Suppose $X$ is a nice enough scheme and $D$ is an effective divisor. Since $\mathcal{O}_X(-D)=\mathcal{I}_D$, there's a natural injection $\mathcal{O}_X(-D)\to\mathcal{O}_X$. On the other hand, $\operatorname{Hom}(\mathcal{O}_X(-D),\mathcal{O}_X)\cong \operatorname{Hom}(\mathcal{O}_X,\mathcal{O}_X(D))\cong \Gamma(X,\mathcal{O}_X(D))$ since $\mathcal{O}_X(D)$ is a line bundle. Which element of this space of global sections is the element corresponding to the natural injection?
My guess, using the description of global sections of $\mathcal{O}_X(D)$ as rational functions which have poles contained in $D$, is that this should be $1$ (what else could it be?). But I'm unsure about how to prove this, and would appreciate some tips.
Suppose that $X$ is irreducible and that it is covered by open affine subsets $U_i$ such that $D|_{U_i}=(f_i)$. We consider $\mathcal O_X(-D),\mathcal O_X,\mathcal O_X(D)$ to be subsheaves of $\mathcal K$ (the constant sheaf associated to the function field of $X$). Then $\mathcal O_X(-D)|_{U_i}$ is generated by $f_i$, $\mathcal O_X|_{U_i}$ is generated by $1$, and your morphism $f:\mathcal O_X(-D)\to\mathcal O_X$, identifying $\mathcal O_X(-D)$ with the ideal sheaf of $D$, maps $f_i$ to $f_i$. Since tensor multiplication of sheaves correponds to multiplication of Cartier divisors, and the Cartier divisor corresponding to $\mathcal O_X(D)$ is given by $f_i^{-1}$, the twisted morphism $f(D):\mathcal O_X\to\mathcal O_X(D)$ maps $f_i\cdot f_i^{-1}$ to $f_i\cdot f_i^{-1}$, i.e. maps $1$ to $1$. And the global section of $\mathcal O_X(D)$ defined by $f(D)$ is precisely the image of $1$, hence your guess is correct.