Exact sequence on $\mathbb{P}^3$ obtained from Euler sequence In the article of title: On the singular scheme of codimension one holomorphic foliations in $\mathbb{P}^3$ the author states that the sequence on $\mathbb{P}^3$ $$0\longrightarrow \mathcal{F}\oplus\mathscr{O}_{\mathbb{P}^3}\longrightarrow\mathscr{O}_{\mathbb{P}^3}^{\oplus4}(1)\longrightarrow\mathcal{I}_Z(d+2)\longrightarrow0 $$is exact,  where $\mathcal{F}$ is a reflexive sheaf of rank two. 
The author states that the sequence above is obtained of the two exact sequences below: $$0\longrightarrow\mathscr{O}_{\mathbb{P}^3}\longrightarrow\mathscr{O}_{\mathbb{P}^3}^{\oplus4}(1)\longrightarrow\mathcal{T}_{\mathbb{P}^3}\longrightarrow0$$  and $$0\longrightarrow\mathcal{F}\longrightarrow\mathcal{T}_{\mathbb{P}^3}\longrightarrow\mathcal{I}_Z(d+2)\longrightarrow0$$where $\mathcal{I}_{Z}$ is the ideal sheaf of the singular scheme $Z$ of foliation.   
How to get the first exact sequence from the last two? 
Suggestions will be welcome. 
Thanks in advance. 
 A: First note that the sheaf $\mathcal{F}$ is the kernel of a map $\omega \colon \mathcal{T}_{\mathbb{P}^3}\rightarrow \mathscr{O}_{\mathbb{P}^3}(d+2)$ defined by a degree $d+1$ homogeneous $1$-form $\omega = F_0dx_0 +F_1dx_1 +F_2dx_2+F_3dx_3$, such that $\sum x_iF_i = 0$. Clearly the image is the ideal sheaf of the singular scheme twisted by $\mathscr{O}_{\mathbb{P}^3}(d+2)$.
From the Euler sequence we see that every (local) vector field is given by the class four (local) sections $(p_0,p_1,p_2,p_3)$ of $\mathscr{O}_{\mathbb{P}^3}(1)$  modulo the relation $$(p_0,p_1,p_2,p_3)\sim (q_0,q_1,q_2,q_3) \Leftrightarrow (p_0-q_0,p_1-q_1,p_2-q_2,p_3-q_3) = \lambda\cdot (x_0,x_1,x_2,x_3)$$ for some (local) holomorphic function $\lambda$.  
Then lift the map given by $\omega$ to the representatives of such vector fields. This gives $\omega \colon \mathscr{O}_{\mathbb{P}^3}(1)^{\oplus4}\rightarrow \mathscr{O}_{\mathbb{P}^3}(d+2)$ which is just $(p_0,p_1,p_2,p_3)\mapsto F_0p_0 +F_1p_1+F_2p_2+F_3p_3$ . Denote the kernel of this map by $\mathcal{F}_0$. Then
$$
0 \rightarrow \mathcal{F}_0 \rightarrow\mathscr{O}_{\mathbb{P}^3}(1)^{\oplus4}\rightarrow \mathscr{I}_{Z}(d+2) \rightarrow 0
$$
We have that $\mathcal{F}_0$ is spanned by the representatives of the elements of $\mathcal{F}$ and we have an exact sequence
$$
0 \rightarrow \mathscr{O}_{\mathbb{P}^3}\rightarrow \mathcal{F}_0 \rightarrow \mathcal{F} \rightarrow 0
$$
where the copy of $\mathscr{O}_{\mathbb{P}^3}$ inside $\mathcal{F}_0$ is spanned by $(x_0,x_1,x_2,x_3)$ which projects to the zero vector field. 
Therefore, to prove that $\mathcal{F}_0 = \mathcal{F}\oplus\mathscr{O}_{\mathbb{P}^3}$ one must prove that this sequence splits. 
In the case of the theorem 3.5 of the paper, $\mathcal{F}$ is locally free. Hence a sufficient condition is $H^1(\mathcal{F}^\vee) = \{0\}$. This condition holds if $\mathcal{F}$ is split. 
If the ideal $(F_0,F_1,F_2,F_3)$ is saturated, we have from the sequence
$$
0 \rightarrow \mathcal{F}_0 \rightarrow\mathscr{O}_{\mathbb{P}^3}(1)^{\oplus4}\rightarrow \mathscr{I}_{Z}(d+2) \rightarrow 0
$$
that $H^1(\mathcal{F}_0(k))=0$,  for every $k$. Using that $\mathcal{F}^\vee = \det(\mathcal{F}^\vee)\otimes \mathcal{F}$ and the sequence
$$
0 \rightarrow \mathscr{O}_{\mathbb{P}^3}\rightarrow \mathcal{F}_0 \rightarrow \mathcal{F} \rightarrow 0
$$
twisted by $\det(\mathcal{F}^\vee) = \mathscr{O}_{\mathbb{P}^3}(d-2)$, we see that $H^1(\mathcal{F}^\vee) = \{0\}$ also holds.
