Adjunctions form a stack - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:20:57Z http://mathoverflow.net/feeds/question/68791 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68791/adjunctions-form-a-stack Adjunctions form a stack Martin Brandenburg 2011-06-25T15:22:47Z 2011-06-28T00:45:13Z <p>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.</p> <p><strong>Question.</strong> 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 $F \to G$?</p> <p>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 <a href="http://mathoverflow.net/questions/39300/pointwise-left-adjoint-yields-a-pseudo-functor" rel="nofollow">former</a> questions.</p> http://mathoverflow.net/questions/68791/adjunctions-form-a-stack/68816#68816 Answer by Finn Lawler for Adjunctions form a stack Finn Lawler 2011-06-25T22:16:53Z 2011-06-25T22:16:53Z <p>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 <a href="http://ncatlab.org/nlab/show/doctrinal+adjunction" rel="nofollow">doctrinal adjunction</a> 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.</p> <p>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.</p> http://mathoverflow.net/questions/68791/adjunctions-form-a-stack/68833#68833 Answer by james-parson for Adjunctions form a stack james-parson 2011-06-26T03:51:48Z 2011-06-27T03:15:11Z <p>(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.)</p> <p>I'll write something very concrete, perhaps more in the style of Vistoli's notes or SGA 1.</p> <p>How certain are you that you need a <em>left</em> adjoint? If you look at <em>right</em> 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.</p> <p>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.</p> <p>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 $\Phi$.</p> <p>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.</p> <p>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 $U$.</p> http://mathoverflow.net/questions/68791/adjunctions-form-a-stack/68882#68882 Answer by Mike Shulman for Adjunctions form a stack Mike Shulman 2011-06-26T22:16:03Z 2011-06-26T22:16:03Z <p>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).</p> <p>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</p> <ol> <li><p>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</p></li> <li><p>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).</p></li> </ol> <p>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.</p> <p>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 <em>colax-idempotent</em>, which means that <em>every</em> 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).</p> <p>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 <em>lax</em> natural transformations between pseudofunctors in terms of their corresponding fibrations.</p> <p>(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 <a href="http://nlab.mathforge.org/nlab/show/generalized+multicategory" rel="nofollow">generalized multicategory</a>).</p> <p>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</p> <ol> <li><p>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.</p></li> <li><p>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.</p></li> </ol> <p>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 $K_1$.</p> <p>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 <em>colax</em> $T_1$-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 <em>pseudo</em> $T_2$-morphism, i.e. it will not be a morphism of fibrations.</p> http://mathoverflow.net/questions/68791/adjunctions-form-a-stack/68991#68991 Answer by David Carchedi for Adjunctions form a stack David Carchedi 2011-06-28T00:45:13Z 2011-06-28T00:45:13Z <p>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.</p> <p>See Borceux's "handbook of categorical algebra 2," 8.4.2.</p> <p>This is also referenced in this preprint by the Kock family:</p> <p><a href="http://arxiv.org/abs/1005.4236v1" rel="nofollow">http://arxiv.org/abs/1005.4236v1</a></p> <p>(See 1.3)</p>