# In which categories is every coalgebra a sum of its finite-dimensional subcoalgebras?

Let $\mathcal V$ be a reasonably nice category — I'm interested in the case when $\mathcal V$ is $\mathbb K$-linear for some field $\mathbb K$, abelian, and has all products and coproducts (hence all limits and colimits, as it is abelian), but I don't mind if you demand similar or weaker properties — with a reasonably nice monoidal structure $\otimes$. Recall that a (counital coassiciative) coalgebra in $\mathcal V$ is an object $A\in\mathcal V$ along with maps $A \to 1$ and $A \to A\otimes A$ that are coassiciative and counital in the sense that certain diagrams commute. The notion of homomorphism of coalgebras is obvious.

When $\mathcal V = \text{VECT}$ is the category of vector spaces, then coalgebras satisfy a particular fundamental property that makes them essentially easy. Namely, any coalgebra is $\text{VECT}$ is the (vector space) sum of its finite-dimensional subcoalgebras. On the other hand, the corresponding statement in $\text{VECT}^{\rm op}$ fails: it is not true that every algebra in $\text{VECT}$ is a pullback of its finite-dimensional quotient algebras. This is in spite of the fact that for many purposes $\text{VECT}$ and $\text{VECT}^{\rm op}$ are equally nice categories.

For a general sufficiently nice category $\mathcal V$, I should replace the word "sum" by "limit" and I should replace "finite-dimensional" by "dualizable". All together, my question is:

For which sufficiently nice monoidal categories is it true that every coalgebra object is a limit of its dualizable subcoalgebra objects?

This is, of course, an open ended question. The very best would be some necessary and sufficient conditions that are easier to check, but that's probably too hard: natural (and naturally occurring) easily-checked sufficient conditions would suffice.

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