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I want to prove that if $T$ is a sufficient statistic for $\theta$, then any transformation one-to-one, lets say $\phi=g(T)$ is also sufficient for $\theta$, I suppose is easier to do with the factorization theorem.

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Let $f_{\theta}(x)$ be the probabilistic density function. By the factorization theorem, there exists nonnegative functions $k_{\theta}$ and $h$ such that $$f_{\theta}(x)=h(x)\,k_{\theta}(T(x)).\tag1\label1$$ Since $g$ is one-to-one, you can make it bijective by restricting the codomain of $g$ to the range of $g$. Then \eqref{1} becomes $$f_{\theta}(x)=h(x)\,k_{\theta}(g^{-1}(g(T(x)))).\tag2\label2$$ Rewrite it as $$f_{\theta}(x)=h(x)\,(k_{\theta} \circ g^{-1})(\phi(x)).\tag3\label3$$ Since $k_{\theta}$ is nonnegative, so $k_{\theta} \circ g^{-1}$ is also nonnegative. Invoke the factorization theorem again to draw the desired conclusion.

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