Suppose we have random samples $X = (X_1, ...,X_n), $ $X_i \sim iid f(x|\theta)$. $T = T(X)$ is a sufficient statistic for $\theta$.
By the definition of sufficiency, we know that $f(x|T=t,\theta) = f(x|T=t)$. Does the following statement hold? $$f(w|T=t,\theta) = f(w|T=t),$$ where $w=W(X)$, a statistic of $X$.