Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

First of all, let me see if i got the 1-categorical version right:

  • Let $\mathcal F:C\to Cat $ be a (pseudo-) functor. The 2-colimit $\mathrm{colim}_C\mathcal F$ is then given by the grothendieck construction $\int_C \mathcal F$ and the 2-limit is given by the category of cartesian sections of the fibration $\int_C \mathcal F\to C$, right?

Can this be transported to the setting of dg$_k$-categories? So:

  1. Is there a notion of fibration of dg categories? I would imagine them to be algebras for a dg-monad $(\mathrm{id}_C,-)$ arising from forming dg-comma categories with the identity-span on $C$.
  2. What about a grothendieck construction for functors $\mathcal F: C\to \mathrm{dgCat}$ from a category $C$ to dg-Categories?
  3. Cartesian sections should then be defined as algebra-morphisms from the identity on $C$ to $\int_C \mathcal F$.

I think there are some problems with what i just said: What are dg-comma categories? What are the right functors $C\to\mathrm{dgCat}$? (I guess one should build a dg-category $C'$ out of $C$ by taking the free $k$-category and then consider it as a dg-category, concentrated in degree 0, and then consider $\mathrm{dgCat}(C',\mathrm{dgCat})$) Same goes for the definition of an algebra morphism: What are the coherences to consider?

So i guess the core question is:

  • What is the right notion of limit for dg categories? (And why?)
share|improve this question
1  
The Grothendieck construction is the "oplax-colimit", to get the 2-colimit, you must invert the Cartesian morphisms. –  Omar Antolín-Camarena Feb 25 '13 at 18:21
    
Hm... wether it is the lax- or oplax-colimit depends on how you define the grothendieck construction: you chose the vertical part of a morphism to be either $f_*x\to x'$ or $x' \to f_*x$; both work equally and we have $\int'_C F = \int_C op \circ F$ –  Garlef Wegart Apr 22 '13 at 19:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.