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This question will fade from very specific to very generic.

Consider an $\lambda$-accessible category $\mathcal{K}$ and let's call $\text{Pres}_{\lambda} \mathcal{K}$ its $\lambda$-presentables. Is the following statement true?

Given a category with directed colimits $ \mathcal{C}$ and a functor $F: \text{Pres}_{\lambda} \mathcal{K}\to \mathcal{C}$, there is an extension $\bar{F} : \mathcal{K} \to \mathcal{C}.$

I do have an idea to prove it.

Consider objects $K, K' \in \mathcal{K}$ and a map $K \stackrel{f}{\to} K'$. I need to give a definition for $\bar{F}(K), \bar{F}(K'), \bar{F}(f)$.

This is my attempt: Choose a directed diagram of $\lambda$-presentables $K_i$ such that $K = \text{colim }K_i$ and same for a diagram $K'_j$ whose colimit will be $K'$.

The map $K_i \to K \to K'$ factors through $K_i \to K \to K'_{\bar{i}}$ because of presentability of $K_i$. We call this map $f_i$. Moreover there is a natural map $$\text{colim}F(K_{\bar i}) \stackrel{i}{\to} \text{colim}F(K_{ i})$$

I pose $$\bar{F}(K) = \text{colim}F(K_i) $$ $$\bar{F}(f) = i \circ \text{colim}(F(f_i)).$$

Probably with this definition composition will not work, but maybe one can make it work.


Now question gets more vague. Consider a category $\mathcal{K}$ and $\mathcal{G}$ the subcategory generated by a strong generator.

Can the following statement be true with some additional hypotesis?

Given a category with directed colimits $ \mathcal{C}$ and a functor $F: \mathcal{G}\to \mathcal{C}$, there is an extension $\bar{F} : \mathcal{K} \to \mathcal{C}.$

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2 Answers 2

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$\mathcal K$ is a free cocompletion of its $\lambda$-presentables under $\lambda$-filtered colimits. Thus $\bar{F}$ exists and preserves $\lambda$-filtered colimits.

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  • $\begingroup$ If $\mathcal{K}$ and $\mathcal{C}$ are locally $\lambda$-presentable (it's stronger than OP's hypothesis), any functor $F:\text{Pres}_{\lambda} \mathcal{K} \to \mathcal{C}$ can be uniquely extended to a $\lambda$-accessible functor $\bar{F}:\mathcal{K} \to \mathcal{C}$. When is $\bar{F}$ colimit-preserving ? A necessary condition is that $F$ must preserve $\lambda$-small colimits. $\endgroup$ Jun 28, 2017 at 16:06
  • $\begingroup$ I can answer my own question thanks to this wonderful answer: mathoverflow.net/a/112206/24563. $\bar{F}$ will preserve the colimits preserved by the functor $\mathcal{K}(X,-)$ where $X$ is $\lambda$-presentable. So in general, only $\lambda$-filtered colimits. $\endgroup$ Jun 29, 2017 at 9:36
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$\require{AMScd} \require{graphicx} \def\K{\mathcal{K}} \def\Pres{\text{Pres}} \def\C{\mathcal C} \def\Lan{\text{Lan}} $ You would like to define $\bar F$ as a suitable Kan extension: in the diagram $$ \begin{CD} \Pres_\lambda \K @>F>> \C\\ @ViVV @.\\ \K \end{CD} $$ (where $i$ is the natural embedding of $\lambda$-presentable objects into $\K$) it is natural to define $\bar F$ as $\Lan_iF$, so that $$ \bar F (K) \cong \varinjlim_{(A,f)\in (i\downarrow K)} FA $$ since of course the comma category $(i\downarrow K)$ is the category of elements of the functor $\hom(i\,\_\, , K)$, a sufficient condition for this colimit to exist in $\C$ is that this latter category is filtered.

But this is well-studied (actually, in a properly dualized form)! For a functor $G : \mathcal{A}\to Set$ the following conditions are equivalent:

  • $G$ commutes with finite limits;
  • The Yoneda extension $\Lan_YG$ commutes with finite limits;
  • the category $\text{Elts}(G)°$ is filtered.

(this is Borceux,1 6.1.3 by the way)

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  • $\begingroup$ But how do you prove that in this case the diagram commutes?! $\endgroup$ Jun 28, 2017 at 8:39
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    $\begingroup$ Snce $i$ is fully faithful, the left Kan extension $Lan_i$ is fully faithful, so that the unit of the adjunction $Lan_i\dashv i^* $ is invertible, and the triangle commutes up to iso. $\endgroup$
    – fosco
    Jun 28, 2017 at 8:52
  • $\begingroup$ (concretely, the pointwise formula for the Kan extension is a colimit indexed by an overcategory. if you start with an element of the full subcategory, this overcategory has a final object and the colimit is easy.) $\endgroup$ Jun 28, 2017 at 16:18

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