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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.

Thoughts about all this?

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    $\begingroup$ 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! $\endgroup$
    – Zhen Lin
    Dec 29, 2020 at 8:49
  • $\begingroup$ 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. $\endgroup$ Dec 29, 2020 at 14:54
  • $\begingroup$ 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. $\endgroup$ Dec 29, 2020 at 15:01

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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.

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  • $\begingroup$ 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 $A^{op} \times B \rightarrow \text{Set}$? $\endgroup$ Dec 30, 2020 at 13:15
  • $\begingroup$ Yes, exactly, your understanding is correct. $\endgroup$
    – Berci
    Dec 30, 2020 at 14:11
  • $\begingroup$ Would it also be correct to say that $F_*$ is left adjoint to $F^*$ in the bicategory of profunctors where $F$ is some functor? $\endgroup$ Dec 31, 2020 at 2:55
  • $\begingroup$ 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). $\endgroup$
    – Berci
    Dec 31, 2020 at 9:49

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