In which categories is the union of subobjects given by the pushout over the intersection? Let $\mathcal C$ be a sufficiently-complete-and-cocomplete category. Let $C \in \mathcal C$, and let $A \rightarrowtail C \leftarrowtail B$ be subobjects. Let $A \cap B = A \times_C B$ be the intersection of these subobjects. Let $AB = A \amalg_{A \cap B} B$ be the pushout of $A,B$ over this intersection. There is an induced map $AB \to C$.
Question: What conditions on $\mathcal C$ ensure that $AB \to C$ is a monomorphism?
It suffices to suppose that $\mathcal C$ is either abelian or a topos, I think. Does it suffice to assume that $\mathcal C$ is exact?
Does it suffice for the category to be adhesive? (and incidentally is every abelian category adhesive?)
 A: As per the comments, I'm giving one possible answer here.
It suffices that $\mathcal{C}$ is a coherent category. The proof is mainly due to Joyal, and it appears e.g. in "Reyes, Gonzalo: From sheaves to logic - Studies in algebraic logic, M.A.A. studies in Math., vol. 9 (1974). The Elephant has also two different proofs, one using the internal language of the category.
A: Theorem 4.7 in the BRICS version of Lack and Sobicisnki's Adhesive categories shows that any adhesive category has the "effective unions" (in Barr's terminology) of the question. However, if you look at the proof, you will find that the full strength of adhesivity is not used. Unwinding the proof, we need substantially less:
Theorem: (Lack and Sobicinski): Let $\mathcal C$ be a category. Suppose that

*

*$\mathcal C$ has pullbacks along monomorphisms;

*$\mathcal C$ has pushouts of monomorphisms along monomorphisms, and the resulting pushout squares consist of monomorphisms;

*Whenever $A \xrightarrow f C \xleftarrow g B$ is a jointly epimorphic pair of monomorphisms in $\mathcal C$, and $C' \xrightarrow h C$ is any map, then the base change $A' \xrightarrow{f'} C' \xleftarrow{g'} B'$ is a jointly epimorphic pair of morphisms.

Then $\mathcal C$ has effective unions.
Notes:

*

*I think this still doesn't cover the abelian case.

