Let $T$ be a sufficient statistic. Suppose $f(T)$ is not a one-to-one function of $T$.
Show $f(T)$ is not a sufficient statistic.
I think this should be proved by contradiction. Since $f$ is not one-to-one, $\exists t_1 \ne t_2 \ni g(t_1)=g(t_2)$. I sough a contradiction in the factorization theorem by writing $f_\theta(x)=h(x)g_\theta(f^{-1}\circ f(T(X)))$. But, I didn't succeed. I tried from the definition of a sufficient statistic, but I don't see a way there too.
I think, the proof can be done without specifying a distribution for the sample. If that's not the case, we may assume, the sample comes from a $\operatorname{Bernouilli}(\theta)$.
Best Answer
The statement as it stands is not true.
If, as you request, we look at the sample $\mathbf{X} =(X_1, X_2,\ldots, X_n)$ representing $n$ i.i.d. $\operatorname{Bernouilli}(\theta)$ random variables taking the values $0$ or $1$, then $\mathbf{X}$ is a sufficient statistic for itself in the sense that $$\Pr(\mathbf{X}=\mathbf{x}\mid \mathbf{X}=\mathbf{x},\theta)=\Pr(\mathbf{X}=\mathbf{x}\mid\mathbf{X}=\mathbf{x})=1$$ but letting $f(\mathbf{x}) = \sum x_i$, then $f(\mathbf{X})$ is also a sufficient statistic for $\mathbf{X}$ as $$\Pr(\mathbf{X}=\mathbf{x}\mid f(\mathbf{X})=f(\mathbf{x}),\theta)=\Pr(\mathbf{X}=\mathbf{x}\mid f(\mathbf{X})=f(\mathbf{x}))=\frac{1}{n \choose f(\mathbf{x})}.$$
Clearly if $f(\mathbf{X})$ is not always either $0$ or $n$, then $f$ is not injective, i.e. is not a one-to-one function.
Perhaps the statement should have been