In his recent post Bartek demonstrated an example of two equivariant sets with atoms that are related by a finitely supported bijection, but not by any equivariant bijection. We show that if the sets are orbit-finite then this situation is impossible.
Proposition. Let and be two orbit-finite equivariant sets with atoms. If there exists a finitely supported bijection then there exists an equivariant bijection .
Let be a finite set of atoms. An -orbit of an element is the set
We introduce some terminology regarding -orbits. An -support of an element contains those elements of the least support of that do not belong to , i.e., . An element which least support has an empty intersection with is called -free.
Recall that the dimension of an element is the size of its least support. We define an -dimension of to be the size of its -support. Each element of a single -orbit has the same -dimension. Therefore, we can also talk about an -dimension of an -orbit.
In equality atoms, every single-orbit set (i.e. a set that consists of one -orbit) is a disjoint union of finitely many -orbits. Exactly one of these -orbits contains all -free elements of .
Example. The single-orbit set of pairs of distinct atoms is a union of three -orbits. The first one consists of -free pairs (it has -dimension 2), the second one consists of pairs with the atom on the first coordinate (it has -dimension 1), and the third one consists of pairs with the atom on the second coordinate (it also has -dimension 1).
Proof. Let be a bijection supported by a finite set of atoms. Note that the function has the following properties:
it maps -orbits of to -orbits of ,
it preserves the -supports of elements (and hence also the -dimension).
Let us first deal with the case when is single-orbit. Take an -free element of . It is not difficult to see that maps to an -free element of of the same -dimension. We define an equivariant function by , where is any permutation of atoms.
The function is well defined. Indeed, suppose that . Since and are disjoint sets of atoms, there exist permutations and which act as and on but are equal on , i.e., . The bijection is supported by , so
Observe that agrees with on the -free arguments, so it is injective on the set of -free elements of . Therefore, is injective on the whole set .
We have shown that is an equivariant bijection between and . Hence, it induces a bijection between the set of -orbits of and the set of -orbits of . But the number of -orbits of is finite and equal to the number of -orbits of , which proves that .
Now suppose that has an arbitrary number of orbits. We will define the equivariant bijection for each orbit of separately, starting from the orbits of the biggest dimension.
Let denote the number of -orbits of with the -dimension . We have that for every . Let be the biggest such that . In both and there are -orbits of -dimension . It is not difficult to see that each of them is -free and belongs to a different orbit of or . Hence, in both and there are orbits of dimension , denoted and , respectively. The function induces a bijective correspondence between these sets. For every pair of corresponding orbits and we define an equivariant bijection similarly as in the single-orbit case above.
We call two sets with atoms -isomorphic if they are related by some -supported bijection. Let and . Both those sets are equivariant. Moreover, and are -isomorphic. This follows from the following characterization: two orbit-finite sets with atoms are -isomorphic if and only if there exists a bijection between the set of -orbits of and the set of -orbits of such that the corresponding -orbits are -isomorphic. Hence, there exists an an -supported bijection and the same construction can be repeated.