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.

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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.

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Equivariant endomorphisms

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Computation with atoms