1. Is it possible to express the full subcategory of colimit-preserving functors between categories $A,B$ as an equalizer of a single pair of maps? More explicitly, I would like to find an equalizer diagram $$ E \to {\bf Cat}(A,B) \underset{\beta}{\overset{\alpha}\rightrightarrows} {\mathcal S}(A,B) $$ such that $E = {\bf Cat}_!(A,B) = \{f : A \to B \mid f \text{ is cocontinuous}\}$ for a suitable pair of maps $\alpha,\beta$ and an object ${\cal S}(A,B)$ depending on $A,B$.
  2. Dual question for continuous functors $A\to B$;
  3. Same question(s) for a generic 2-category $\cal K$ and for the subcategory ${\cal K}_!(A,B)$ of cocontinuous 1-cells.

I have a guess on how the answer might be "yes", but I also incur in all sorts of (set-theoretic) problems.

Everything starts from the remark that $f : A\to B$ is cocontinuous if and only if $$ f(\text{Lan}_gh) \cong \text{Lan}_gfh $$ or, in a more evocative notation, $(g_!h)^*(f) \cong g_!h^*(f)$.

So $f$ is cocontinuous if and only if it equalizes all pairs $((g_!h)^*, g_!h^*)$, when $g,h$ sun over spans $Y \overset{g}\leftarrow X \overset{h}\to A$, if and only if it preserves left extension of $h$ along $g$, for every $h,g$, if and only if it equalizes the pair $$ E \to {\bf Cat}(A,B) \underset{(g_!h)^*}{\overset{g_!h^*}\rightrightarrows} {\bf Cat}(Y,B) $$ Somehow, it seems that then $f$ is cocontinuous if and only if it belongs to the limit of the following diagram: define a category $\mathfrak B$ having as class of objects a point, $a$, disjoint a proper class of other objects $\{b_\lambda\}$. Morphisms are defined in this way:

  1. $\mathfrak B(a,a) = \{1_a\}$
  2. $\mathfrak B(a,b_\lambda) = \{u_\lambda,v_\lambda\}$
  3. $\mathfrak B(b_\lambda,b_\mu) = \{1_{b_\lambda}\}$ if $\lambda = \mu$ and empty otherwise.

Whatever it is, $E$ is the limit over $\mathfrak B$.

Of course, all this is very naive and needs some extra set-theoretic care to work properly, as $\mathfrak B$ is not a small diagram...

  • $\begingroup$ Why not explain your idea? $\endgroup$ Sep 16 '18 at 18:26
  • $\begingroup$ Done! Hope it's more clear now $\endgroup$
    – fosco
    Sep 16 '18 at 19:39

I don't follow what connection your category $\mathfrak B$ has to any of the other data in the problem, but you can certainly say that the category of functors strictly preserving some choice of colimits is the equalizer of the two functors that send $F:A\to B$ to $(F(\mathrm{colim} D))_D$ and $(\mathrm{colim}F(D))_D$, where $D$ runs over small diagrams in $A$. However, a cocontinuous functor in the usual sense is one for which the canonical comparison map $\lambda_D\mathrm{colim}F(D)\to F(\mathrm{colim} D)$ is an isomorphism, for all $D$, and similarly for your expression in terms of preserving left Kan extensions. It's not enough to ask for the mere existence of an isomorphism, which would produce in any case produce a pseudo-equalizer rather than an equalizer. Since it's a given map that must be an isomorphism, the category of cocontinuous functors is naturally seen as an inverter, a certain weighted limit of the diagram which augments the diagram we took the equalizer of above with a natural transformation encoding all the $\lambda_D$. Any weighted limit can be expressed as an equalizer, but not of quite the form you want, since the canonical comparison map is naturally part of the data. I don't see any way to get around that.

  • $\begingroup$ > "I don't follow what connection your category $\mathfrak B$ has to any of the other data in the problem" strictly preserving a single shape of colimit can be expressed as an equalizer, and preserving it up to a prescribed iso is an inverter. Then, preserving all shapes is the "wide equalizer" of the whole family of parallel arrows indexed by the various $a\rightrightarrows b_\lambda$. $\endgroup$
    – fosco
    Sep 18 '18 at 15:52
  • $\begingroup$ Ok, makes sense. I'd say you can just pack all the $b_\lambda$ into a single product, though, which recovers my answer. Since Cat(A,B) is already large, I see no reason to be squeamish about size here. $\endgroup$ Sep 18 '18 at 17:24

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