Let $f: \mathbf{C} \to \mathbf{D}$ be a functor between small categories and $\mathbf{M}$ a cocomplete category. Then we have the coend formula $$ (f_! X)(d) = \int^{c \in \mathbf{C}} \mathbf{D} (f(c),d) \otimes X(c) \quad (*) $$ for the left Kan extension of a functor $X : \mathbf{C} \to \mathbf{M}$ along $f$.

Now assume that all categories and functors involved here are enriched over simplicial sets (for example). Then the same formula holds for the enriched left Kan extension when interpreted correctly (in terms of tensoring over simplicial sets).

In the next step assume that $\mathbf{C}, \mathbf{D}$ and $\mathbf{M}$ are actually a simplicial model categories. This is the setup discussed in:


I would like to see a formula like $(*)$ for the $derived\ enriched$ left Kan extension, preferrably using the derived hom and maybe a replacement of $X$?

For the case where $\mathbf{C}$ and $\mathbf{D}$ are just enriched over simplicial sets, but not treated as model categories, I found a model for the derived enriched left Kan extension in Riehl's $Categorical\ Homotopy\ Theory$ using the bar construction, but can we maybe express it directly using the appropriate homotopical analogon of $(*)$?

Thank you for any hints.

  • $\begingroup$ I would say that the bar construction is the "appropriate homotopical analogon of (*)". $\endgroup$ Mar 14, 2018 at 18:48
  • $\begingroup$ Thank you for your comment. So, but if I use the bar construction, do I see the derived hom somewhere (instead of $\mathbf{D}(f(c),d)$ in (*)? For the applications I have in mind, this is crucial. $\endgroup$ Mar 15, 2018 at 18:41
  • $\begingroup$ If $\mathbf{D}$ is just an ordinary category, what do you mean by "derived hom"? $\endgroup$ Mar 15, 2018 at 21:27
  • $\begingroup$ Sorry, I would like $\mathbf{C}$ and $\mathbf{D}$ to be also simplicial model categories. I fixed this. $\endgroup$ Mar 16, 2018 at 8:15
  • $\begingroup$ The nLab page does not discuss the possibility of C and D being model categories. I don't really even know what the concept means in this case. For one thing, an arbitrary functor between model categories doesn't even induce a functor between their induced $(\infty,1)$-categories. There are two canonical ways that it might, its left and right derived functors. So when you talk about $f:C\to D$ and $X:C\to M$ and the "derived enriched" Kan extension, do you mean a representative for the $(\infty,1)$ Kan extension of their left derived functors? Or right? $\endgroup$ Mar 16, 2018 at 11:21

1 Answer 1


I guess I have to take the challenge. :-)

I think the best approximation to what you are looking for is the calculus of homotopy coherent co/ends developed in

Cordier, Jean-Marc, and Timothy Porter. "Homotopy coherent category theory." Transactions of the American Mathematical Society 349.1 (1997): 1-54.

where the authors define a notion of "coherent end" for a functor $T : A^o\times A\to B$ of $\bf Kan$-enriched categories. But if you look deeply into it, you will notice that the definition relies on the same idea of "resolution of the diagonal" that the bar construction was engineered for (or rather, I can't see how to avoid it). Good news is that $\oint_A T$ (and the things that stem from it: natural transformations, Kan extensions...) are actually intuitive generalizations of what you expect them to be knowing the 1-categorical "calculus".

There are however other meanings in which your question can be interpreted: there are many ways to regard a coend as a functor. You can ask the legitimate question "when is $\int_A$ a Quillen functor?" (it has an adjoint, under assumptions you have for free when your categories carry a model structure) and there's an answer for that. Or more precisely, you can ask the equally legitimate question "when is it possible to derive the weighted limit functors?"; and there's an answer also for that. Have a look at the $n$Lab page on homotopy coends!

  • $\begingroup$ First of all, thank you. I will look at these references and see whether this helps. $\endgroup$ Mar 15, 2018 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.