Some bijections cannot be smoothed

It can happen that two equivariant sets with atoms are related by a finitely supported bijection, but not by an equivariant bijection.

It is easy to see that, for equivariant sets with equality atoms $X$ and $Y$ , a finitely supported function $f:X\to Y$ may exist even if an equivariant function from $X$ to $Y$ does not. For example, let $X$ be a singleton and $Y$ the set of atoms $\mathbb{A}$ ; one can map the unique element to an atom, but one needs to choose the atom first.

It is less clear what happens if the finitely supported function $f$ is a bijection: does an equivariant bijection exist as well?

The answer is negative. For example, consider

$X = \mathbb{A}\times \mathbb{N} + \{*\}$
$Y = \mathbb{A}\times\mathbb{N}$ ,

choose an atom $a\in\mathbb{A}$ and define a function $f:X\to Y$ by:

$f(*) = (a,0)$
$f(a,n) = (a,n+1)$
$f(b,n) = (b,n)$ for $b\neq a$ .

It is easy to see that $f$ is a bijection supported by $\{a\}$ . However, no equivariant bijection (or even an equivariant function) from $X$ to $Y$ exists, since $X$ contains an element with empty support and $Y$ does not.

Interestingly though, if $X$ and $Y$ are orbit-finite, then finitely supported bijections can be smoothed into equivariant ones. This was proved by Joanna, who in the near future will have written a post about it.

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Some bijections cannot be smoothed

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Computation with atoms