Adjunctions form a stack - MathOverflow

Adjunctions form a stack

2011-06-25T15:22:47Z

Let $C$ be a base category, $F,G$ be two categories fibered over $C$ and $F \to G$ be a morphism. The following criterion is used very often: If all the fiber functors $F_U \to G_U$ ($U \in C$) are equivalences, then $F \to G$ is an equivalence. Although most books seem to use this as it is was trivial, a little work has to be done; see for example the notes by Angelo Vistoli on fibered categories, Section 3.5.

Question. Assume that all fiber functors $F_U \to G_U$ have a left adjoint $G_U \to F_U$. Is it possible to extend these to a left adjoint $G \to F$ of $F \to G$?

Somehow this feels like the assertion "adjunctions form a stack". I don't know if this is true at all: Somehow we have to arrange that the units $1 \to F_U G_U$ and counits $G_U F_U \to 1$ are compatible in $U$, but is this possible? Any references (at last, this should be well-known) are welcome. Note the similarity to one of my former questions.

Answer by Finn Lawler for Adjunctions form a stack

2011-06-25T22:16:53Z

Following up on my comment above, I think you do indeed get a lax transformation $G \to F$ that won't be pseudo in general. In fact, this seems to be a case of doctrinal adjunction for the 2-monad on $[\operatorname{ob} C, \mathrm{Cat}]$ whose algebras are Cat-valued functors $C \to \mathrm{Cat}$. Doctrinal adjunction says that if you have a 2-monad T on a 2-category K, and an adjunction $f \dashv g$ in K, then there is a bijection between 2-cells that make f a colax morphism of T-algebras, on the one hand, and 2-cells that make g a lax morphism on the other; moreover, the entire adjunction lives in T-Alg if and only if f is a pseudo morphism of algebras and the 'colax' part of f's structure 2-cell is the mate of u's (lax) structure 2-cell.

In your situation, $F \to G$ is a pseudo morphism of algebras, and the adjunctions $F_U \to G_U \dashv G_U \to F_U$ form a single adjunction in $[\operatorname{ob} C, \mathrm{Cat}]$, so that by the above the right adjoints make up a lax transformation, but the data you have isn't enough to make it pseudo.

Answer by james-parson for Adjunctions form a stack

2011-06-26T03:51:48Z

(Edit: As Mike Shulman points out, I switched my left and right in what follows. Furthermore, since I don't produce a morphism of fibrations, I probably do not even address the original question, regardless of the chirality confusion. Sorry! His answer is clearly a much better response.)

I'll write something very concrete, perhaps more in the style of Vistoli's notes or SGA 1.

How certain are you that you need a left adjoint? If you look at right adjoints instead, then the result proves itself. Unless I am misreading Finn Lawler's response, it looks as if he also switched to right adjoints, but since he doesn't mention it, perhaps I am misunderstanding. I'll write out the details to produce the right adjoint, but it is probably easier to work it out yourself than to read what follows.

I'm guessing that when you write "morphism $F\to G$," you mean a cartesian functor over $C$. (I looked up your reference to Vistoli's exposition, and this seems to be his usage.) If that guess is wrong, then what I say below may not be relevant, since I use the cartesian property repeatedly.

Let $\Phi: F \to G$ be a cartesian functor between categories fibered over $C$. Assume that for each object $U$ of $C$, the functor $\Phi_U:F_U \to G_U$ on fiber categories admits a right adjoint. We wish to show that $\Phi$ admits a right adjoint. Let $y$ be an object of $G$ over the object $U$ of $C$. We need to produce a universal arrow $y\to \Phi x$ from $y$ to $\Phi$. The obvious candidate is a universal arrow $y\to \Phi_U x = \Phi x$ to $\Phi_U$, which exists by hypothesis (after switching your "left" for my "right"). We must prove that this arrow is universal to $\Phi$.

Let $y\to \Phi x'$ be a morphism in $G$ covering $\rho: U\to U'$ in $C$. Let $\rho^* x'\to x'$ be a pullback in $F$. Since $\Phi$ is cartesian, $\Phi(\rho^* x') \to \Phi x'$ is a pullback in $G$, and so $y\to \Phi x'$ factors uniquely as the composition of a morphism $y \to \Phi(\rho^* x')$ in $G_U$ with $\Phi(\rho^* x')\to \Phi x'$. By the universal property of $y \to \Phi x$ over $U$, there is a unique morphism $x \to \rho^* x'$ in $F_U$ such that $y \to \Phi(\rho^* x')$ factors as the composition of $y\to \Phi x$ and $\Phi x\to \Phi(\rho^* x)$. Composing $x\to \rho^* x'$ and $\rho^* x'\to x$ gives us a morphism $x\to x'$ covering $\rho$ such that $y \to \Phi x'$ factors as the composition of our candidate universal arrow $y \to \Phi x$ and $\Phi x \to \Phi x'$. It remains to see that such a $x\to x'$ is unique.

Suppose we have two morphisms $x\to x'$ in $F$ such that $y\to \Phi x'$ factors as the composition of $y\to \Phi x$ with either of the corresponding $\Phi x \to \Phi x'$. Since $y\to \Phi x$ is vertical, we find that $\Phi x \to \Phi x'$ covers $\rho$. Since $\Phi$ is a functor over $C$, we find that both morphisms $x\to x'$ cover $\rho$. Thus it suffices to show that the two morphisms $x\to \rho^* x'$ through which our morphisms $x\to x'$ factor are equal. Using the fact that $\Phi(\rho^* x')\to \Phi(x')$ is cartesian (as I am assuming $\Phi$ is cartesian), one sees that the $y/\Phi(x)/\Phi(x')$ picture (with morphisms covering $\rho$) pulls back to a $y/\Phi(x)/\Phi(\rho^* x')$ picture in the fibers over $U$. Now the desired uniqueness follows from the universal property of $y\to\Phi(x)$ over $U$.

Answer by Mike Shulman for Adjunctions form a stack

2011-06-26T22:16:03Z

Finn is absolutely right that this is a case of doctrinal adjunction, but the situation is complicated by the fact that there are two 2-monads we might be considering. First, there is the 2-monad $T_1$ on the 2-category $K_1=[\mathrm{ob} C, \mathrm{Cat}]$, whose (pseudo)algebras are (pseudo)functors $C^{\mathrm{op}} \to \mathrm{Cat}$. A lax/colax/pseudo morphsim of $T_1$-algebras is precisely a lax/colax/pseudo natural transformation (which, by the way, is one way to remember the correct meanings of lax vs. colax in the latter case).

In this case, doctrinal adjunction tells us that if F and G are (pseudo) $T_1$-algebras and $\Phi\colon F\to G$ is a pseudo $T_1$-morphism, then

If $\Phi$ has a left adjoint in the underlying 2-category $K_1$, then this left adjoint automatically has a structure of colax $T_1$-morphism (this only requires $\Phi$ to be lax), while
If $\Phi$ has a right adjoint in the underlying 2-category $K_1$, then this right adjoint automatically has a structure of lax $T_1$-morphism (this only requires $\Phi$ to be colax).

Note that having an adjoint in $K_1$ is precisely the hypothesis you proposed: that each fiber functor $\Phi_U$ has an adjoint. In general, it is not automatic in either case above that the adjoint $T_1$-morphism is pseudo; the invertibility of its (co)lax structure map is an additional condition to be imposed.

Secondly, there is a different 2-monad $T_2$ on the 2-category $K_2 = \mathrm{Cat}/C$, whose (pseudo)algebras are (cloven) fibrations. This 2-monad is what's called colax-idempotent, which means that every morphism in $K_2$ between $T_2$-algebras admits a unique structure of colax $T_2$-morphism. It follows (not entirely obviously) that any lax $T_2$-morphism is actually pseudo (its unique colax structure is necessarily an inverse to any lax structure one might give it).

One can check that pseudo $T_2$-morphisms are exactly morphisms of fibrations (i.e. functors over $C$ preserving cartesian arrows), and therefore correspond (under the equivalence between fibrations and pseudofunctors) to pseudo natural transformations, i.e. to pseudo $T_1$-morphisms. Similarly, colax $T_2$-morphisms (that is, arbitrary functors over $C$) correspond to colax natural transformations (colax $T_1$-morphisms); but there is no way to represent lax natural transformations between pseudofunctors in terms of their corresponding fibrations.

(As an aside, the passage from $T_1$ to $T_2$ is an instance of a general construction which, given a well-behaved 2-monad, produces a new (co)lax-idempotent one with the same algebras. The key word to look for is generalized multicategory).

Now, doctrinal adjunction applied to $T_2$ says that if $\Phi\colon F\to G$ is a pseudo $T_2$-morphism (that is, a morphism of fibrations), then

If $\Phi$ has a left adjoint in the underlying 2-category $K_2$, then that left adjoint automatically becomes a colax $T_2$-morphism, which is no condition at all since $T_2$ is colax-idempotent.
If $\Phi$ has a right adjoint in the underlying 2-category $K_2$, then that right adjoint automatically becomes a lax $T_2$-morphism, and therefore a pseudo one.

This second case is the only one in which we can deduce that the adjoint is actually a pseudo morphism. Note, though, that the hypothesis that $\Phi$ has an adjoint in $K_2$ is stronger than that it has an adjoint in $K_1$.

So what about james-parson's answer? First of all, although he mentions right adjoints, his argument is actually about left adjoints: a universal arrow $y\to \Phi x$ means that x is the value at y of a left adjoint to $\Phi$ (the universal arrow being the adjunction unit). He assumes a left adjoint in $K_1$, which by the general nonsense above should only allow us to deduce the existence of a colax $T_1$-structure on this adjoint. But this is equivalent to a colax $T_2$-structure, which (since $T_2$ is colax-idempotent) is equivalent to just saying that we have an adjoint in $K_2 = \mathrm{Cat}/C$. And in fact, that is exactly what he constructs. In general, this adjoint will not be a pseudo $T_2$-morphism, i.e. it will not be a morphism of fibrations.

Answer by David Carchedi for Adjunctions form a stack

2011-06-28T00:45:13Z

There is a a simple criterion for such a fiber-wise adjunction to extend to an internal adjunction in the 2-category of fibered categories (over your fixed base category). I believe this is what you should be asking.

See Borceux's "handbook of categorical algebra 2," 8.4.2.

This is also referenced in this preprint by the Kock family:

http://arxiv.org/abs/1005.4236v1

(See 1.3)