# Do the adjoint functor theorems usefully dualise?

The special and general adjoint functor theorems exist to construct left adjoints to particular functors given certain conditions on them. However, I've not been able to find much mention – at least, not in my lecture notes nor in a quick Google search – of using them or something like them to construct right adjoints.

I suppose that if $F \dashv G$ then $G^\mathrm{op} \dashv F^\mathrm{op}$, so I can just apply the adjoint functor theorems to the opposites of the categories I'm interested in. Perhaps something like:

General Adjoint Functor Cotheorem: suppose $\mathcal C$ is locally small and cocomplete. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint if and only if it preserves small colimits and, for each object $A$ of $\mathcal C$, the comma category $G \downarrow A$ has a weakly terminal set.

Special Adjoint Functor Cotheorem: suppose $\mathcal C, \mathcal D$ both locally small, $\mathcal C$ cocomplete, co-well-powered1, and with a separating set. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint iff $F$ preserves small colimits.

Is that right? Is it useful?

1 Not sure if this is standard terminology, but I hope it's clear what I mean: for each object $A$, there is a category of epimorphisms with domain $A$, and a category would be co-well-powered if all these epimorphism categories were equivalent to a partially ordered set (they are already preorders, since if $fg = h$ with $g$ and $h$ epimorphisms, then $f$ is unique)

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I'm not sure what the footnote is meant to say, but the answer to your question is 'yes, it dualizes (just like anything else in category theory)'. – Ittay Weiss May 21 '13 at 11:08
Right – the real substance of my question is (a) have I got the dualisations right and (b) are they ever actually used? (why haven't I heard about them?) – Ben Millwood May 21 '13 at 11:12
a) the dualisations are correct. b) which uses of the adjoints functor theorems do you know? – Ittay Weiss May 21 '13 at 11:17
The Stone-Cech compactification from the SAFT, the construction of free groups (and other algebraic structures) from the GAFT (although I'm told that you pretty much need to already know how to build free groups to do that). I know there are more but I can't recall them off the top of my head. I have however just found this mathoverflow.net/questions/17409/… which is very similar to my question. – Ben Millwood May 21 '13 at 11:30
These are theorems, not cotheorems! A theorem starts from some premises and reaches some conclusions. A cotheorem starts from some nclusions and reaches some copremises. – Qiaochu Yuan Dec 12 '15 at 19:42

The dualizations are correct and well-known at least for category theorists. Note that the assumptions in Co-Saft are clearly satisfied if $\mathcal{C}$ is presentable. Thus, left adjoint functors $\mathcal{C} \to \mathcal{D}$ coincide with cocontinuous functors $\mathcal{C} \to \mathcal{D}$. This is a basic and important fact of presentable categories. Here is a nice application: Every presentable monoidal category (i.e. a monoidal category whose underlying category is presentable, and the tensor product should be cocontinuous in each variable) is automatically closed monoidal. In other words, as soon we have a tensor product, we get inner homs for free. In practice, often it is vice versa, but not always.

In my work cocomplete monoidal categories appear where I still have no idea how to describe the inner homs. Nevertheless, they are useful, for example in order to apply Kock's results about monoidal monads. More specifically, if we have a monoidal monad on a cocomplete monoidal category, we can quite easily write down a tensor product on the category of modules, but the proof of associativity is not trivial and actually it seems that we have to impose additional assumptions such as closedness (see arXiv:1205.0101 for details). You surely have seen this for tensor products of modules over a ring in the usual sense: The proof for $M \otimes_R (N \otimes_R K) \cong(M \otimes_R N) \otimes_R K$ only using the definition of the tensor product as a classifying object for bilinear maps is a bit tricky, because first we have to fix some element of $M$ etc, and this implicitly uses the closed structure.

Here is another application: Let $f : X \to Y$ be a morphism of schemes. Then $f^* : \mathsf{Mod}(Y) \to \mathsf{Mod}(X)$ has a right adjoint $f_*$. The functor restricted to quasi-coherent modules $f^* : \mathsf{Qcoh}(Y) \to \mathsf{Qcoh}(X)$ also has a right adjoint. When $f_*(\mathsf{Qcoh}(X)) \subseteq \mathsf{Qcoh}(Y)$, the right adjoint is $f_*$. This happens, for example, when $f$ is quasi-compact and quasi-separated. But even when this is not the case, $f^*$ has a right adjoint, since it is cocontinuous and $\mathsf{Qcoh}(Y)$ is presentable by a result of Ofer Gabber. On the other hand, it is not so easy to write down the right adjoint and probably it's not really useful. Related to that: The inclusion $\mathsf{Qcoh}(X) \hookrightarrow \mathsf{Mod}(X)$ is cocontinuous, hence has a right adjoint. This means that every sheaf of modules has a "quasi-coherator". For example, if $X$ is affine, then the quasi-coherator of $M \in \mathsf{Mod}(X)$ is $\widetilde{\Gamma(X,\mathcal{O}_X)}$. See here for more about that.

If $f : X \to Y$ is a morphism of schemes, which is affine, then $f_* : \mathsf{Qcoh}(X) \to \mathsf{Qcoh}(Y)$ is cocontinuous (actually the converse is also true by a classical result of Serre, generalized by Rydh), hence has a right adjoint. But it seems that you can only write it down when $f$ is finite. Then the right adjoint is $f^{!}(M):=\underline{\hom}_{\mathcal{O}_Y}(f_* \mathcal{O}_X,M)$. If $f$ is not finite, we have to take the quasi-coherator of $f^!(M)$.

It seems that in each application of Co-Saft, the right adjoint is either "pathological" or we can write it down directly anyway and showing the adjunction directly.

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Thanks, this is very useful, will accept if nothing else shows up in the next few days. Although I wonder if in your concluding line, "pathological" means "we cannot write it down directly" in which case it's something of a tautology :) – Ben Millwood May 22 '13 at 13:48

as a sort of meta answer: there are more left adjoints than right adjoint (this is of course nonsense!!). More to the point, quite often we are interested in finding left adjoints to functors (in order to create free objects (which include all of the uses you mentioned)). Also quite often, the functors that typically interest us do not have right adjoints at all. This is related to the fact of life that mathematics is not self-dual. For whatever reason, many of the naturally occurring functors we meet tend to have left adjoint but often they lack right adjoints.

So, while category theory is self-dual, practiced mathematics is not. A similar phenomenon is the common concept of a cofibrantly generated model category, while there are very very few examples of naturally occurring categories that support the dual notion of a fibrantly generated model category.

Notice that this lack of duality in practiced mathematics starts very early on. The concept of cartesian product of sets is much more important than the dual notion of disjoint union. And while noticing this, it is the cartesian product of set that is at the heart of the definition of category. So the bias towards one concept over its dual is very deeply rooted.

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