Equivariant endomorphisms

Fact. Suppose that  X is a one-orbit set and

    \[f : X \to X\]

is an equivariant function. Then f is a bijection.

Proof. Let M be the set of  equivariant endomorphisms from  X to X. This is a finite monoid, which contains f. By a standard result on finite monoids, there must be some natural number n such that  f^n is idempotent, i.e.

    \[f^n = f^{2n}\]

Let x be any element in the range of f^n. By idempotency of f^n, this  x is a fixpoint, i.e.

    \[f^n (x) = x.\]

By equivariance of f^n, it follows that 

    \[f^n(\pi(x)) = \pi(x).\]

holds for every atom automorphism \pi, and therefore f^n is the identity on X by the assumption that X has one orbit. Since f^n is the identity, it follows that f has to be a bijection. \Box

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. 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 X, and this requires the finite support assumption.

Leave a Reply

Your email address will not be published.

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>