I recently noticed that there are two senses in which colimits are functorial, and I'm curious about their interplay.

Let $C$ be a cocomplete category. Then, on the one hand, for any diagram category $I$ we have a functor $\mathrm{colim} : \mathrm{Fun}(I,C) \to C$ (left adjoint to the "constant $I$-shaped diagram" functor). But then also, colimits are functorial for maps of categories over $C$: for any pair of composable arrows $$ I \xrightarrow{F} J \xrightarrow{G} C , $$ we obtain an induced map $$\mathrm{colim}_I(GF) \to \mathrm{colim}_J(G)$$ in $C$ (by the universal property of $\mathrm{colim}_I(GF)$).

These two situations can be unified: ignoring set-theoretic issues (and I guess maybe coherence issues too), there is a functor $$ \mathrm{Cat}^{op} \xrightarrow{\mathrm{Fun}(-,C)} \mathrm{Cat} , $$ whose corresponding Grothendieck fibration $$ X_C \to \mathrm{Cat} $$ has as its fiber over $I \in \mathrm{Cat}$ the functor category $\mathrm{Fun}(I,C)$, while its cartesian arrows select pullbacks. Thus, I would expect that "colimit" should assemble to a functor $$ X_C \xrightarrow{\mathrm{colim}} C $$ whose restriction to each fiber is a left adjoint.

Has anyone studied this sort of thing?

  • 7
    $\begingroup$ The short answer is: derivators. $\endgroup$
    – Zhen Lin
    Mar 27, 2015 at 23:08
  • $\begingroup$ @ZhenLin Great! If you expand this into an answer that gives some explanation, then I'll be happy to accept it. $\endgroup$ Mar 28, 2015 at 19:11

1 Answer 1


This observation has now been codified (in the $\infty$-categorical setting) in section 3 here: http://arxiv.org/pdf/1510.03525v1.pdf


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