Can finite presentability be tested with respect to sequential colimits?

Question

Let $\mathcal C$ be a locally finitely-presentable category, and let $C \in \mathcal C$ be an object such that for all sequential colimits, the map $$\varinjlim \operatorname{Hom}(C, X_i) \to \operatorname{Hom}(C, \varinjlim X_i)$$ is an isomorphism. Then does it follow that the analogous map is an isomorphism for an arbitrary filtered colimit, i.e. that $C$ is finitely-presentable?

Notes:

• Standard reductions tell us it suffices to check when the indexing category is a directed poset, or even just when the indexing category is a regular cardinal $\kappa$, even under the additional assumption that the colimit is a transfinite composition, i.e. it is smooth, i.e. for all limit $\alpha < \kappa$ we have $X_\alpha = \varinjlim_{i < \alpha} X_i$.)
• Perhaps it’s also worth asking the following: suppose that $\operatorname{Hom}(C,-)$ commutes with sequential colimits and also that any map from $C$ to a coproduct factors through a finite subcoproduct. Then is $C$ finitely presentable?

Answer 1

It does not follow in full generality. There might be extra assumptions you can add to make it work, but here is a counterexample:

Let $R= \mathbb Z[x_\alpha, y_{\alpha\beta}, \alpha\in\kappa,\alpha<\beta\in\kappa]/(x_\alpha y_{\alpha\beta}= x_\beta)$, so a polynomial ring on a bunch of generators $x_\alpha$ equipped with divisibility relations (realized by polynomial generators $y_{\alpha\beta}$, and let $\mathcal C$ be the category of commutative $R$ algebras. This is locally finitely presentable for general reasons, and I'm going to look at $C= R[x_\alpha^{-1}, \alpha\in\kappa]$.

I claim that this is not compact, yet it is sequentially compact if $\kappa$ is regular and uncountable.

Lemma 1: $C$ is not compact.

Proof: Consider, for $\beta <\kappa$, the ring $C_\beta := R[x_\alpha^{-1}, \alpha<\beta]$. We have $C=colim_{\beta <\kappa} C_\beta$, but $C$ is not a retract of any $C_\beta$, since in $C_\beta$, $x_\beta$ is not invertible.

Lemma 2: $C$ is sequentially compact if $\kappa$ is regular and uncountable.

Proof: $C$ is a localization of $R$, so it sub-initial, in the sense that $\hom(C,-)$ is a subfunctor of the constant point functor. Thus, sequential compactness amounts to : for a given sequential diagram $X_n$, if there exists a map $C\to colim_n X_n$, then there exists a finite $n$ and a map $C\to X_n$.

So let $X_n$ be such a sequential diagram, $h: C\to colim_n X_n$ a map of commutative $R$ algebras, and consider the function $f:\kappa\to \omega, \alpha\mapsto \min\{n\mid x_\alpha$ is invertible in $X_n\}$. Firstly, $f$ is well defined: for any fixed $\alpha$, there exists such an $n$, and so there exists a minimal one.

Second, since $x_\alpha$ divides $x_\beta$ whenever $\alpha<\beta$, $f$ is nondecreasing.

Thus $\kappa= \bigcup_n f^{-1}(\{\leq n\})$. It follows from cofinality considerations (this is where regular and uncountable come into play) that one of those sets is unbounded, and therefore the whole of $\kappa$, i.e. there exists $n$ such that for all $\alpha$, $x_\alpha$ is invertible in $X_n$. For this $n$, there exists a (necessarily unique) map $C\to X_n$.

I think a positive answer to the question is related to some kind of noetherian-ness of $\mathcal C$, and so one should look for a way to axiomatize this, i.e. for a way to remove the kind of phenomenon outlined in the example above.

I just realized that there is a much simpler example. Since I already wrote the previous one I'll leave it as is, but let me point out the following : for an uncountable regular cardinal $\kappa$, the poset $\kappa + 1$ viewed as a category is cocomplete, each successor ordinal is finitely presented. From this is it follows that this category is locally finitely presentable, and that $\kappa$ in there is not compact, yet sequentially compact.