On the definition of maximal invariant

Question

On page 184 of Statistical Models by A. C. Davison, he states that

It (an invariant $a(Y)$) is maximal invariant if every other invariant statistic is a function of it, or equivalently, $$a(y) = a(y') \text{ implies that } y' = g(y) \text{ for some } g \in \mathcal{G}. \tag{1}$$

A little more background: here the topic is about group transformation models. $\mathcal{G}$ is the group of actions on the sample space $\mathcal{Y}$. A statistic $B = b(Y)$ is said to be invariant if $b(y) = b(g(y))$ for every $g \in \mathcal{G}$.

I understand that condition $(1)$ implies$^\dagger$ that "every other invariant statistic is a function of it" but have difficulty in proving the other direction. I also checked several other references (e.g., Theorem 2.3 in Group Invariance Applications in Statistics by M. Eaton) and found that only the direction that I understood is mentioned. So my question is, is the statement by Davison really an equivalence (or just half direction is true)? If it is, how to prove "every other invariant statistic is a function of it" implies $(1)$?

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Proof of $\dagger$: Let's first clarify that by its primitive definition, an invariant $s$ is a mapping from $\mathcal{Y}$ to $\mathbb{R}$. If the range of $s$ is denoted by $\mathcal{R}_s \subset \mathbb{R}$, we can define a mapping $\tilde{s}$ from $\mathcal{Y}/\mathcal{G}$ to $\mathbb{R}$ by $\tilde{s}([y]) = s(y)$, where its domain $\mathcal{Y}/\mathcal{G}$ consists of orbits of $\mathcal{Y}$ under $\mathcal{G}$. Because $s$ is invariant, $\tilde{s}$ is well-defined (i.e., the value $\tilde{s}([y])$ does not rely on the choice of $y \in [y]$) and the range of $\tilde{s}$ is exactly $\mathcal{R}_s$.

With the above notations, if $a$ is a maximal invariant, the by definition $\tilde{a}$ is a bijection between $\mathcal{Y}/\mathcal{G}$ and $\mathcal{R}_a$, hence its inversion $\tilde{a}^{-1}: \mathcal{R}_a \to \mathcal{Y}/\mathcal{G}$ is well-defined, and satisfies $\tilde{a}^{-1}(\tilde{a}([y])) = [y]$ for any $[y] \in \mathcal{Y}/\mathcal{G}$ and $\tilde{a}(\tilde{a}^{-1}(v)) = v$ for any $v \in \mathcal{R}_a$. It then follows for any invariant $b$ and $y \in \mathcal{Y}$ that \begin{align*} b(y) = \tilde{b}([y]) = \tilde{b}(\tilde{a}^{-1}(\tilde{a}([y]))) = (\tilde{b} \circ \tilde{a}^{-1})(a(y)) =: f(a(y)), \tag{2}\label{2} \end{align*} where $f: \mathcal{R}_a \to \mathbb{R}$ is a real-valued function defined on $\mathcal{R}_a$. $\eqref{2}$ shows that $b = f \circ a$ is a function of $a$.

Answer 1

Suppose $a(y) = a(y')$ but $y'$ cannot be expressed as a transformation of $y$ for any $g \in \mathcal{G}$. In other words, the equivalence class $[y']$ that $y'$ belongs (aka, the "orbit") and the equivalence class $[y]$ that $y$ belongs are disjoint, which allows us to choose an invariant $b^*$ such that \begin{align} b^*(y) \neq b^*(y'). \end{align} By condition, $b^*$ is a function of the maximal invariant $a$, i.e., $b^* = f(a)$ for some function $f$, but this then leads to contradiction: \begin{align*} b^*(y) = f(a(y)) = f(a(y')) = b^*(y'). \end{align*}

Therefore, $a(y) = a(y')$ must imply that $y' = g(y)$ for some $g \in \mathcal{G}$. This proves the other direction of the equivalence.