$f_{\theta}(x)$, then $T$ is sufficient for $\theta$ if and only if nonnegative functions $g$ and $h$ can be found such that
$f_{\theta} = h(x)g_{\theta}(T(x)) $
The statement is: if $F(t)$ is a one to one function and $T$ is a sufficient statistic, then $F(T)$ is a sufficient statistic.
I can't see this from the factorization criterion, how is $F(T)$ a sufficient statistic?
It's quite clear that if $F$ is one-to-one, then an inverse $F^{-1}$ exists and is also one-to-one, thus if we are given $F(T)$ we can recover $T$ through inversion, and since $T$ is sufficient, $F(T)$ is as well.
For instance, if I know that the sum of the observations in the sample is sufficient for a parameter $\theta$, then I also know that twice the sum is also sufficient. I also know that the sample mean is sufficient. The square of the sum is not necessarily sufficient. It could be if the support of the distribution from which the observations were drawn is nonnegative, because then the transformation is one-to-one on the support; but if the support is, say, $(-\infty, \infty)$, then no, such a transformation is not one-to-one on this interval and therefore will not result in a sufficient statistic.