Very often one has the feeling that set-theoretic issues are somewhat cheatable, and people feel like they have eluded foundations when they manage to cheat them. Even worse, some claim that foundations are irrelevant because each time they dare to be relevant, they can be cheated. What these people haven't understood is that the best foundation is the one that allows the most cheating (without falling apart).
In the relationship between foundation and practice, though, what matters the most is the phenomenology of every-day mathematics. In order to make this statement clear, let me state the uncheatable lemma. In the later discussion, we will see the repercussion of this lemma.
Lemma (The uncheatable).
A locally small, large-cocomplete category is a poset.
The lemma shows that no matter how fat are the sets where you enrich your category, there is no chance that the category is absolutely cocomplete.
Example. In the category of sets, the large coproduct of all sets is not a set. If you enlarge the universe in such a way that it is, then some other (even larger) coproduct will not exist. This is inescapable and always boils down to the Russel Paradox.
Remark. Notice that obvious analogs of this lemma are true also for categories based on Grothendieck Universes (as opposed to sets and classes). One can't escape the truth by changing its presentation.
Excursus. Very recently Thomas Forster, Adam Lewicki, Alice Vidrine have tried to reboot category theory in Stratified Set Theory in their paper Category Theory with Stratified Set Theory (arXiv: https://arxiv.org/abs/1911.04704). One could consider this as a kind of solution to the uncheatable lemma. But it's hard to tell whether it is a true solution or a more or less equivalent linguistic reformulation. This theory is at its early stages.
At this point one could say that I haven't shown any concrete problem, we all know that the class of all sets is not a set, and it appears as a piece of quite harmless news to us.
In the rest of the discussion, I will try to show that the uncheatable lemma has consequences in the daily use of category theory. Categories will be assumed to be locally small with respect to some category of sets. Let me recall a standard result from the theory of Kan extensions.
Lemma (Kan). Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span where $\mathsf{A}$ is small and $\mathsf{C}$ is (small) cocomplete. The the left Kan extension $\mathsf{lan}_f g$ exists.
Kan extensions are a useful tool in everyday practice, with applications in many different topics of category theory. In this lemma (which is one of the most used in this topic) the set-theoretic issue is far from being hidden: $\mathsf{A}$ needs to be small (with respect to the size of $\mathsf{C})$! There is no chance that the lemma is true when $\mathsf{A}$ is a large category. Indeed since colimits can be computed via Kan extensions, the lemma would imply that every (small) cocomplete category is large cocomplete, which is not allowed by the uncheatable. Also, there is no chance to solve the problem by saying: well, let's just consider $\mathsf{C}$ to be large-cocomplete, again because of the the uncheatable.
This problem is hard to avoid because the size of the categories of our interest is as a fact always larger than the size of their inhabitants (this just means that most of the time Ob$\mathsf{C}$ is a proper class, as big as the size of the enrichment).
Notice that the Kan extension problem recovers the Adjoint functor theorem one, because adjoints are computed via Kan extensions of identities of large categories, $$\mathsf{R} = \mathsf{lan}_\mathsf{L}(1) \qquad \mathsf{L} = \mathsf{ran}_\mathsf{R}(1) .$$ Indeed, in that case, the solution set condition is precisely what is needed in order to cut down the size of some colimits that otherwise would be too large to compute, as can be synthesized by the sharp version of the Kan lemma.
Sharp Kan lemma. Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span where $\mathsf{B}(f-,b)$ is a is small presheaf for every $b \in \mathsf{B}$ and $\mathsf{C}$ is (small) cocomplete. Then the left Kan extension $\mathsf{lan}_f g$ exists.
Indeed this lemma allows $\mathsf{A}$ to be large, but we must pay a tribute to its presheaf category: $f$ needs to be somehow locally small (with respect to the size of $\mathsf{C}$).
Kan lemma Fortissimo. Let $ \mathsf{A} \stackrel{f}{\to} \mathsf{B} $ be a functor. The following are equivalent:
- for every $g :\mathsf{A} \to \mathsf{C}$ where $\mathsf{C}$ is a small-cocomplete category, $\mathsf{lan}_f g$ exists.
- $\mathsf{lan}_f y$ exists, where $y$ is the Yoneda embedding in the category of small presheaves $y: \mathsf{A} \to \mathcal{P}(\mathsf{A})$.
- $\mathsf{B}(f-,b)$ is a is small presheaf for every $b \in \mathsf{B}$.
Even unconsciously, the previous discussion is one of the reasons of the popularity of locally presentable categories. Indeed, having a dense generator is a good compromise between generality and tameness. As an evidence of this, in the context of accessible categories the sharp Kan lemma can be simplified.
Tame Kan lemma. Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span of accessible categories, where $f$ is an accessible functor and $\mathsf{C}$ is (small) cocomplete. Then the left Kan extension $\mathsf{lan}_f g$ exists (and is accessible).
Warning. The proof of the previous lemma is based on the density (as opposed to codensity) of $\lambda$-presentable objects in an accessible category. Thus the lemma is not valid for the right Kan extension.
References for Sharp. I am not aware of a reference for this result. It can follow from a careful analysis of Prop. A.7 in my paper Codensity: Isbell duality, pro-objects, compactness and accessibility. The structure of the proof remains the same, presheaves must be replaced by small presheaves.
References for Tame. This is an exercise, it can follow directly from the sharp Kan lemma, but it's enough to properly combine the usual Kan lemma, Prop A.1&2 of the above-mentioned paper, and the fact that accessible functors have arity.
This answer is connected to this other.
Best Answer
One can prove that for any non-zero ring $R$ the category $R$-Mod$^{op}$ is not a category of modules. Indeed any category of modules is Grothendieck abelian i.e., has exact filtered colimits and a generator. So for $R$-Mod$^{op}$ to be a module category $R$-Mod would also need exact (co)filtered limits and a cogenerator. It turns out that any such category consists of just a single object.
I believe this is stated somewhere in Freyd's book Abelian Categories but I am not sure exactly where off the top of my head. Edit it is page 116.
Further edit: For categories of finitely generated modules here is something else, although it is more in the direction of the title of your question than what is in the actual body of the question. Suppose we let $R$ be a commutative noetherian regular ring with unit and let $R$-mod be the category of finitely generated $R$-modules. Then we can get a description of $R$-mod$^{op}$ using duality in the derived category.
Since $R$ is regular every object of $D^b(R) \colon= D^b(R-mod)$ is compact in the full derived category. The point is that $$RHom(-,R)\colon D^b(R)^{op} \to D^b(R) $$ is an equivalence (usually this is only true for perfect complexes, but here by assumption everything is perfect). So one can look at the image of the standard t-structure (which basically just "filters" complexes by cohomology) under this duality. The heart of the standard t-structure is $R$-mod sitting inside $D^b(R)$ so taking duals gives an equivalence of $R$-mod$^{op}$ with the heart of the t-structure obtained by applying $RHom(-,R)$ to the standard t-structure.
In the case of $R =k$ a field then this just pointwise dualizes complexes so we see that it restricts to the equivalence $k$-mod$^{op}\to k$-mod given by the usual duality on finite dimensional vector spaces.
As another example consider $\mathbb{Z} $-mod sitting inside of $D^b(\mathbb{Z})$ as the heart of the standard t-structure given by the pair of subcategories of $D^b(\mathbb{Z})$ $$\tau^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i>0\}$$ $$\tau^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i\leq 0\} $$ It is pretty easy to check that $RHom(\Sigma^i \mathbb{Z}^n, \mathbb{Z}) \cong \Sigma^{-i} \mathbb{Z}^n$ and $RHom(\Sigma^i \mathbb{Z}/p^n\mathbb{Z}, \mathbb{Z}) \cong \Sigma^{-i-1}\mathbb{Z}/p^n\mathbb{Z}$ so that this t-structure gets sent to $$\sigma^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i> 0, \; H^{0}(X) \; \text{torsion}\}$$ $$\sigma^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i< 0, H^0(X) \; \text{torsion free}\} $$ using the fact that objects of $D^b(\mathbb{Z})$ are isomorphic to the sums of their cohomology groups appropriately shifted.
So taking the heart we see that $$\mathbb{Z}-mod^{op} \cong \sigma^{\leq 0} \cap \Sigma\sigma^{\geq 1} = \{X \; \vert \; H^{-1}(X) \; \text{torsion free}, H^0(X) \; \text{torsion}, H^i(X) = 0 \; \text{otherwise}\} $$ with the abelian category structure on the right coming from viewing it as a full subcategory of $D^b(\mathbb{Z})$ with short exact sequences coming from triangles. It is the tilt of $\mathbb{Z}$-mod by the standard torsion theory which expresses every finitely generated abelian group as a torsion and torsion free part.