# Defining a $2-$category of "partial functors" to define a colimit functor even though all colimits don't exist.

Maybe a crazy idea but hear me out:

Not every category has all colimits/limits of a certain shape so it doesn't make sense to ask for a functor $$\text{colim}: C^I \rightarrow C$$. But what if we define a strict $$2-$$category $$\text{PartialCat}$$, which consists of categories and "partial functors" $$F : A \rightarrow B$$ defined only on a subcategory of $$A$$. So a partial functor between $$A$$ and $$B$$ is a diagram $$A \leftarrow S \rightarrow B$$ where $$S$$ is a subcategory of $$A$$.

Composition of $$A \leftarrow S \xrightarrow F B$$ and $$B \leftarrow S' \xrightarrow G C$$ is defined by the diagram $$A \leftarrow F^{-1}(S') \rightarrow S' \xrightarrow G C$$, giving us the diagram $$A \leftarrow F^{-1}(S') \rightarrow C$$ i.e a partial functor $$A \rightarrow C$$.

If we find a suitable notion of $$2-$$morphisms between partial functors we can define $$\text{colim}$$ as the left adjoint of the diagonal functor $$C \xleftarrow{id} C \xrightarrow \Delta C^I$$ in the $$2-$$category $$\text{PartialCat}$$.

$$\text{colim}$$ would then ideally be defined on the full subcategory of $$C^I$$ consisting of the diagrams $$I \rightarrow C$$ which have a colimit in $$C$$ and take those diagrams to their colimit.

This would allow us to bypass a lot of the diagram manipulation used to prove simple results such as "Left adjoint functors preserve colimits" for categories where all colimits don't exist and instead use $$2-$$categorical machinery. We generally prove results about colimits by using their universal property but we could bypass this and just use facts about adjunctions in $$2-$$categories basically.

We could also generalize this construction to define colimits of diagrams in $$(\infty,1)-$$categories easier since these are generally harder to understand just in terms of their universal property. Proving that $$(\infty, 1)-$$left adjoints preserve $$(\infty, 1)-$$colimits is not an easy result to prove analytically but becomes easy if we assume that there is a $$(\infty,1)-$$functor $$\text{colim}: C^I \rightarrow C$$ for quasicategories $$C$$ and $$I$$. This is just a formal consequence of the theory of adjunctions in a $$2-$$category. Specifically the homotopy $$2-$$category of quasicategories.

As for what the $$2-$$morphisms in $$\text{PartialCat}$$ should be I have no idea, maybe if $$A \leftarrow S \rightarrow B$$ and $$A \leftarrow S' \rightarrow B$$ are two partial functors called $$F$$ and $$G$$ respectively we can define $$\text{Hom}(F,G)$$ as the set of natural transformations between the two functors but restricted to $$S \cap S' \rightarrow B$$ so they have a common domain.

Explicitly: $$\text{Hom}(F,G) = \text{Nat}(\left.F\right|_{S \cap S'} , \ \left.G\right|_{S \cap S'})$$

I don't know how these natural transformations are supposed to compose though so it's probably wrong. Can't quite figure it out.

• Your "suitable notion of 2-morphism" is probably circular: I imagine one of your "suitability" requirements is that the left adjoint of the diagonal is the colimit functor! Dec 29, 2020 at 8:49
• Not really, all I'm asking is if there is some reasonably natural notion of 2-morphism so that what I claim happens, happens. For example the existence of a faithful $2-$functor $\text{Cat} \rightarrow \text{ PartialCat}$ seems to make sense as a requirement. Dec 29, 2020 at 14:54
• Furthermore, is it wrong to look for a definition of an object so that it explicitly satisfies some condition anyway? Mathematicians chase definitions all the time. See the entire field of $(\infty, 1)-$ categories. Dec 29, 2020 at 15:01

I suggest to use profunctors induced by (or simply instead of) partial functors.

First of all, a profunctor $$H:A^{op}\times B\to {\rm Set}$$ is said to be functorial if $$H(a,-):B\to {\rm Set}$$ is a representable functor for each object $$a\in A$$. This is equivalent to saying that $$H\simeq \hom_B(F-,\, -)\,=:F_*$$ for a functor $$F:A\to B$$.
And actually the profunctor of the colimit functor can be defined for any category, even if it lacks some colimits of the given shape: namely, consider $$H:(C^I)^{op}\times C\to {\rm Set}\quad (\underset{I\to C}D,\,a)\mapsto \{D\to a\text{ cocones in }C\}\,,$$ where a $$D\to a$$ cocone can be thought of as a natural transformation $$D\to a$$ to the constant functor.

Note that a diagram $$D\in C^I$$ has a colimit, by the very definition, if and only if $$H(D,-)$$ is representable.

Secondly, not only partial functors but also all categorical relations (that is, subcategories of $$A\times B$$), moreover any span of categories $$A\overset{P}\leftarrow R\overset{Q}\to B$$ do determine a profunctor in a natural way, namely, $$P^*\otimes Q_*:A\not\to R\not\to B$$ where $$P^*=\hom_A(-,\,P-);\ \ Q_*=\hom_B(Q-,\,-)$$ and $$\otimes$$ means composition (tensor product) of profunctors, which is a colimit/coend construction that yields a quotient of $$\displaystyle\bigsqcup_{r\in R}P^*(a,r)\times Q_*(r,b)$$.

I guess, the $$2$$-category you are defining would naturally embed into the bicategory of profunctors this way.

• I think I understand. So a profunctor $A \not \rightarrow B$ is a functor $A^{op} \times B \rightarrow \text{Set}$? And I guess 2-morphisms of profunctors are just natural transformations of the functors $A^{op} \times B \rightarrow \text{Set}$? Dec 30, 2020 at 13:15
• Yes, exactly, your understanding is correct. Dec 30, 2020 at 14:11
• Would it also be correct to say that $F_*$ is left adjoint to $F^*$ in the bicategory of profunctors where $F$ is some functor? Dec 31, 2020 at 2:55
• Yes, that's right, moreover, if I remember correctly, a profunctor is functorial iff it has right adjoint. (Though I'm not 100% sure on left/right, as both directions of profunctor composition are in use in different texts). Dec 31, 2020 at 9:49