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Suppose two sets $B = \mathbb R$, and $A$.

$F$ is a set of mappings from$A$ to $B$, such that $\forall f_1, f_2 \in F$, there exists a bijection $g: B \to B$ , such that $f_2 = g(f_1)$, equivalently, $f_1 = g^{-1}(f_2)$.

What algebraic structures does $F$ have?

An example is here, where $A$ is a probability space, $B$ is $\mathbb R$ with its Borel sigma algebra, and $F$ is a set of random variables with the above property, and $g$'s are Borel measurable bijections. But I am curious if there is a pure algebraic structure on $F$?

We may consider $A=B$, though this is not true in general.

Thanks!

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Tim
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1 Answers1

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For any $f_0:A\to B$ define $F_{f_0}$ as the set of all functions "compatible with $f_0$", that is to say $$F_{f_0}=\{g(f_0):\ g\in \text{bijections}(B)\}$$

For any set $F$ as in the question, and for any $f_0\in F$, clearly $F\subset F_{f_0}$. So we can look at the algebraic nature of $F_{f_0}$.

The groupg $G=\text{bijections}(B)$ naturally acts transitively on $F_{f_0}$.

Let $H$ be the stabilizer of $f_0$. We have that $H$ is isomorphic to the product of the groups $\text{bijections}(f_0^{-1}(b))$ when $b$ varies in $B$.

$F_{f_0}$ is now identified with the cosets of $G/H$. (so if $H$ is normal then $F_{f_0}$ is a group --- e.g. if $f_{0}$ is injective then $F_{f_0}$ is isomorphic to $G$)

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