Consider the following claim:
Let $(\mathcal{X},\mathcal{M})$, $(\mathcal{Y},\mathcal{N})$ and $(\mathcal{Z},\mathcal{O})$ be measurable spaces, and let $f \colon \mathcal{X} \to \mathcal{Y}$ and $g \colon \mathcal{Y} \to \mathcal{Z}$ be measurable functions. Then $\sigma(g \circ f) = \sigma(f)$ if and only if $g$ is bijective, where $\sigma(g \circ f) = (g \circ f)^{-1}(\mathcal{O}) = f^{-1}(g^{-1}(\mathcal{O}))$ and $\sigma(f) = f^{-1}(\mathcal{N})$.
This is false: for the “if” part of the statement, just consider $\mathcal{X} = \mathcal{Y} = \mathcal{Z}$, $\mathcal{M} = \mathcal{N}$, $\mathcal{O} = \{ \emptyset, \mathcal{X} \}$ and $f = g = \mathrm{id}_{\mathcal{X}}$.
I was going through Shao’s Mathematical Statistics, and on page 100 it is claimed that if $X$ is a random vector ($\mathbb{R}^{n}-$valued) and $T(X)$ is a statistic (which to my understanding means that $T$ is a measurable function from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$, both endowed with the Borel $\sigma-$algebra), then $\sigma(T(X)) = \sigma(X)$ if and only if $T$ is bijective. What makes this statement true, when the previous was not?
