Equivariant endomorphisms

Fact. Suppose that is a one-orbit set and is an equivariant function. Then is a bijection.

Proof. Let be the set of  equivariant endomorphisms from to . This is a finite monoid, which contains . By a standard result on finite monoids, there must be some natural number such that is idempotent, i.e. Let be any element in the range of . By idempotency of , this is a fixpoint, i.e. By equivariance of , it follows that holds for every atom automorphism , and therefore is the identity on by the assumption that has one orbit. Since is the identity, it follows that has to be a bijection. Comment. The fact also follows from the representation theorem, but we do not know if the representation theorem always holds.

One thought on “Equivariant endomorphisms”

1. Bartek says:

It should be clarified that although the statement of this result looks like it holds for arbitrary transitive G-sets, it does not. A key step in the proof is that there are only finitely many equivariant endofunctions on , and this requires the finite support assumption.