Sheaf homomorphism $\mathcal{O}_X \to \mathcal{O}(D)$ and dually to $\mathcal{O}(-D) \to \mathcal{O}_X$.

Question

I am reading Daniel Huybrechts Complex Geometry. I don't understand how these two maps are defined.

The description of the sheaf $\mathcal{O}(D)$ is given in the answer

https://math.stackexchange.com/a/1749343/414708

My knowledge of sheaf theory and algebraic geometry is limited. Could you please tell me how these maps are defined.

Answer 1

First of all you might check this. There are many posts about exact sequences of hypersurfaces. Two small remarks:

1. the map $s: \mathcal{O}_X \rightarrow \mathcal{O}(D)$ is uniquely determinant by the image s$(1)$. This defines a global section of the line bundle $\mathcal{O}(D)$. $Y$ is defined as the zero locus $Z(s)=(s)_0 \subset X$ of this section.
2. Dually, the map $\mathcal{O}(-D) \rightarrow \mathcal{O}$ can be seen as an embedding as the ideal (sheaf) $I$ in $\mathcal{O}_X$ that cuts out $Y$. then $Y$ has sheaf of sections $\mathcal{O}_D\cong\mathcal{O}/I$.

Edit: Explaining every detail is 2 chapters in Hartshorne but: The line bundle $\mathcal{O}(D)$ has global sections $s_i$ sucht that their zero locus $(s_i)_0$ define hypersurfaces $Y_i\subset X$ which is "linear equivalent" to $Y$. If you're new to all of this, don't go too hard on this and think of $Y_i \sim Y$ like homotopic equivalence in algebraic topology. The two hypersurfaces are somehow equivalent in an algebraic geometric sense (they differ by multiplying with a rational function).