Let $X$ be complex manifold, the image of $\text{Div}(X)\to
\text{Pic}(X)$ is generated by those line bundles with $H^{0}(X,L) \ne
0$ ( i.e. it admits a global non trivial holomorphic section).
The proof for the image of a divisor $D$, gives a non trivial global section goes as follows:
Proof: Any divisor $D=\sum a_i\left[Y_i^i\right] \in \operatorname{Div}(X)$ can be written as $D=\sum a_i^{+}\left[Y_i^{+}\right]-\sum a_j^{-}\left[Y_j^{-}\right]$with $a_k^{\pm} \geq 0$. Hence, $\mathcal{O}(D) \cong \mathcal{O}\left(\sum a_i^{+}\left[Y_i\right]\right) \otimes$ $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)^*$ and the line bundles $\mathcal{O}\left(\sum a_i^{+}\left[Y_i\right]\right)$ and $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)$ are both associated to effective divisors and, therefore, admit non-trivial global sections.
My question is we have proved that $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)$ admits non trivial global section however $\mathcal{O}(D)$ tensors a dual bundle of it, it's not clear that the dual bundle admits non trivial global section?
Best Answer
It is enough to consider effective divisors. Write a divisor $D$ as $D_1-D_2$, with $D_1,D_2$ both effective, and let $L_1:=\mathcal O(D_1)$, $L_2:=\mathcal O(D_2)$ be the associated bundles. Then $\mathcal O(D)=\mathcal O(D_1)\otimes \mathcal O(-D_2)= L_1\otimes L_2^* \in \langle L_1,L_2\rangle$, with $H^0(X,L_1)\neq 0$ and $H^0(X,L_2)\neq 0$, because $D_1,D_2$ are effective. Therefore, the image of the map $\mathrm{Div}(X)\to \mathrm{Pic}(X)$ is generated by line bundle having global sections, i.e. arising from effective diviors.
For the question in the comment, it is also correct: if both $L=\mathcal O(D)$ and $L^*= \mathcal O(-D)$ have non trivial global section, there exist divisor $A\in |D|$ and $B\in |-D|$. But $A+B$ is a principal divisor, and $A$ and $B$ are bot effective, whence $A=B=0$.