Let X be a projective variety. We consider two Cartier divisors $D,E$ such that $E\geq D$ and the relative morphisms
$\phi_D: X - - -> \mathbb{P}(H^0(X, O_X(D))^*)$ and $\phi_E: X- - -> \mathbb{P}(H^0(X, O_X(E))^*)$
Is there a relation between them?
I think that the map $\phi_D$ can be factorized via the map $\phi_E$ in the following sense:
It's clear that $L(D)\subseteq L(E)$. If $v$ and $w$ are two rational sections of $O_X(D)$ and $O_X(E)$ respectively, then $H^0(X, O_X(D))\cong L(D)$ via the tensor map $\otimes v$ and $H^0(X, O_X(E))\cong L(E)$ via the tensor map $\otimes w$. So we get a monomorphism
$0\to H^0(X, O_X(D))\to H^0(X, O_X(E))$
which sends a section $s$ to a section $\{(U_j, w_j\cdot \frac{s_j}{v_j})\}$. This means $h^0(X, O_X(D))\leq h^0(X, O_X(E))$.
Now we can consider the dual morphism
$H^0(X, O_X(E))^*\to H^0(X, O_X(D))^*$
and then we would consider the morphism induced to the projective spaces
$F: \mathbb{P}(H^0(X, O_X(E))^*)---> \mathbb{P}(H^0(X, O_X(D))^*)$
that it is not well defined everywhere because in general the dimension of $H^0(X, O_X(E))^*$ is greater or equal than the dimension of $H^0(X, O_X(D))^*$ and so the kernel of our map will be not trivial. In particular $F$ is not defined on $\{[f]: f|_{H^0(X,O_X(D))}=0\}$. Thus we can consider the morphism
$F\circ \phi_E: X---> \mathbb{P}(H^0(X, O_X(D))^*)$
that takes a point $p\in U_j$ and sends it to
$[s\to w_j(p)\frac{s_j}{v_j}(p)]$
This morphism will be not defined on $Bs(E)\cup \phi_E^{-1}\{[f]: f|_{H^0(X,O_X(D))}=0\}$ that corresponds to $Bs(D)\cup supp(E-D)$, right?
Moreover we observe $F\circ \phi_E=\phi_D$ on $X\setminus (Bs(D)\cup supp(E-D))$, right?
Why is it interesting for me this question?
In this way we could study the morphism $\phi_D$ through the morphism $\phi_E$. For example, if $D$ is an effective globally generated divisor, then $mD\geq (m-1)D$ and this permit us to say
$h^0(X, O_X((m-1)D))\leq h^0(X, O_X(mD))$
Moreover, if $D$ is also ample, then we have
$\phi_D=F\circ \phi_{mD}$
where $\phi_{mD}$ is an embedding. If I prove that $F$ has finite fiber, then $\phi_D$ has finite fiber, that is the usual property of ample divisors.
I know this property can be proved by contradiction taking a curve $C$ on a fiber $\phi_D^{-1}(p)$, and so
$0<deg((mD)|C)=(mD).C=mD.C=m\phi_D^*H.C=mH.\phi_*C=0$
but I would understand if my argument is good to prove it in another way.
