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 $\operatorname{Bernouilli}(\theta)$.