If not, are there any interesting subcategories that can be concertized? If I am not mistaken, the category of reduced finite type varieties over the complex numbers would be an example, where the forgetful functor to sets would be given by looking at the underlying map of points.
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The category of schemes is not small-concrete. Let S be a generating set. Let U be the set of all rings A such that Spec A is an open subset of a scheme in S. Let X be a set whose cardinality is larger than any element of U, for example, 2^{\bigsqcup_{A \in U} A}. Let K be the field Q(t_x)_{x \in X}, where t_x are a collection of algebraically independent generators indexed by X. So |K| is larger than |A| for any A in U. Since ring maps from a field are always injective, Hom(Spec A, Spec K)={} for every A in U, and therefore Hom(s, Spec K)={} for every s in S. There is only one map from the empty set to itself. But Spec K has nontrivial isomorphisms, coming from permuting the generators. So Hom(Spec K, Spec K) ---> SetHom( \bigsqcup_{s \in S} (Spec K)(S), \bigsqcup_{s \in S} (Spec K)(S)) is not injective. |
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I'd like to suggest that this isn't quite the right question. At least, it seems to me that modifying the question (in a direction that Theo was hinting) would be more interesting. The problem with the question as asked is that, for a given category C, the mere existence of a faithful functor C --> Set tells you very little indeed. Perhaps you have some reason for wanting to know that I can't see. But a condition that seems to have more bite is 'small-concreteness', defined as follows. Let C be a category. A set-valued functor U: C --> Set is small if it can be expressed as a small colimit of representables. Call a category C small-concrete if there exists a small, faithful functor C --> Set. In the special case that C is small, all set-valued functors on C are small and small-concrete = concrete. It's not too hard to show that a category is small-concrete if and only if it admits a generating set. (A generating set in a category C is a [small] set S of objects such that, for any distinct maps f, g: a --> b in C, there exist s in S and q: s --> a such that fq \neq gq.) The existence of a generating set is one of the conditions in the Special Adjoint Functor Theorem: see Categories for the Working Mathematician. You can exploit this as follows. Suppose you want to show that the category of affine schemes is not small-concrete (which would imply that the category of all schemes isn't either). Assuming for a contradiction that it is small-concrete, the category Ring of commutative rings has a cogenerating set. Since Ring is locally small and small-complete, the Special Adjoint Functor Theorem tells us that every limit-preserving functor from Ring to a locally small category has a left adjoint. I guess it's possible to cook up (or look up) an example of a limit-preserving functor out of Ring that doesn't have a left adjoint. That would produce the desired contradiction. |
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Here is a concretization of Ring^{op}: For any ring R, let 2^R be the set of subsets of R. Given a map f: R --> S , we define 2^f : 2^S --> 2^R by I --> f^{-1}(I). I claim that this is faithful. Proof: let f and g be two different maps R --> S. So there is some r in R with f(r) \neq g(r). But then 2^f( {f(r)} ) contains r and 2^g( {f(r)} ) does not contain r, so 2^f \neq 2^g. I want to say that I can extend the functor 2^* to schemes by taking a directed limit over all open affines, but I am nervous about the details. |
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I don't have an answer, but by Yoneda, it reduces to the question of whether the category of functors from commutative rings to sets has a faithful functor to sets. |
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Ok, ignoring my deleted answer where I misread the question, I don't have a full solution (working on some details), but to extend the functor you describe much beyond actual, literal varieties over C, you'll need to handle the case of Spec(k[x]/x^2), which has the property that a map from it to a variety is a tangent vector, so you'll need to find some set S such that hom(S,V), as sets, has a map for every tangent vector to V, for any V. I don't see how to satisfy it, but I've not yet got a proof that you can't, and it doesn't say anything about functors not being the forgetful functor on C-varieties, though it gives a possible way to show that it can't extend much. |
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