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From Analysis and Its Foundations By Eric Schechter

A concrete category consists of a collection of objects and a collection of morphisms.

I am curious what an "abstract category" is? I found the Wikipedia page for category theory seems only about concrete categories.

Thanks and regards!

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Do these help abc-of-categories-abstract-vs-concrete and The Joy of Cats? –  Amzoti Jan 8 '13 at 15:38
    
@Amzoti: Thanks! So a concrete function is an abstract function with a faithful function to the category of sets. What is called "category" is the same concept as "abstract category". –  Tim Jan 8 '13 at 15:52
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For simplicity/beginners, you may even drop the faithful functor and say that in a concrete category the objects are sets (endowed with additional structuer) and the morphisms are functions between these sets (with possible discussion what "are" means here). Example: Groups, topological spaces etc. (More strictly, these are note conrete, but concretizable) The standard example for an properly abstract (i.e. non-concretizable) category, i.e. homotopy, essentially works because one cannot take suitable representatives of homotopy classes, so to speak. –  Hagen von Eitzen Jan 8 '13 at 15:58
    
@HagenvonEitzen: The homotopy category hTop is not concretizable. Does this imply we can't even formalize these kinds of categories in a strong set theory with Grothendieck universes? I.e. does it say there cannot be a set-formulation of these categories, or is the non-concretizable problematic only to be understood to be a restriction regarding functors to the specific concrete category set. I assume the latter is the case, as an equivalence "class" is also just a set after all. –  NikolajK Apr 2 '13 at 9:07
    
@NickKidman I agree with your last observation. $\operatorname{Mor}_{\mathbf{hTop}}(X,Y)$ is a quotient of the set(!) $\operatorname{Mor}_{\mathbf{Top}}(X,Y)$, hence clearly a set. But we cannot functorially assign an "underlying" set to each space and a corresponding map for each morphism (homotopy class of continuous map). Note that we'd be allowed to take something else instead of the "really underlying" sets of the topological spaces, e.g. add arbitrary additional set-like info, and still cannot make this work. I don't know the detailed proof, though. –  Hagen von Eitzen Apr 2 '13 at 16:12

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