If $X$ is a normal complex variety, and $D$ is an effective $\mathbb{Q}$-Cartier Weil divisor, then there is a natural map $\mathcal{O}(D) \otimes \mathcal{O}(-D) \rightarrow \mathcal{O}_X$. My question: under what conditions is this map surjective? Obviously this is true if $X$ is smooth or $D$ is Cartier; I am interested in the singular, non-Cartier case. I am happy to assume $X$ is $\mathbb{Q}$-factorial and log terminal, if that helps at all.

Here is what I know:

- It is not surjective in general. For instance, if $D$ is a ruling of the quadric cone, then this map is not surjective.

**EDIT: Statement 2 is not correct:**

*I had convinced myself that the exact sequence below implies surjectivity, but this only holds if $\mathcal{O}_D(D)$ is isomorphic to $\mathcal{O}_D \otimes \mathcal{O}_X(D)$, which is not true in general, hence statement 2 is false!*

- If $X$ is log terminal, and $D$ is Cartier in codimension 2 (meaning the non-Cartier locus of $D$ has codimension at least 3 in $X$), this map is surjective. This follows because the sequence $0 \rightarrow \mathcal{O}_X \to \mathcal{O}_X(D) \rightarrow \mathcal{O}_D(D) \to 0$ is exact in this case.

Are there any other conditions that guarantee surjectivity of this map? I am particularly interested in the case when $D$ happens to be the (reduced, but possibly not irreducible) exceptional locus of a birational morphism $f: X \rightarrow Y$.

**EDIT: in the comments below, it appears that this only holds if $D$ is, in fact, a Cartier divisor.**