I am reading through Rick Miranda's Algebraic Curves and Riemann Surfaces. There is a proposition that is left as an exercise for the readers to check. However, I am not sure how to verify it.
$\textbf{Set up:}$ In the following, $X$ is a compact Riemann surface. $D$ is a base point free divisor, i.e. for any $p\in X$, there is some divisor $E'$ in the complete linear system $|D|=\{E:E\sim D,\, E\geq 0\}$ so that $E'(p)=0$. It was shown in the text that any such divisor $D$ gives rise to a holomorphic map $\phi_D:X\to\mathbb P^n$, by choosing any basis $\{f_0,\dots,f_n\}\subset L(D)$ and take $\phi_D=[f_0:f_1:\cdots:f_n]$. Picking another basis would only affect $\phi_D$ up to a linear change of coordinates. We have also seen that there is a natural identification $|D|\cong \mathbb P(L(D))$; and $L(D-p)$ is a hyperplane in $L(D)$ when $D$ is base point free.
$\textbf{Proposition:}$ The projectivization $\mathbb P(L(D-p))$ corresponds to the subspace $\{E:E\sim D,\,E\geq p\}\subset |D|$. The map $\phi:X\to |D|^*$ defined by $p\mapsto \{E:E\sim D,\,E\geq p\}$, with a suitable coordinates, is the map $\phi_D$, here $|D|^*$ is the dual projective space.
$\textbf{My thoughts:}$ I don't know how should I identify $\mathbb P^n$ with $|D|^*$. So I tried to fix a point $p\in X$ first, choose a basis $\{f_1,\dots,f_n\}$ of $L(D-p)$, and complete it to a basis $\{f_0,\dots,f_n\}$ of $L(D)$. Then $\phi_D(p)=[f_0(p):\dots:f_n(p)]=[1:0:\cdots:0]$ since $\operatorname{div}(f_0)(p)$ is smaller than that of other $f_i$'s. So I figured maybe the point $[1:0:\cdots:0]$ should be identified with $\{E:E\sim D,\,E\geq p\}$. However if I choose another point $q$, then $\phi_D$ would adjust linearly. It is not so clear to me how $\phi_D$ would change.
I would like to be given hints on how to prove this proposition. Thanks in advance.