Can orbit-finite semigroups be straightened?

A straight set is an equivariant set which is equivariantly isomorphic to a disjiont union of sets of the form $\atoms^{(n)}$ . A straight semigroup is an equivariant semigroup whose universe is a straight set. We raise the following:

Question 1: is every equivariant, orbit-finite semigroup an image of some orbit-finite, straight semigroup under an equivariant mapping?

We show some simple preliminary observations towards the above question.

First, let us see why the answer to the question is not trivially true. Let $S$ be an equivariant semigroup. It is true that there is a straight set $X$ , equivariant mappings $f:X\to S$ and $m:X\times X\to X$ , such that $f(m(x,y))=f(x)\cdot f(y)$ for all $x,y\in X$ . However, it is not clear that the mapping $m$ can be chosen to be associative.

Example. We show an example which is not orbit-finite, but demonstrates some problems with the naive approach. Let $P$ be the set of all finite subsets of $\atoms$ , equipped with the operation $\triangle$ , where $X\triangle Y$ is the symmetric difference of $X$ and $Y$ . $(P,\triangle)$ is an equivariant semigroup, although not an orbit-finite one. There is the obvious surjection $f:\atoms^{(\ast)}\to P$ which maps a tuple of distinct atoms to its support. The set $\atoms^{(\ast)$ is straight (although not orbit-finite). The question is whether there is an associative operation $\cdot$ on it which lifts the operation $\triangle$ , so that $f(\bar x\cdot \bar y)=f(\bar x)\triangle f(\bar y)$ . A natural candidate attempt would be to define $\bar x\cdot \bar y$ as the tuple of those elements which appear an even number of times in the tuple $\bar x\bar y$ , ordered according to their order of appearance in $\bar x\bar y$ . However, this operation is not associative, as verified by the triple $(\atom 1)$ , $(\atom 1)$ , $(\atom 2,\atom 1)$ . Similarly, one can show that there is no associative operation $\cdot$ on $\atoms^{(\ast)}$ such that $f(\bar x\cdot \bar y)=f(\bar x)\triangle f(\bar y)$ .

Example. Any equivariant semigroup $S$ (such as $(P,\triangle)$ above) can be straightened, by a non orbit-finite set, as follows. Choose any straight set $X$ and equivariant surjective mapping $f:X\to S$ . Then $X^+$ with concatenation is a straight semigroup, and the mapping $f':X^+\to S$ given by $f'(a_1\cdots a_n)=f(a_1)\cdots f(a_n)$ is a surjective homomorphism from $X^+$ to $S$ .

Question 2. Does the following property hold. If $S$ is an image of a straight semigroup under an equivariant mapping, then it is an image of a straight semigroup under an equivariant surjection $f:S'\to S$ which preserves least supports, i.e. $\sup x=\sup {f(x)}$ .

In particular, for the semigroup $(P,\triangle)$ , is there an equivariant epimorphism which preserves least supports from a straight semigroup?

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Can orbit-finite semigroups be straightened?

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Computation with atoms