Can someone give me an example of a category (telling me what the objects and morphisms are) where the objects are not some construction on sets or classes?
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2$\begingroup$ This seems like a slightly different question from the one you asked on math.SE (math.stackexchange.com/questions/260623/…), mostly because you used "are not" instead of something like "cannot be considered as." $\endgroup$– Qiaochu YuanCommented Dec 17, 2012 at 12:28
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3$\begingroup$ @Salvo: as far as I can tell, the OP wants to ask a question similar to "can you give me an example of a category which is not concretizable?" but something more like "can you give me an example of a category which cannot be constructed from a concretizable category?" except I am not really sure how to make this precise. $\endgroup$– Qiaochu YuanCommented Dec 17, 2012 at 12:30
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5$\begingroup$ Even after the OP agreed with Qiaochu's formulation, the question is still, as Qiaochu said, not easily made precise. Having no clear idea what "some construction on" (in the original question) and "constructed from" (in Qiaochu's reformulation) mean, I vote to close as "not a real question". The OP's comment on Tim Porter's answer leads me to think that he wants a category whose objects and morphism don't constitute sets or classes, which contradicts the (usual) definition of category; then the OP should tell us what alternative definition of category (s)he has in mind. $\endgroup$– Andreas BlassCommented Dec 17, 2012 at 15:03
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4$\begingroup$ Is there any mathematical object at all which is not a construction on sets of classes? I don't know what to think of this question. Either it is not correctly phrased or the person who asks doesn't know anything about the foundations of mathematics. $\endgroup$– Fernando MuroCommented Dec 17, 2012 at 16:42
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3$\begingroup$ A category has some sort of collection of objects and some sort of collection of morphisms. If you want an example of a category where neither of these is "a construction on sets or classes", it would help if you gave an example of a collection of things that you do not consider to be "a construction on sets or classes". If we had such a collection $C$, maybe the discrete category whose objects are $C$ (and whose morphisms are only identities) would satisfy you. $\endgroup$– Omar Antolín-CamarenaCommented Dec 17, 2012 at 20:08
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4 Answers
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The free topos (rather over the top, but still). See the book of Lambek and Scott. In categorical logic and the computer science associated to functional programming, the issue is really the other way round. There a category is often given via "syntax" and finding some model in which the objects are sets with possible structure is one of the basic aims of the theory. (If morphisms are some sort of program, or proof, objects are not so transparent.)
answered Dec 17, 2012 at 12:59
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$\begingroup$ And this sort of game can be played with any syntactic category. $\endgroup$– David Roberts ♦Commented Dec 17, 2012 at 23:14
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You say: where the objects are not some construction on sets or classes. What do you actually mean here? I will give several examples of my interpretation of this, in some you may think it violates that criterion.
Let $G$ be any group and form a category $G[1]$ with exactly one object which will be denoted $*$. The set of morphisms from * to * will be $G$ and the composition will be the multiplication in $G$ with the identity morphism on * being the identity element of $G$. That gives a category (in fact a groupoid). (You really only need a monoid not a group of course.)
If $X$ is a topological space, the fundamental groupoid of $X$ is a category in which the objects are points in $X$ (does this violate the conditions that you imposed?), and the morphisms/arrows are homotopy classes of paths between the points.
Let $(P,\leq)$ be a partially ordered set, and think of it as a category, i.e. objects are points, arrows are pairs $(x,y)$ with $x\leq y$. Similarly any equivalence relation on a set gives a category, in fact a groupoid, but again your question is not precise enough on what you want so you may feel this is cheating.
(Just in case the question resulted from a problem set in a category theory course, I have left you to write down ALL the details! In any case, it is a good idea to do so.)
answered Dec 17, 2012 at 12:46
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1$\begingroup$ Thanks Tim for the answer, but to me- all these things you have mentioned are constructions from sets. The "elements" of the group are morphisms in a category with one point. But these morphisms form a set. In the second example, the objects are points of a topological space, and a topological space is constructed from sets. I can say the same thing for the third example. I know that I am not being precise with what I am asking, but it is difficult if sets seem to loom over everything I see in mathematics. $\endgroup$– jd94j39Commented Dec 17, 2012 at 12:53
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3$\begingroup$ As someone else pointed out in the usual DEFINITION of a category you are required to have a SET of morphisms between objects, so from that point of view, your search is DOOMED! If you take a different
foundation' for mathematics, what one do you want? You can follow Lawvere's idea of using categories as the basic things from which to build things... note this is not a foundational exercise as such, rather a pragmatic one. Here is an idea: take categories as basic, then the 2-category of categories (no size to be mentioned since set theory isanathema') and functors might pass must. :-) $\endgroup$ Commented Dec 17, 2012 at 18:20 -
1$\begingroup$ DOOOOOOOOMED!!!11! - even if you don't demand local smallness! (which Tim is assuming) Because there might be a class of arrows between two objects! $\endgroup$– David Roberts ♦Commented Dec 17, 2012 at 23:11
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I do not know exactly what is asked. For me all objects of Mathematics are sets. But perhaps what you want is what I used as an example to illustrate the concepts of Category Theory. (I call it the Toy Category).
Objects = Natural numbers.
Morphism: $\alpha\colon n\to m$ such that $m\ \vert\ \alpha \gcd(n,m)$ (here $\alpha\in{\bf Z}$). Two morphisms $\alpha\colon n\to m$ and $\beta\colon n\to m$ are considered the same if $\alpha\equiv \beta\bmod m$.
For example $35\colon36\to60$ is a morphism.
The composition of two morphisms $\alpha\colon n\to k$ and $\beta\colon k\to m$ is by definition $\alpha\beta\colon n\to m$. You may check it is well defined ( and the compositions of equivalents gives equivalents ).
You may check this is a Category, and for example every isomorphism is an automorphism and the Group of automorphisms of n is isomorph to $({\bf Z}/n{\bf Z})^*$.
Is this what you want?
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$\begingroup$ To demystify this category: it's equivalent to the category of finite cyclic groups. $\endgroup$ Commented Dec 18, 2012 at 12:40
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$\begingroup$ @Tom Leinster It is my attempt to give a sense the question of the OP. Presented as I have done it, you may insert this category under Peano axioms of Natural Numbers (as formulated for example in E. Mendelson, Introduction to Mathematical Logic, van Nostrand 1964). In this way you may think that it is part of a theory previous to Set Theory. $\endgroup$– juanCommented Dec 18, 2012 at 15:40
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$\begingroup$ Yes, I agree, and I see why you mentioned it in response to the original question. It's just that when I saw your answer, I thought to myself: "what is this category?" It took me a little time (pleasant enough) to figure it out, so I thought I'd leave a comment to help anyone else who was puzzled by where your example came from. Sorry if my comment sounded dismissive. $\endgroup$ Commented Dec 19, 2012 at 0:41
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Category theory does not base on sets. More, there's no such thing as "the" Set Theory. Each theory has its own collection of axioms. If your specific category does not involve any particular set theory, it does not rely on set theory.
You may think of modeling categories in certain set theories; but again, it depends on how you define sets. Say, you can (kind of frivolously) define sets as discrete categories.
