I have a question about a step in the proof of the Existence of Flattening Stratification I found in Nitsure's paper here: https://arxiv.org/abs/math/0504590 This question is closely related to Local freeness of $\pi_*F(r)$ from flatness of $F$ where the proof of Lemma 3.2 was discussed which we are probably going to apply here.

That's the converse implication of part (ii) (page 22):

If $\phi : T \to S$ is a morphism such that $\mathcal{F}_T$ is flat, then the Hilbert polynomial is locally constant over $T$. Let $T_f$ (this notation is introduced in part (i) of the proof) be the open and closed subscheme of $T$ where the Hilbert polynomial is $f$. Clearly, the set map $|T_f | \to |S|$ factors via $|S_f |$.

Now the step I not understand:

But as the direct images $(\pi_T)_*\mathcal{F}_T(i)$ are locally free of rank $f(i)$ on $T_t$, it follows in fact that the schematic morphism $T_f \to S$ factors via $S_f$ ...

**Question**: Why flatness of $\mathcal{F}_T$ implies that all
twisted direct images $(\pi_T)_*\mathcal{F}_T(i)$ are locally
free?

Seemingly, Nitsure used implicitely here Lemma 3.2 + a certain exercise (p 15):

Lemma + Exercise imply that $\mathcal{F}$ is flat if and only if there exist some integer $N$ such that for all $r ≥ N$ the direct image $\pi_*\mathcal{F}(r)$ is locally free.

But, it says only that for flat $\mathcal{F}_T$ the
twisted direct images $(\pi_T)_*\mathcal{F}_T(r)$ are locally free
**IF** the $r$ twist is big enough! On the other hand in the proof is clamed
that $(\pi_T)_*\mathcal{F}_T(i)$ are locally free for **ALL** $i$'s.
This aspect confuses me. Does anybody know why assuming
$\mathcal{F}_T$ flat, implies that all twisted direct images
$(\pi_T)_*\mathcal{F}_T(i)$ are locally
free independend of $i$?