# Todd Trimble Stuff on coalgebras

## Local presentability and cartesian closure

We establish the following result:

###### Theorem

If $V$ is locally presentable and is given a symmetric monoidal closed structure, then the category of cocommutative comonoids in $V$ is cartesian closed and locally presentable.

Let $Cocom(V)$ be the category of commutative comonoids in $V$, and $U: Cocom(V) \to V$ the forgetful functor. There are three main tasks:

• Show that $Cocom(V)$ is accessible. Then, since cocomplete accessible categories are locally presentable, they are also complete (Adamek and Rosicky, …). In particular, $Cocom(V)$ has equalizers. (Finite products are given by tensor products in $V$.)

• Show that $U$ has a right adjoint $Cof: V \to Cocom(V)$.

• Given the fact that $Cocom(V)$ has equalizers and $U \dashv Cof$, the category $Cocom(V)$ is cartesian closed. Products in $Cocom(V)$ are given by tensor products in $V$.

I am roughly following Porst here; see particularly the proposition on page 8 and remarks on page 9. If the first task is completed, then the second is easy: using the fact that $Cocom(V)$ is locally presentable and the fact that $U: Cocom(V) \to V$ preserves colimits, it has a right adjoint by the special adjoint functor theorem. In fact, $U$ is comonadic. The third task is by a direct construction of the exponential in $Cocom$, using the symmetric monoidal closed structure of $V$, the right adjoint $Cof$, and equalizers in $Cocom(V)$.

### Accessibility of $Cocom(V)$

The first task is divided into two subtasks: first is to show that the category of coalgebras for the finitary endofunctor

$F(v) = (v \otimes v) \times I$

is accessible. Porst says (fact 5, page 6):

• If $F$ preserves directed colimits (and it does), then $Coalg(F)$ is accessible.

A detailed proof is given by Adamek and Porst (see Theorem IV.2, page 16). The relevant and more general results are given in the following lemma and theorem.

###### Lemma

Let $k$ be a regular infinite cardinal. If $F: \mathbf{C} \to \mathbf{C}$ is $k$-accessible and $(C, \gamma: C \to F C)$ is a coalgebra whose underlying object is $k$-presentable in $C$ (i.e., $\hom(C, -): C \to Set$ preserves $k$-filtered colimits which exist in $C$), then $(C, \gamma)$ is $k$-presentable in $Coalg(F)$. Conversely, the underlying object of a $k$-presentable $F$-coalgebra is $k$-presentable in $V$.

###### Proof

Suppose given a $k$-filtered diagram $J \to Coalg(F)$ with colimit $(B, \beta)$; let $(B_j, \beta_j: B_j \to F(B_j))$ denote nodes of the diagram. We must show that $\hom(C, -): Coalg(F) \to Set$ preserves this colimit. That is, any coalgebra map $C \to B$ must be shown to factor through one of the $B_j \to B$ in $Coalg(F)$, and one should check that any two factorings resolve at a later stage.

The underlying functor $Coalg(F) \to \mathbf{C}$ preserves and reflects colimits, so $B$ is the colimit of the $B_j$ in $\mathbf{C}$. Since $C$ is $k$-presentable in $\mathbf{C}$, any map $f: C \to B$ factors through one of the cone components $g_i: B_i \to B$ in $\mathbf{C}$, say $f = g_i h$. Suppose now $f: C \to B$ is a coalgebra map. Then

$F(g_i) \beta_i h = \beta g_i h = \beta f = F(f)\gamma = F(g_i) F(h)\gamma$

using the fact that $g_i$ and $f$ are coalgebra maps. Since $F$ and $\hom(C, -)$ preserve $k$-filtered colimits, the comparison map

$colim_j \hom(C, F B_j) \stackrel{(\hom(C, F g_j))}{\to} \hom(C, F B)$

is a bijection; therefore the equality of elements $F(g_i)\beta_i h = F(g_i)F(h)\gamma$ in $\hom(C, F B)$ implies that $\beta_i h$ and $F(h)\gamma$ belong to the same equivalence class in $\colim_j \hom(C, F B_j)$. In other words, there exists an arrow $u: B_i \to B_j$ in the diagram (a coalgebra map) such that

$F(u)\beta_i h = F(u)F(h)\gamma$

whence $\beta u h = F(u h)\gamma$, making $u h: C \to B_j$ a coalgebra map. We moreover have

$f = g_i h = g_j u h$

so $f$ factors in $Coalg(F)$ through one of the nodes $B_j$. The fact that any two factorings resolve at a later stage is clear.

• Example: Let $k$ be a field. Then $Vect_k$ is locally finitely presentable, since every vector space is a filtered colimit of its finite-dimensional subspaces. The functor
$Vect_k \to Vect_k: v \mapsto \otimes^n v$

is finitary; therefore so is $F(v) =$

To be continued…

###### Theorem

If $\mathbf{C}$ is locally finitely presentable and $F: \mathbf{C} \to \mathbf{C}$ is finitary, then $Coalg(F)$ is locally $\omega_1$-presentable (and often locally finitely presentable). If $k$ is an uncountable regular cardinal, $C$ is locally $k$-presentable, and $F$ is $k$-accessible, then $Coalg(F)$ is locally $k$-presentable.

Next, Porst quotes a result from Adamek and Rosicky (Locally Presentable and Accessible Categories, 2.76) which figures in the theorem that the 2-category $Acc$ of accessible categories and accessible functors is closed under lax limits in $Cat$. If $F, G: \mathbf{C} \to \mathbf{D}$ are 1-cells of $Cat$ and $\eta, \theta: F \to G$ are 2-cells, the equifier $Eq(\eta, \theta)$ is the full subcategory of $\mathbf{C}$ whose objects $c$ satisfy $\eta c = \theta c$.

If $\mathbf{C}$ and $\mathbf{D}$ are locally presentable, $F$ and $G$ are $k$-accessible functors $\mathbf{C} \to \mathbf{D}$, and $\eta, \theta: F \to G$ are transformations, then $Eq(\eta, \theta)$ are locally presentable.

## Interlude: monos and equalizers in $Cocom(Vect_k)$

Suppose $U: \mathbf{C} \to Set$ is faithful.

###### Proposition

If $U(f)$ is monic, then so is $f$.

The proof is trivial.

Now suppose $f: C \to D$ is monic in $Cocom(Vect_k)$. We know subcoalgebra inclusions $D' \hookrightarrow D$ are monic by Proposition 1. Hence the pullback $f^{-1}(D') \to C$ is monic as well, as is of course the pullback $f^{-1}(D') \to D'$.

###### Lemma

A morphism $f: C \to D$ in the category of finite-dimensional cocommutative $k$-coalgebras is monic iff $U(f)$ is monic.

###### Proof

By taking linear duals, this is equivalent to the statement that epimorphisms $f: A \to B$ in the category of finite-dimensional commutative $k$-algebras are surjective. Suppose otherwise, so we have an exact sequence of modules

$A \stackrel{f}{\to} B \to B/A \to 0.$

with $B/A$ nonzero. Apply the functor $- \otimes_A B$ to arrive at an exact sequence

$A \otimes_A B \to B \otimes_A B \to B/A \otimes_A B \to 0.$

That the first arrow is an iso is just the characterization of epis in terms of pushouts. Hence $B/A \otimes_A B = 0$, and therefore $B/A \otimes_A B/A = 0$. On the other hand, considering $B$ to be finitely generated as an $A$-module, with a minimal set of generators $b_1, \ldots, b_n$, then we have surjections

$B/A \to B/A\langle b_1, \ldots, b_{n-1}\rangle \leftarrow A$

where the second arrow takes $1 \in A$ to $b_n$. By minimality, the second arrow is nonzero, with kernel an ideal $I$ of $A$, so we have a surjection

$B/A \to A/I$

and hence a surjection $0 = B/A \otimes_A B/A \to A/I \otimes_A A/I$. Hence $A/I \otimes_A A/I = 0$. But because $A \to A/I$ is an epi in the category of commutative rings, the pushout projection $A/I \otimes_A A \to A/I \otimes_A A/I = 0$ is an iso, and this contradicts properness of $I$.

Now consider the directed system of inclusions of finite-dimensional subcoalgebra inclusions $D' \hookrightarrow D$ (whose colimit is $D$). The pulled-back system $f^{-1}(D') \hookrightarrow C$

Revised on August 28, 2015 at 20:01:01 by Todd Trimble