Is every dg-coalgebra the colimit of its finite dimensional dg-subcoalgebras?

Question

I saw this result in A Model Category Structure for Differential Graded Coalgebras by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $\mathbb{Z}$-graded?.

Thanks.

Answer 1

For coassociative dg-coalgebras over any field $k$ the answer is positive, because:

1. Let $C$ be a $\mathbb Z$-graded coalgebra and $D\subset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^{gr}\subset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $D\subset D^{gr}$ and $D^{gr}$ is a finite-dimensional graded subcoalgebra of $C$.

2. Let $(C,d)$ be a dg-coalgebra and $D\subset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^{dg}=D+d(D)\subset C$. Then $D\subset D^{dg}$ and $D^{dg}$ is a finite-dimensional dg-subcoalgebra of $C$.

Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $\mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.

Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.

EDIT: I've realized that the preceding paragraph is problematic for the following reason: given a $k$-submodule $D$ is a $k$-module $C$, the tensor product $D\otimes_k D$ is not a submodule of the tensor product $C\otimes_k C$, generally speaking. So the very notion of a $k$-subcoalgebra for a commutative ring $k$ is problematic, or at least requires extra care with nonexact tensor products. So, I retract the preceding paragraph.

One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=\mathfrak g=k[[z]]\,d/dz$, the Lie algebra of vector fields on the formal disk.

Answer 2

Yes. The category of $\mathbb{Z}$-coalgebras is locally presentable, and objects are filtered colimits of finite dimensional subobjects. See the appendix to Coalgebraic models for combinatorial model categories by Ching and Riehl. See also Lemma 5.2 of Model Structures for Coalgebras by Drummond-Cole and Hirsh. This paper of Adamek and Porst might also be helpful.

By the way, the main result of the Getzler-Goerss paper you cite is generalized in Corollary 6.3.5 of A necessary and sufficient condition for induced model structures by Hess, Kedziorek, Riehl, and Shipley. It works for any $\mathbb{Z}$-graded coalgebras over any commutative ring $R$.