Timeline for Does this concept have a name, and what are some of its applications?
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18 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 30, 2010 at 18:55 | comment | added | Qiaochu Yuan | I don't think there is a natural definition of the "smallest" integral domain whose field of fractions is, say, C(t). This is a lot of integral domains, e.g. C[t] or C[1/t] but even C[t^2, t^3], the order relation on them as subrings of C(t) seems very unnatural to me, and I don't see a natural order relation. | |
| Jun 30, 2010 at 18:34 | history | edited | Robin Saunders | CC BY-SA 2.5 |
Add followup question for the case where the smallest such Y is X
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| Jun 30, 2010 at 10:16 | comment | added | Peter LeFanu Lumsdaine | ...however, @Theo: don't worry, I don't think there are any coherence conditions to ask for! The “associativity” you mention is just a naturality square, and similarly for higher powers of f. So I'm pretty sure that just a natural isomorphism F ≅ F^2 is enough. | |
| Jun 30, 2010 at 10:12 | comment | added | Peter LeFanu Lumsdaine | @Robin: never underestimate how often when we think of things as equal, we mean isomorphic! :-) If you define "field of fractions" in the usual/natural/obvious way, then Frac_(_k) is isomorphic to k, but not equal! Of course, you can redefine the action of Frac by cases as “check if R is a field; if so, leave it unchanged; if not, form the field of fractions”. Then it really is the identity on D; but everything you do with it will have to go by cases, and generally this seems more work, and less, um, natural, than just accepting that natural isomorphism is as good as equality! | |
| Jun 29, 2010 at 14:20 | answer | added | Tom Boardman | timeline score: 2 | |
| Jun 29, 2010 at 14:13 | comment | added | Mariano Suárez-Álvarez | Hmm, how is the «algebraic closure» a functor? Suppose $f:K\to L$ is an inclusion of fields: there are in general several maps $\overline K\to\overline L$, even several which extend $f$. | |
| Jun 29, 2010 at 11:43 | answer | added | Tom Boardman | timeline score: 1 | |
| Jun 29, 2010 at 11:31 | comment | added | Robin Saunders | Sorry, I was being a bit loose in my use of language. Hopefully my first comment above explicitly shows what I meant: F is a functor on C which is the identity on D. If you prefer, it's a functor from C to itself which is the identity on its image. The illustrative examples I gave in the question itself should give an idea of what I'm talking about. I'm not too worried about the general case of functors onto a subcategory D which are the identity on D; I'm more interested in cases where the "minimal preimage" concept I outlined is naturally defined. | |
| Jun 29, 2010 at 10:04 | answer | added | Peter LeFanu Lumsdaine | timeline score: 3 | |
| Jun 29, 2010 at 8:19 | comment | added | Theo Johnson-Freyd | You'll need to precisify a few definitions. Namely, it just never happens in the real world that functors, objects, etc are ever equal to each other, only isomorphic, and you can hope naturally, uniformly, or canonically isomorphic. So what is an "idempotent functor"? Presumably you mean a functor $F$ along with a natural isomorphism $\phi: F^2 \cong F$, which satisfies some conditions. My best guess would be that the compatibility conditions are, at the minimum, an associativity constraint that the two ways of applying $\phi$ to get $F^3 \overset\sim\to F$ are the same. | |
| Jun 29, 2010 at 4:22 | comment | added | Kiochi | Ah, rereading your questions I see what you meant. | |
| Jun 28, 2010 at 23:18 | comment | added | Robin Saunders | And yes, closure of subsets of a topological space would be an example of the sort of functor I'm talking about, although in general there wouldn't be a "minimal" subset Y whose closure was a given closed subset X. | |
| Jun 28, 2010 at 23:16 | comment | added | Robin Saunders | Kiochi: For my first example, given an algebraically closed field X [∈ D], what is the smallest field Y [∈ C] whose algebraic closure F(Y) = X? For my second example, given a field X [∈ D], what is the smallest ring Y [∈ C] whose field of fractions F(Y) = X? Since F is a closure operator, I should really have stipulated that not only is F idempotent, but in fact F(X) = X ∀ X ∈ D. So the smallest Y containing X such that F(Y) = Y would be X itself. Daniel: I do not mean C to equal D; D is a subcategory of C, so F can certainly be composed with itself. | |
| Jun 28, 2010 at 22:26 | comment | added | Kiochi | Do you mean F(Y) = Y and Y contains X? Otherwise it doesn't seem to match your examples. | |
| Jun 28, 2010 at 21:48 | comment | added | Daniel Litt | Another example might be the (poset) category of subsets of a topological space, with the functor being "closure." Was this what motivated your question? I also assume you mean C to equal D, as otherwise idempotency ($F\circ F=F$) does not make sense, as one cannot compose $F$ with $F$. In any case, one might try to take the limit of e.g. the diagram given by all objects that map to $X$ and all arrows that map to the identity. Or all objects that map to something isomorphic to $X$ and all arrows that map to an isomorphism. | |
| Jun 28, 2010 at 21:44 | history | edited | Robin Saunders | CC BY-SA 2.5 |
Advantage of considering concrete categories is that objects have a natural order by isomorphic inclusion
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| Jun 28, 2010 at 21:37 | history | asked | Robin Saunders | CC BY-SA 2.5 |