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The following question eats my brain: The standard definition of a "category" and a list of examples following this definition confuses me. My question is simple:

(Q1)If someone write "the category of finite groups" what are the objects of this category? Surely, $\mathbb{Z}_6$ is in this category. What about the other instances of $\mathbb{Z}_6?$ What prohibits to add extra copies of $\mathbb{Z}_6$ into objects? There is no "equality" of objects on "objects" other than the "standard equality class" of the object which models its theory. But the theory of an "object" is not included in the category. Let us formalize the theory of $\mathbb{Z}_6$ groups by adding extra conditions to the axioms of groups so that if a usual group satisfies these extra conditions then its isomorphic to $\mathbb{Z}_6.$ Now, is every model of these abstract conditions in our category? Or not? Do you include model theoretic semantic into category or not?

What does category theory give different than the model theory then?

(Q2) If someone chooses objects up to isomorphisms then why a "functor" (should be morphism) is called isomorphism? I saw somewhere that there is a mono which is not a monomorphism in the usual sense. So are there , for instance finite groups, which are isomorphic in the categoric sense but not isomorphic in the normal sense? Please note that finite group category is just an example, you are welcome to add interesting examples.

Thank you.

Edit: "functor" of Q2 should be "morphism".

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I don't understand Q2. –  Qiaochu Yuan Jul 5 '11 at 22:56
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You may want to have a look at Lawvere's thesis. –  t.b. Jul 5 '11 at 23:05
    
@Theo: thank you for the link. But could you also clarify how does this thesis related to the question? Is it for Q2? Could you also answer Q1 please? When do people say "category of groups" do they take iso. classes? Why they don't take 2 copies of the same group? –  categoryboy Jul 5 '11 at 23:34
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Two things: First: There is a categorical description of monomorphisms: $m$ is a monomorphism if $mf = m g$ implies that $f = g$. Now it turns out that even if $m$ is a map of underlying sets a monomorphism $m$ need not be injective. Dually there are epimorphisms ($fe = ge$ implies that $f = g$, which are a bit easier to come up with: e.g. the inclusion $\mathbb{Z} \to \mathbb{Q}$ is an epimorphism in the category of rings, or a map between topological spaces is an epimorphism if and only if it has dense range. Second: I gave you that link because Lawvere discusses among other things... –  t.b. Jul 6 '11 at 0:06
    
...how different axioms of algebraic structures give rise to different categories (which mostly turn out to be equivalent). It may not be the best thing to read at this point, though. –  t.b. Jul 6 '11 at 0:07

3 Answers 3

up vote 12 down vote accepted

The natural notion of isomorphism of categories is not isomorphism (the existence of two functors which are inverses) but equivalence. A good intuition for equivalence is that it behaves like homotopy equivalence of spaces. In particular, just as many different spaces with different sets of points (in particular, sets of points with different cardinalities) can be homotopy equivalent, many categories with different sets of objects can be equivalent. Thus in the category of categories, "the set of objects" is not well-defined as it is not invariant under equivalence, just as in the homotopy category, "the set of points" is not well-defined as it is not invariant under homotopy equivalence.

When someone says "the category of groups," they are refraining from specifying a particular set of objects and morphisms because any reasonable choice gives the same category up to equivalence. For example, you can take

  • The category whose objects are sets $G$ (say in ZFC) equipped with maps such that etc. and whose morphisms are group homomorphisms, or
  • The category whose objects are, roughly speaking, isomorphism classes of groups and whose morphisms are group homomorphisms

and the corresponding categories are equivalent; the latter is just the skeleton of the former.

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I will reread your post, but, it does not matter to take a copy of $D_4$ or two copies of $D_4$ in the objects. Is it correct? –  categoryboy Jul 5 '11 at 23:01
    
@categoryboy: more or less. @Theo: okay, fair enough. I added a "roughly speaking" disclaimer. –  Qiaochu Yuan Jul 5 '11 at 23:21
    
Thank you very much for your answer. When people say "category of groups" do they take "iso classes of groups" or not?. What does "roughly" mean? –  categoryboy Jul 5 '11 at 23:30
    
ok "skeleton" solves the question. I will reask Q2 later, namely the morphisms which don't have real world analogs later. Thank you. –  categoryboy Jul 5 '11 at 23:44
    
@Qiaochu: Hmm, I'm not sure I believe your assertion about what happens when someone says "the category of groups". When I say "the category of groups" I mean the category whose objects are...groups! Namely, the first thing you say (or almost: I honestly don't think about sets in terms of ZFC set theory, although I might if pressed: in reality I am more Platonic / lazy [not in a good way!] than that). The second thing you describe leads, as you say, to a different and even non-isomorphic category. I don't think we should pretend that equivalent categories are actually the same, right? –  Pete L. Clark Jul 6 '11 at 7:35

Necessary context: In my opinion, it is not obligatory to base mathematics on set theory, although, of course, comparisons and discussions are interesting. Similarly, I do not think it is obligatory to compare categorical "foundations" with set-theoretic "foundations". At one end, Grothendieck found it necessary to postulate the existence of very many large cardinals (see Weibel's "Homological..."). At another end, in fact, by this point I definitely do not think that set theory reflects the practice of mathematics. Exaggerated category theory does not, either. I do not worry about "the category of sets", either.

The category of finite groups includes all finite groups... :) Yes, it includes many different copies of the dihedral group $D_4$, including different versions painted blue, red, or yellow. Or copies which are the same color, but are "distinct".

After some years/decades, I worry much less about the alleged set of sets that don't include themselves. I do not want a list of prohibitions that also prevents me from "forming" this set. I also do not want a prohibition against sharp knives when I am cutting up vegetables, even tho' I may cut myself. I do not want a prohibition against water, even tho' I may drown myself in my own bathtub by lying face down and breathing it in. Perversities.

So, yes, there are all those different copies of the same isomorphy-class of a group. Yes, if one thinks of ways to squeeze out a paradoxical-seeming something, probably one can. But why should one? :)

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Is it just me or does your penultimate paragraph here have hints of paraconsistent logic? I'm not sure I can articulate this properly, but it seems that allowing such "[p]erversities" requires ignoring the contradictions they might create :) –  Shaun Mar 8 at 23:12
    
Correction: "allowing such [p]erversities in Mathematics". –  Shaun Mar 8 at 23:21
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@Shaun, yes, in effect, I sincerely reserve the right to ignore some (seeming?) contradictions. My reference to "perversities" is that it is perverse to try to form the set of sets that don't contain themselves, for example, although it might have some sort of "academic interest". –  paul garrett Mar 8 at 23:43

Q1: You indeed include all finite groups as objects. Any two copies of the same finite group are isomorphic, but not equal. Note that in the definition of a category, the class of objects is not required to be a set, but only the class of morphisms from one object to another is required to be a set.

You absolutely do not include any model theoretic notion whatsoever. Given a language and a theory, the category of models of that theory (with morphisms homomorphisms) is a category. That is the only relationship between these two notions.

Model theory and category theory are meant to express completely different things. It is true that both give a way to talk about mathematics, but for both this is not their ultimate role. Model theory is meant for one to express notions like provability, decidability, and to prove magnificently nontrivial theorems like the compactness theorem, and the Löwenheim-Skolem.

Category theory is mainly meant as a formal way of generalizing topological ideas. Thus, one gets extensions of the classical notion of homology, homotopy, and so forth. This is where it differ considerably from model theory: model theory is completely helpless when it comes to topology because the definition of a topology requires second order logic.

Q2: The definition of an isomorphism in a category is a morphism $r:A\rightarrow B$ that has a morphism $s:B\rightarrow A$ such that $r\circ s=id_B$ and $s\circ r=id_A$. In many cases, this coincides with your preconceived notion of what an isomorphism "should be" (indeed if what you want from your morphism to be an isomorphism for it just to be one to one and onto, then this is true). This is just a definition, and it only sometimes jibes with what you would want it to mean. As Qiauchu observed, in the category of categories isomorphism of categories is not what you really want to be considered an isomorphism. But life's tough that way. Same goes for mono and epic -- it often works with your intuition, but many times it doesn't.

You should note that not all categories are "concrete" (meaning that their objects are sets, and that their morphisms are functions between these sets): some may just be points with a bunch of arrows, and a rule for composition. So in these non-concrete categories, your intuition for what mono and epic mean (for example) has no meaning! So the categorical definitions have the advantage that they are very general.

Hope that helps in some way.

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Thank you. Combining your first paragraph with "skeleton" of the category, we can use iso. classes or the collection of all groups as the "category of groups".(I am just verifying myself) Related to Q2, are there any iso s of a well-known category which is not "obtainable" by means of the usual techniques? Because I think it is possible to construct such a mono. –  categoryboy Jul 5 '11 at 23:54
    
Regarding Q1 you are right, if you care about categories up equivalence rather than up to isomorphism of categories (I have never met anyone who cared about iso.'s of categories) then you can pick however many of any copy. Strictly speaking, when people say "all finite groups" you take all of them. About Q2: surely if you take a non-concrete category then there is no "usual technique". But for concrete examples (maybe someone can come up with an easier one) if you look at the category of topological spaces, with morphisms continuous maps where you identify any two homotopic ones, then an –  Nicole Jul 6 '11 at 0:20
    
isomorphism in that category will not imply a homeomorphism. But as you see from this example, the whole point is that "usual technique" is not a well defined term. What you're asking is: is a categorical isomorphism always what you expect? But you can expect many things. As I said in the answer, if you are in a concrete category (as groups) and you want to define an isomorphism as a morphism which is 1-1 and onto (as do want to define for groups) then the categorical isomorphism is indeed what you expect. –  Nicole Jul 6 '11 at 0:22
    
About topology requiring second-order logic, compare math.stackexchange.com/questions/46656/… –  Carl Mummert Jul 6 '11 at 1:31
    
@Wesley Thank you Wesley for this wonderful info. –  categoryboy Jul 6 '11 at 1:48

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