The background is as follows:
Let $S$ be a locally noetherian scheme, $E$ a vector bundle over $S$ of rank $N+1$. Let $X=\mathbb{P}(E)$ be the projective bundle and $\pi:X\longrightarrow S$ be the projection. Let $T$ be a divisor on $S$ and $Y$ be a closed subscheme of $X_T=X\times_S T$ and $\pi_{T}|_Y:Y\longrightarrow T$ (which is the restiction of $\pi_T:X_T\longrightarrow T$)induces a $\mathbb{P}^n$ bundle on $T$ such that $(\pi_T|_Y)^{-1}(t)$ is a linear subspace of $\pi^{-1}(t)$ for all $t\in T$.
a) Let $I_Y$ be the ideal defining $Y$, and let $E'=\pi_*(I_Y\otimes\mathcal{O}_X(1))$. Then $E'$ is a locally free $\mathcal{O}_S$ module. And $\mathbb{P}(E')$ is called the elementary transformation of $X$ with center $Y$
b) $(\pi_T|_Y)_*(\mathcal{O}_Y\otimes\mathcal{O}_X(1))$ is a locally free $\mathcal{O}_T$ module of rank $n+1$. So if $F$ is any quotient bundle of $E_T=E\otimes\mathcal{O}_T$ of rank $n+1$, then $ker\phi=E'$ where $\phi:E\longrightarrow E_T\longrightarrow F$.
The theorem is that (a) implies (b)
The following are my doubts
1) Why is $(\pi_T|_Y)_*(\mathcal{O}_Y\otimes\mathcal{O}_X(1))$ locally free?
2) How can we see that $E'=\pi_*(I_Y\otimes\mathcal{O}_X(1))$ from (a) is the kernel of $\phi$?
3) Here $F$ is arbitrary, why is the kernel not changing then?
4) What role is $Y$ playing part (b). The statement does not seem to talk about $Y$ at all.
This has been puzzling me for a while. Any help will be very useful. Thanks in advance!