copy of H1.

1.1givenF2

Ordinary embeddings cannot exist in both directions between the two graphs, since otherwise the rigidity fact would make them isomorphic, contrary to the hypothesis. Hence at least one direction has no ordinary embedding.

2.1givenF1F3step 1.1∎

Label the source of a direction without an ordinary embedding H1 and its target H2. By the copy-embedding equivalence, H2 contains no copy of H1. Relabelling preserves the equal-order hypothesis.

Source notes

The cited article uses this orientation in the equal-order case. The argument above derives it from mutual-embedding rigidity; in fact that rigidity gives the orientation even without the equal-order hypothesis.

Depends on

Used by · 1

Dependency tree · 2 levels

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Sources