Proving an estimator is sufficient?

Question

If theta-hat is sufficient for theta, show that any one to one function of theta-hat is also sufficient for theta

Answer 1

If theta-hat is sufficient, then E(theta-hat) = theta. So then f(theta-hat), being one-to-one, assigns each value of theta-hat to a specific value of theta. Thus, for each theta-hat, there exists a theta such that theta-hat is a sufficient estimator for theta.