# Construction Of Opposite Category as a Structure

I am delving a bit into category theory and something has me curious about opposite categories. I have checked several books and I can't seem to find an answer.

Given a category C, the opposite category is just the abstract category with the objects of C and with the arrows of C reversed. However, the opposite category can sometimes be realised (is equivalent to) a category where the objects are sets with additional structure, and the arrows are homomorphisms of these structures.

For instance one of the examples on Wikipedia is that the opposite category of commutative rings is equivalent to the category of affine schemes.

Question. How would one in general, given a category C, find a category where the objects are mathematical structures with underlying sets that satisfy additional axioms, and the morphisms are homomorphisms of those structures, and which is the opposite of the original category C?

For instance, if we take the category where the objects are groups, and the morphisms are group homomorphisms, what is its opposite category in the above sense? Is there some way to find this from the first-order axiomatization of groups?

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There was a notion previous of that of category which was introduced byy (I think) Bourbaki and it was that of set with structure (in a precise mathematical sense). But these have been almost forgotten, since one of the advantages of abstract categories is really that you do not have to worry about how they relate with the category Set explicitly. – Andrea Ferretti May 3 '10 at 18:30
@Andrea: Concrete categories still have a rich theory! Check out joy of cats. @CrazyHorse: Very good question, 1+. I think you don't ask for a general method how to concretize the dual of a concrete category (perhaps this is impossible at all!), but rather for a big list of examples where this is possible. I'm very curious about the answers! So far I have the impression that duals of algebraic categories became concrete after adding geometry to it. You may be interested in Johnstone's book Stone spaces. – Martin Brandenburg May 3 '10 at 18:58
Charles Staats brings up an excellent point in the comments to Martin Brandenburg's answer that bears repeating: depending on your definition of "mathematical structures with underlying sets that satisfy additional axioms," affine schemes may fail to fit this definition! – Qiaochu Yuan May 4 '10 at 2:42
(More precisely, a map between affine schemes is not determined by what it does to points; one also needs to know what happens to the structure sheaf. This should be clear if you think about what morphisms between fields look like in the affine scheme picture.) – Qiaochu Yuan May 4 '10 at 2:56
@Qiaochu: I think you misunderstood Charles' comment. It says two things: a) The concretization of $Rng^{op}$ reuses $Rng$. b) A scheme is not determined by the underlying topological space. Nevertheless, of course, the category of affine schemes is a category of mathematical structures with underlying sets that satisfy additional axioms! The sets are ordered pairs $(X,O_X)$, and so on ... – Martin Brandenburg May 4 '10 at 8:06

One way of formalizing the desired categories is given by concrete categories. A category is called concrete (more precise: "concretizable") if it has a faithful functor to the category of sets $Set$. Thus the question is: Is the dual of a concrete category concrete again? The answer is yes: Since a composition of faithful functors is faithful and a dual of a faithful functor is also faithful, it suffices to show that $Set^{op}$ is concrete. But it is not hard to see that the contravariant hom-functor $Hom(-,2)$ (i.e. the power set) yields the desired faithful functor $Set^{op} \to Set$.

However, this solution is somewhat useless. If we apply the proof to $Grp^{op}$, we get sets of the form $P(X)$, where $X$ is a group and morphisms $P(X) \to P(Y)$ should be induced by group morphisms $Y \to X$.

Perhaps we should demand of our concretization that it does not reuses the given category? But this seems to be hard to formalize. Anyway, in the category of groups it would be interesting ...

EDIT: What about the following: If $k$ is a field, then $(k-Vect)^{op}$ is equivalent to the category of pairs $(X,p)$, where $X$ is an affine $k$-scheme and $p$ is a rational point of $X$ such that the corresponding maximal ideal $a$ satisfies $a^2=0$. :-) If $R$ is a ring, then $(R-Mod)^{op}$ is equivalent to the category of pairs $(X,p)$, where $X$ is an affine $R$-scheme, $p$ is a $R$-valued point of $X$ such that the closed image of $X \to Spec(R)$ is $Spec(R)$ and and the closed image of $p : Spec(R) \to X$ is cut out by an ideal $a \subseteq \mathcal{O}_X(X)$ with $a^2=0$. What about dropping the affine-condition, do we get "global modules"? Abstract-Nonsense!

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I think a more specific interpretation of the OP's question is something like "is the dual of the category of models of a Lawvere theory also the category of models of a Lawvere theory?" – Qiaochu Yuan May 4 '10 at 0:44
If you demand that the concretization does not reuse the given category, then, arguably, you are also excluding the example of affine schemes. Come to think of it, the usual functor from affine schemes to Set, given by taking the underlying topological space, is not faithful, so it's not precisely even a concretization. – Charles Staats May 4 '10 at 2:14
@Qiaochu: I didn't know Lawvere theories, thank you for mentioning them. I like your interpretation of the original question. @Charles: I should not have said "reuse". Perhaps I just meant that you just don't rephrase the definition and use duals which do not fit to your category (yes, this is not precise...). – Martin Brandenburg May 4 '10 at 8:09
@Qiachu: Typically, the answer is negative. A category of models of a Lawvere theory is always regular and it has finite limits, but these properties are not easily preserved by passing to the opposite (which turns limits to colimits, monos to epis, etc.) – Andrej Bauer May 4 '10 at 8:38
Martin, you qualify your suggestion as possibly useless. But that's exactly what should be expected of such a general construction. It's in fact a rather nice answer that helps people overcome the feeling that concrete categories are somehow "better". – Andrej Bauer May 4 '10 at 8:40

I am going to spell out Martin's construction with minimal use of category-theoretic terminology (such as "faithful" and "representable") because it's exactly what, uhm, CrazyHorse, asked for. (I can't believe I am talking to a crazy horse.)

Take a concrete category $\mathbf{C}$. Its objects are of the form $(X,S_X)$ where $X$ is a carrier set and $S_X$ is some additional structure on $X$. Morphisms are functions between carrier sets that are "structure preserving", whatever that means. Its opposite $\mathbf{C}^\mathrm{op}$ is equivalent to the following concrete category $\mathbf{D}$:

• an object of $\mathbf{D}$ is a pair $(P(X), (X,S_X))$ where $P(X)$ is the powerset of $X$ and $(X,S_X)$ is an object of $\mathbf{C}$. That is, the additional structure of an object in $\mathbf{D}$ is an object of $\mathbf{C}$.

• a morphism $f : (P(X), (X,S_X)) \to (P(Y), (Y,S_Y))$ in $\mathbf{D}$ is a function $f : P(X) \to P(Y)$ for which there exists a morphism $g : (Y,S_Y) \to (X,S_X)$ in $\mathbf{C}$ such that $f = g^{-1}$. (Note: for any given $f$ there exists at most one such $g$.)

The moral is: a general answer to a general query is generally not very useful. Of course, in particular cases there will be other, more useful, categories which are equivalent to $\mathbf{C}^\mathrm{op}$.

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Qiaochu proposed models of Lawvere theories to interpret the desired categories. I think they are (up to equivalence) the same as finitary algebraic categories. Here is my proof that $Set^{op}$ is not such a category:

Assume $Set^{op}$ is equivalent to some finitary algebraic category $C$. Then $C$ is complete and there is $x \in C$ such that every object in $C$ is isomorphic to some power $x^I$. Besides, morphisms $f : x^I \to x^J$ are induced by a unique map $\sigma : J \to I$ such that $f(a)=(a_{\sigma(j)})_{j \in J}$. Since $C$ is finitary algebraic, $x^I$ is the usual cartesian product of copies of $x$. In particular, if $d=|x|$, then every object of $C$ has cardinality $d^p$ for some cardinal number $p$. Observe that $d$ cannot be $0$ or $1$ since otherwise $Set^{op}$ and thus $Set$ had a finite skeleton. In particular, there are lots of infinite objects in $C$. However, their cardinality is restricted!

Now the key tool is the Theorem of Löwenheim-Skolem (upwards). It yields the existence of an infinite cardinal number (depending on the signature of $C$), such that every cardinal number above arises as the cardinality of an object in $C$. In particular, every sufficient large infinite cardinal number $x > 2^d$ has the form $d^p$ for some infinite cardinal number $p$. If $p < d$, then $d^p \leq 2^{dp} \leq 2^d \aleph_0 < x$, a contradiction. Thus $d \leq p$ and thus $x = d^p = 2^p$. Now, assuming GCH (which is probably not needed, but it makes the proof work), $2^p = p^+$ is regular. Thus every sufficient large cardinal number is regular, which is certainly false, since $\aleph_{\alpha + \omega}$ is singular for every ordinal number $\alpha$.

I hope it's correct.

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Here is another reason why the opposite of a category of finitary algebras isn't itself such a thing: mathoverflow.net/questions/29442/… – Peter Arndt Sep 26 '10 at 14:31

In your request you suppose that the category is concrete i.e. has a faithul functor on $Set$ (for example particular make of structured sets and morphisms, or models (of a theory) on Set or on topological space or else), but this isn't true in general (countrexample are from Homotopy). Anyway the Yoneda lemma (and immersion) is "like a extension" that relate a category $C$ by $Set$, in this way a model of the dual $C^{op}$ as the essentially some objects, but you can "concretizing" the morphisms: in $C^<=CAT(C, Set)$ consider the subcategory generated by $h^X,\ X\in C$. However the dual of a concrete category is still concrete (considering the covariant Power-set functor $Set\to Set^{op}$.

A more hard quation is: Gived a theory $T$ and its category of model $C(T)$ in Set, there is a "dual" theory $T'$ such that $C(T)^{op}$ is embeddable in $C(T')$ ? But from elementary example (Pontriagjn duality ecc) we need also a "more high" category for the base of $T'$ models. In this philosophy there is a large study of how make categorical model of a theory, see P. Johnstone "Stone Spaces" or "Sketches of an Elephant: A Topos Theory Compendium".

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