$\DeclareMathOperator\Set{Set}\DeclareMathOperator\Vect{Vect}\DeclareMathOperator\Coalg{Coalg}\DeclareMathOperator\ProVect{ProVect}\DeclareMathOperator\prolim{prolim} $Let $K$ be a field and $F: \Set \to \Vect_K$ the free $K$-vector space functor, which naturally lifts to a functor $\bar{F}: \Set \to \Coalg_K$, where $\Coalg_K$ is the category of cocommutative coalgebras over $K$.
Question: Does the functor $\bar{F}: \Set \to \Coalg_K$ preserve filtered limits?
I will give some remarks to address this question:
Let $I$ be a filtered category and $X: I^\text{op} \to \Set$ a functor. I ask if the coalgebra $\bar{F}(\lim(X))$ is the limit of the functor $ \bar{F} \circ X : I^\text{op} \to \Set \to \Coalg_K$ in the category of cocommutative coalgebras over $K$.
Therefore I will explain how to construct the filtered limit in the category of cocommutative coalgebras over $K$:
Let $\ProVect_K$ be the category of pro-vector spaces over $K$, which are formal filtered diagrams of vector spaces. There is a natural embedding $\Vect_K \subset \ProVect_K$ considering a vector space as the constant filtered diagram. This embedding preserves finite limits but does not preserve filtered limits. Moreover this embedding admits a right adjoint that takes the maximal subvector space within a given pro-vector space.
The tensor product of $K$-vector spaces extends to the category $\ProVect_K$ of pro-vector spaces, and the extended tensor product preserves filtered limits component-wise. Let $\Coalg_K^\text{pro}$ be the category of cocommutative coalgebras in the category of pro-vector spaces over $K$. The embedding $\Vect_K \subset \ProVect_K$ induces an embedding $\Coalg_K \subset \Coalg_K^\text{pro}$ that admits a right adjoint $R$ that takes the maximal coalgebra within a given coalgebra in pro-vector spaces.
Since the tensor product of pro-vector spaces preserves filtered limits component-wise, the forgetful functor $\Coalg_K^\text{pro} \to \ProVect_K$ creates filtered limits. Therefore the limit of the functor $ \bar{F} \circ X : I^\text{op} \to \Set \to \Coalg_K \subset \Coalg_K^\text{pro}$ is formed in the following steps: First one takes the limit of the functor $ F \circ X : I^\text{op} \to \Set \to \Vect_K \subset \ProVect_K$, which we denote by $\prolim(F \circ X)$. The pro-vector space $\prolim(F \circ X)$ is naturally a coalgebra over $K$ and the maximal subcoalgebra $R( \prolim(F \circ X)) $ in $\prolim(F \circ X)$ is the limit in coalgebras over $K$.
So I ask if the map $$ F(\lim(X)) \to \lim(F \circ X) \subset \prolim(F \circ X)$$ exhibits $F(\lim(X))$ as the maximal subcoalgebra in $\prolim(F \circ X)$.
I can see so far that the linear map $$F(\lim(X)) \to \lim (F \circ X) $$ is injective for the following reason: Let $k_1 x_1 +\dotsb+k_n x_n \in F(\lim X)$ with $n \geq 1$, $k_1,\dotsc,k_n \in K$ and pair-wise different elements $x_1,\dotsc,x_n \in \lim(X)$ such that for every $\ell \in I$ we have that $k_1 x^\ell_1 +\dotsb+k_n x^\ell_n=0 \in F(X_\ell) $ where $x_i^\ell$ is the image of $x_i$ in $ X_\ell$. Since the elements $x_1,\dotsc,x_n \in \lim(X)$ are pair-wise distinct, for any $1 \leq i \neq j \leq n$ there is a $r_{i,j} \in I$ such that $x^{r_{i,j}}_i \neq x_j^{r_{i,j}}$. As $I$ is directed, there is an $r \in I$ such that for every $1 \leq i \neq j \leq n$ there is a morphism $r_{i,j} \to r$ in $I$. So $x^r_1,\dotsc,x^r_n$ are pair-wise distinct. Thus the equation $k_1 x^r_1 +\dotsb+k_n x^r_n=0 \in F(X_r) $ implies that $k_1 =\dotsb= k_n =0$.