Talk:Coproduct
From Wikipedia, the free encyclopedia
 |
|||||||||||||||||||||||
| Mathematics rating: | B Class | Mid Priority | Field: Algebra | ||||||||||||||||||||
|
|||||||||||||||||||||||
Please excuse my bad ASCII art. I'll do a real pic as soon as I figure out how. -- Fropuff 06:13, 2004 Jul 10 (UTC)
(Ah, those were the days. Geometry guy 21:36, 16 May 2007 (UTC))
[edit] Canonical injections monic?
Following discussion didn't provide a solution; any other input? -- Baarslag 16:57, 2 April 2006 (UTC)
The canonical injections are not necessarily injective, so maybe the link is confusing. Are they always mono though? -- Baarslag 13:36, 1 April 2006 (UTC)
- They are not necesarily mono. I'd love to give a counterexample, but I didn't come up with one. Anyone else? -lethe talk + 15:44, 1 April 2006 (UTC)
- Thanks for the speedy reply. I think you are right, but what is wrong with the following proof of ij : Xj → X being monic?
-
- Take two arrows, say p and q, to Xj with ij O p = ij O q and choose fj : Xj → Y to be the identity. There now exists a unique morphism f from X to Y such that fj = f O ij.
- Now ij O p = ij O q, so
- f O ij O p = f O ij O q, so
- fj O p = fj O q, so
- p = q. Thus, ij is monic.
-
- Again, it's probably faulty since I couldn't find this in any textbook. -- Baarslag 17:49, 1 April 2006 (UTC)
- Well I don't know if it's faulty, but it's certainly a little fishy; you're supposed to have a collection of maps Xj → Y. You want all those maps to be the identity, which can only happen if each Xj=Y. Thus all your components must be equal. So perhaps what you've proved that if all the components are equal, then the canonical injection is mono. Lemme look at it a little more. -lethe talk + 18:17, 1 April 2006 (UTC)
- Sure take your time, maybe a counterexample will pinpoint the mistake in the 'proof' above. As for some more clarification: the choice of Y and each fj : Xj → Y is free, and the choice Y = Xj and fj = id (for just one j) seems to do the trick. Cheers. -- Baarslag 18:56, 1 April 2006 (UTC)
- Oh I see, you're only assuming that one of the fj is an identity map, not all of them. Still, this seems like a loss of generality to me; what if no such collection exists? There is no guarantee that there are morphisms Xk → Xj. Is it so? Anyway, yes, we need a counterexample. I've got two texts that say that it is the case that the canonical injection/projection need not be monic/epic, but neither provides a counterexample. -lethe talk + 19:05, 1 April 2006 (UTC)
- Right, only one. Ofcourse, the proof must be flawed somewhere as you've pointed out, but I still wonder what exactly is wrong. To be sure, my reasoning was the following: the definition states
- For any object Y and any collection of morphisms fj : Xj → Y, there exists a unique morphism f from X to Y such that fj = f O ij.
- And the 'proof' above says that since the coproduct must handle any fj : Xj → Y (no constraints here), it must certainly handle fj = id, and this special case already seemed to force ij : Xj → X to be monic. --Baarslag 21:16, 1 April 2006 (UTC)
- Well like I said, assuming the existence of this family of morphisms entails a loss of generality. Yes, the universal property guarantees the equations to hold for any such family of morphisms. But you choose a particular family of morphisms which may not exist. If this family of morphisms exists, then the equations hold, and therefore the injections are monic. But it may be that there is no such family of morphisms. More explicitly, you choose Y to be equal to one of the Xj. Thus you are assuming that there exist morphisms fk: Xk→Xj for all k. For the case k=j, we certainly know such a morphism exists, it's the identity morphism. But what about the case i≠j? How do you know you have a morphism Xk→Xj in this case? If you can find one, then the proof goes through, but there's no guarantee of such a morphism. This is the flaw in the proof, or at least, it's the only flaw that I can see. The proof will still go through in some cases, like if all the Xk are equal, but not in general. Do you see my point? Is it correct? -lethe talk + 03:44, 2 April 2006 (UTC)
- Lethe, thanks for your patience with me. I think you're right. Staring at your argument, maybe some specially crafted finite category with carefully chosen arrows might provide the counterexample we are looking for. I will try the most obvious ones. -- Baarslag 16:57, 2 April 2006 (UTC)
- Well like I said, assuming the existence of this family of morphisms entails a loss of generality. Yes, the universal property guarantees the equations to hold for any such family of morphisms. But you choose a particular family of morphisms which may not exist. If this family of morphisms exists, then the equations hold, and therefore the injections are monic. But it may be that there is no such family of morphisms. More explicitly, you choose Y to be equal to one of the Xj. Thus you are assuming that there exist morphisms fk: Xk→Xj for all k. For the case k=j, we certainly know such a morphism exists, it's the identity morphism. But what about the case i≠j? How do you know you have a morphism Xk→Xj in this case? If you can find one, then the proof goes through, but there's no guarantee of such a morphism. This is the flaw in the proof, or at least, it's the only flaw that I can see. The proof will still go through in some cases, like if all the Xk are equal, but not in general. Do you see my point? Is it correct? -lethe talk + 03:44, 2 April 2006 (UTC)
- Oh I see, you're only assuming that one of the fj is an identity map, not all of them. Still, this seems like a loss of generality to me; what if no such collection exists? There is no guarantee that there are morphisms Xk → Xj. Is it so? Anyway, yes, we need a counterexample. I've got two texts that say that it is the case that the canonical injection/projection need not be monic/epic, but neither provides a counterexample. -lethe talk + 19:05, 1 April 2006 (UTC)
- Sure take your time, maybe a counterexample will pinpoint the mistake in the 'proof' above. As for some more clarification: the choice of Y and each fj : Xj → Y is free, and the choice Y = Xj and fj = id (for just one j) seems to do the trick. Cheers. -- Baarslag 18:56, 1 April 2006 (UTC)
- Well I don't know if it's faulty, but it's certainly a little fishy; you're supposed to have a collection of maps Xj → Y. You want all those maps to be the identity, which can only happen if each Xj=Y. Thus all your components must be equal. So perhaps what you've proved that if all the components are equal, then the canonical injection is mono. Lemme look at it a little more. -lethe talk + 18:17, 1 April 2006 (UTC)
- Again, it's probably faulty since I couldn't find this in any textbook. -- Baarslag 17:49, 1 April 2006 (UTC)
Some googling turned up this paper which says on page 19 that in an extensive category, coproduct injections are mono. This suggests not only that the result is not true in general, but gives us a criterion for when it does hold. Now I just have to figure out what an "extensive category" is. Then it should be easy to find a counterexample. -lethe talk + 19:14, 1 April 2006 (UTC)
- The paper of course has a definition of extensive categories, but no nonexamples. -lethe talk + 19:20, 1 April 2006 (UTC)
- I've read the paper, and you're right, it does suggest that coprojections are not mono. Prior to my question here, I've done some googling myself and got the impression that coprojections are mono iff the coproduct is really disjoint. Maybe it rings a bell somewhere. I hope we can crack this one. -- Baarslag 21:01, 1 April 2006 (UTC)
- Well I don't know what you mean by "disjoint". Does it mean for concrete categories that the underlying sets of the images of the injection functions are disjoint as sets? Does it have any meaning for generic categories? Certainly I see that for some categories, like groups or pointed spaces, the "components" of the coproduct are not disjoint, so the injections are not injective. But they're still monic, so that doesn't help, and even if it did, I don't know any category-theoretic way to characterize this notion of "disjointness". -lethe talk + 03:44, 2 April 2006 (UTC)
- Oh, now I see that that paper defines "disjointness" for coproducts: the pullback of the injections is the initial object. I think both groups and pointed spaces satisfy this, even though, as sets, they are not disjoint. So OK, nevermind my above comment. -lethe talk + 04:07, 2 April 2006 (UTC)
- OK, so the paper says that if the category is extensive, then its coproducts are disjoint, and its injections are monic. You suggest that perhaps monic injections and disjoint coproducts are equivalent. It sounds believable, I guess. Let's play around with the diagrams. And I'd still really like to know an example of a category that is not extensive. -lethe talk + 04:10, 2 April 2006 (UTC)
- Extensiveness is a little out of my league, but I will try to sketch some finite diagrams (see comment above) today. -- Baarslag 16:57, 2 April 2006 (UTC)
- OK, so the paper says that if the category is extensive, then its coproducts are disjoint, and its injections are monic. You suggest that perhaps monic injections and disjoint coproducts are equivalent. It sounds believable, I guess. Let's play around with the diagrams. And I'd still really like to know an example of a category that is not extensive. -lethe talk + 04:10, 2 April 2006 (UTC)
- Oh, now I see that that paper defines "disjointness" for coproducts: the pullback of the injections is the initial object. I think both groups and pointed spaces satisfy this, even though, as sets, they are not disjoint. So OK, nevermind my above comment. -lethe talk + 04:07, 2 April 2006 (UTC)
- Well I don't know what you mean by "disjoint". Does it mean for concrete categories that the underlying sets of the images of the injection functions are disjoint as sets? Does it have any meaning for generic categories? Certainly I see that for some categories, like groups or pointed spaces, the "components" of the coproduct are not disjoint, so the injections are not injective. But they're still monic, so that doesn't help, and even if it did, I don't know any category-theoretic way to characterize this notion of "disjointness". -lethe talk + 03:44, 2 April 2006 (UTC)
- I've read the paper, and you're right, it does suggest that coprojections are not mono. Prior to my question here, I've done some googling myself and got the impression that coprojections are mono iff the coproduct is really disjoint. Maybe it rings a bell somewhere. I hope we can crack this one. -- Baarslag 21:01, 1 April 2006 (UTC)
[edit] Possible solution
Define the following category:
Two unequal arrows p, q: D → A, v: A → C, w: B → C. And define vp = vq. C is a coproduct and v, w the canonical injections, since it's the only limiting cocone on A and B, but v is not mono by definition. Is this correct? -- Baarslag 23:17, 2 April 2006 (UTC)
- Yes, that looks correct. Nice work. So I guess you just made a category such that the family of morphisms you use in the proof doesn't exist. I was thinking about looking for something like that in concrete categories. The problem is, in most concrete categories, there are lots of morphisms between most objects. The category of fields has lots of objects with no morphisms between them, but no coproducts. Ho hum. -lethe talk + 23:44, 2 April 2006 (UTC)
- Sorry for putting it on top, I was afraid you'd miss it. You inspired the proof, so it's your 'nice work' as well. I'm still shaky after my forelast 'proof', so I'm not completely sure the current proof is solid. Did you triple-check it? -- Baarslag 00:19, 3 April 2006 (UTC)
- The proof, with the assumption of the existence of those morphisms, looks correct to me. I guess this provides a way of understanding why the canonical injections usually are monic. In most categories, we can find such morphisms. Any category with zero morphisms, for example. -lethe talk + 01:19, 3 April 2006 (UTC)
- Sorry for putting it on top, I was afraid you'd miss it. You inspired the proof, so it's your 'nice work' as well. I'm still shaky after my forelast 'proof', so I'm not completely sure the current proof is solid. Did you triple-check it? -- Baarslag 00:19, 3 April 2006 (UTC)
