Join of two elements

Let a,b be two elements with atoms. The pair (a,b) satisfies the following universal property: (a,b) supports both a and b, and if y is another element with this property, then y supports (a,b).

We demonstrate the existence of an element denoted a\land b, with a dual property: a\land b is supported both by a and by b, and if y is another element with this property, then y is supported by a\land b.  The construction does not require any assumptions on the symmetry.

Recall that an element x supports an element y if for any permutation \pi of the atoms, \pi(x)=x implies \pi(y)=y. Equivalently, there is an equivariant mapping which maps x to y. If x supports y and y supports x, then x and y are equivariantly equivalent. The element a\land b satisfying the  above universal property is unique, up to equivariant equivalence.

Example. If a,b,c are distinct atoms, then (a,b)\land (b,c) is equivariantly equivalent to b. On the other hand, \set{a,b}\land\set{b,c} is equivariantly equivalent to \emptyset.

We now present the construction of a\land b in general. It is vaguely inspired by the construction of a pushout using coproducts and coequalizers.

Let a and b be two elements. Let A denote the orbit of a and B denote the orbit of b. There is a smallest equivariant relation \sim on the disjoint union A\coprod B such that a\sim b. Let a\land b denote the \sim-equivalence class of the element a.

Claim. a\land b is supported both by a and by b, and if y is another element with this property, then y is supported by (a\land b).

Proof. Since \sim is equivariant, it defines an equivariant mapping f from A\coprod B to P_{fs}(A\coprod B), which assigns to an element its equivalence class. In particular, f(a)=a\land b is supported by a. Similarly, a\land b is supported by b.

Now, let y be any element which is supported both by a and by b. Let Y be the orbit of y. Then there is a unique equivariant mapping f:A\to Y such that f(a)=y, and a unique equivariant mapping g:B\to Y such that g(b)=y. Consider the equivalence relation \approx on A\times B such that u\approx v for u\in A,v\in B iff f(u)=g(v).

Observe that there is an equivariant mapping h from the set of \approx-equivalence classes to Y , and that this mapping maps the \approx-equivalence class of a to y.

This relation is equivariant and since a\approx b, it follows that it is coarser than the relation \sim. In particular, there is an equivariant mapping which maps the \sim-equivalence class of an element u to its \approx-equivalence class.

It follows that there is an equivariant mapping which maps the element a\land b, i.e., the \sim-equivalence class of a, to y.

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