Are fibrations coreflective in a 2-category?

Question

The notion of a Grothendieck fibration in the 2-category $CAT$ of small categories, can be written down to make sense for any 2-category, and such a morphism in a 2-category is called a fibration:

http://ncatlab.org/nlab/show/fibration+in+a+2-category

Now, in the case of $CAT$, consider a category $C$, then I believe that the 2-category of Grothendieck fibrations over $C$ is a (non-full) coreflective 2-subcategory of the slice 2-category $CAT/C$ (where morphisms from $D \stackrel{f}{\rightarrow}C$ to $D' \stackrel{g}{\rightarrow}C$ are pairs $(h,\alpha)$, such that $\alpha:gh \Rightarrow f$). At least, this seems to be true if $C$ is a groupoid.

My argument is to describe explicitly the right-adjoint which sends a functor $f:D \to C$ to the Grothendieck construction of the presheaf of categories $c \mapsto Hom_{Cat/C}\left(C/c\stackrel{}{\searrow}C,D\stackrel{f}{\searrow}C\right).$

If this doesn't work in general, it should at least if $C$ is a groupoid.

Now, let $E$ be a 2-category (if need be, we can assume its a (2,1)-category). Is is true that for every $e$ in $E$, that the 2-category of fibrations over $e$ is coreflective in $E/e$?

P.S. If this is not true in general, when is it true?

Answer 1

First of all, I don't think the phrase "non-full coreflective subcategory" has any meaning. A reflective or coreflective subcategory is full, by definition. Since fibrations are not a full subcategory, the question to ask is instead whether they are comonadic, and as Finn says this would follow from the fact that they are monadic (which is true whenever the 2-category E has finite limits) as soon as the monad has a right adjoint.

(You do need to use the usual slice 2-category, not the lax version, although of course the two are the same in a (2,1)-category or even just when C is groupoidal. And in fact, in that case fibrations are a full subcategory, so you can call them reflective or coreflective.)

However, for a general 2-category, the monad will not have a right adjoint. (It would, indeed, be a right Kan extension in Span(E), but such an extension might not exist.) It does, however, if fibrations and opfibrations in E are exponentiable; see e.g. here.

Answer 2

Sketch of an answer: Remember that $E/e$ is equivalent to $\operatorname{Span} E(e,1) $, and that the (lax-idempotent) 2-monad whose algebras are fibrations is given by composition with the span $\Phi e \colon e^{\mathbf{2}} \rightrightarrows e $ where the domain is the cotensor of e with the arrow category $\mathbf{2} = \bullet \to \bullet$. This functor $\Phi e \otimes -$ should have a right adjoint given by the right Kan extension (in $\operatorname{Span} E$) $\mathsf{ran}_{\Phi e} p$ of p along $\Phi e$. Now what you want is an Eilenberg--Moore-type theorem to tell you that a right adjoint to a 2-monad is a 2-comonad, and that their categories of (co)algebras are equivalent, which will give you your right adjoint to the forgetful functor.

Edit: Actually, this is all only true for E/e the usual slice 2-category, with (essentially) commuting triangles as morphisms, not the 'lax' version you mention, where the triangles contain a non-invertible 2-cell.

Answer 3

Your problem (where the $Hom$ in your definition is restricted of functors mapping any morphism in cartesians) is in G. Giraud "Le metode de la descente" t.5.9 p. 72 (2">http://www.numdam.org/numdam-bin/feuilleter?id=MSMF_1964_2) where is showed that your funtor (restricted as above) has a left-adjoint.