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Why is the pullback not just defined as in the case of the category of sets

Perhaps there are some issues with categories that are not small, but what about small categories? My first thought was: perhaps all small categories do not have products. Do they? If not, why doesn't the cartesian product just do the trick?

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What does "cartesian product" mean in an arbitrary category? – Chris Eagle Jun 20 '11 at 19:57
You probably just mean 'product' (categorical product) instead of Cartesian product. Anyway, I don't understand the question: the pullback is defined as it is. In SET it happens that we can describe it as a quotient of a product. Perhaps you mean something like: can we do the same in any category which has products? But I don't know what a 'relation' in an arbitrary category is. – wildildildlife Jun 20 '11 at 20:08
@wildildildlife: it's a subobject of a product, not a quotient. – Chris Eagle Jun 20 '11 at 20:25
up vote 10 down vote accepted

There are a few problems here:

  1. Not every category is sets with additional structure. Many familiar examples are: Groups, topological spaces, and just about every other "elementary" examples of categories that you came across when you first learned what a category was. These categories are all examples of concrete categories, which are categories equipped with a faithful functor to sets. Even though concrete categories are more general than "sets with structure", not every category is concrete. For example, the category of topological spaces where morphisms are homotopy classes of continuous maps cannot be made into a concrete category. Also, there is no faithful functor from $Set^{op}$ to $Set$ which is the identity on objects, and so if there is a way to turn $Set^{op}$ into a concrete category (which I do not believe there is), it is at least non-obvious.

  2. Even if a category was just sets with additional structure, the pullback of two sets might not have the structure required. If you have a concrete category and the forgetful functor is right adjoint to some "free" functor, then because right adjoints preserve limits, you will in fact have that the underlying set of the pullback is the pullback of the underlying sets. However, colimits won't work out so well.

  3. Even if you are in a category of structured sets, and even if the set theoretic pullback is still in your category, you might not have that universal property that you want to have, which would take away some of the usefulness. Of course, this doesn't happen when the forgetful functor is part of an adjunction as above, and I am hard pressed to think of any examples (because it is much more likely that things break in spectacular ways than in subtle ways).

  4. Not every category even has products. For example, in the category of fields, all morphismsare field extensions, and so to have projection map, the product of two fields would have to be a common subfield. However, unless you consider the field with one element to be a field, there are no common subfields of two fields of different characteristic. From a different perspective, the ring that you get by taking the Cartesian product of two fields is NOT a field.

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$Set^{op}$ is concretized by the functor that is powerset on objects and inverse image on morphisms. – Chris Eagle Jun 20 '11 at 21:03
@Chris Eagle: Thank you, that's very nice. – Aaron Jun 20 '11 at 21:14

Your intuition is right in the sense that pullbacks can be constructed from products and equalizers, see relation between pullbacks and other categorical limits, proposition 2. Products are generalizations of Cartesian products of sets. Equalizers are generalizations of specific subsets; I do not understand what is “modulo a relation”, but I feel equalizers are what you need. :)

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