TITLE: Pull back of twisted sheaf under a regular map associated to a base point free linear system
QUESTION [5 upvotes]: Let $D$ be a divisor on a normal projective variety $X$ and $V$ be a subspace of the global section of $\mathscr O_X(D),$ L is a base point free linear system and $\phi_L:X\overset{(g_0:\cdots:g_n)}\longrightarrow \mathbb P^n$ the regular map associated to L.
I have to show 
$\phi^*_L(\mathscr O_{\mathbb P^n}(m))\cong O_X(mD)$ for all $m\in\mathbb Z.$
PS. I thought about it and I have the following idea.
Let $E=1.Z(x_0)$ be the divisor in $\mathbb P^n.$ Then $\mathscr O_{P^n}(E)|_{U_i}=(X_i/X_0)\mathscr O_{U_i}$ where $U_i=\{x_i\neq 0\}.$ Then I know there exists a divisor $\tilde E$ in $X$ such that $\mathscr O_{X}(\tilde E)|_{V_i}=(g_i/g_0)\mathscr O_{X}|{V_i}$ where $V_i=\phi_L^{-1}(U_i).$
I do not understand how to relate $O_{X}(\tilde E)$ with $O_{X}(D)$
Any kind of hint or suggestion will be extremely helpful.

REPLY [1 votes]: Let $Z(x_0)$ be a divisor of $\mathbb{P}^n$. Then a section $s\in \phi^*_L(O_{\mathbb{P}^n}(mZ(x_0)))$ has the property that for each couple of  open neighborhoods $(\phi_L^*)^{-1}(U_i)$ and  $(\phi_L^*)^{-1}(U_j)$ we have 
$s_i=(\frac{(x_0)_i}{(x_0)_j})^m\circ \phi_Ls_j=\frac{(x_0)_i^m\circ \phi_L}{(x_0)_j^m\circ \phi_L}s_j=(\frac{g_{0i}}{g_{0j}})^ms_j=g_{ij}^ms_j$ 
where $g_0$ is a global section of $D$ and the first component of the map $\phi_L$ while $g_{ij}$ is the cocycle of the divisor $D$. Then $g_{ij}^m$ is the cocycle of the divisor $mD$ so you have that $s$ is a section of the divisor $mD$, that it  means $s\in O_X(mD)$
{"set_name": "stack_exchange", "score": 5, "question_id": 3312384}