Is the following known? (I don't know if I have actually proved it, but it looks good and I don't want to waste effort on the details if it is known, as seems fairly likely.) Suppose C is a category with a factorization system E/M and T is a triple on C, with category of algebras C^T, free and underlying functors F^T and U^T. Under reasonable hypotheses (complete and M-well-powered) there is a factorization system E^T/M^T on C^T in which M^T is U^{-1}(M) and E^T is the complementary class. Then there is a 1-1 correspondence between full subcategories of C^T closed under M^T-subobjects, products, and split epimorphisms; and triple maps T --> S for which F^T --> F^S lies in E^T. It is crucial, BTW, that the E's be epic and the M's monic.
Hi, I haven't really looked at your conclusions in detail, but I want to point out that as far as I know, in order for the factorization system E^T/M^T to exist, T (the functor of the triple T) has to send arrows in E into E. Bill Rowan
No. I didn't say that E^T = U^{-1}(E). If, for example, M consists of all monics, then so will M^T (which is defined to be U^{-1}M) and then E^T consists of all extremal epics. For example, in Cat, the image of a functor is not a subcategory, but it certainly generates a subcategory. Under reasonable conditions (arbitrary intersections of M-subobjects) every class M has a complementary class of epics. Michael Barr On Sun, 6 Oct 2002, Bill Rowan wrote:
Hi,
I haven't really looked at your conclusions in detail, but I want to point out that as far as I know, in order for the factorization system E^T/M^T to exist, T (the functor of the triple T) has to send arrows in E into E.
Bill Rowan
Dear Michael, your result on factorizations of algebra homomorphisms is, in my opinion, new and very nice. It does n o t require the category C to be complete, nor the factorization system to be proper (i.e., it holds even if E-morphisms are not epis). All it requires is a complete factorization system (E,M) , i.e., such that every cone of M-morphisms has a multiple pullback and pullbacks along M-morphisms exist. Then for every monad T the category C^T has the factorization system you desribe: M^T = preimage of M under U, and E^T = the complementary class. The proof is based on Max Kelly's observation that complete systems have M included in Mono. Best regards, Jiri xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx alternative e-mail address (in case reply key does not work): J.Adamek@tu-bs.de xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 10-Oct-2002 13:00:05 -0300,1719;000000000000-00000000
Well, it may be that completeness is not required, but I am not now interested in hypothesis chopping. Maybe later. As for whether it is necessary that E consist of epis, if it doesn't, then not every regular monic is in M and then the category of S-algebras is not even closed under equalizers, therefore not reflective. While this case, might have some interest, I am trying to get a clean statement with a direct proof. I have come to the conclusion that a better theorem might just use an E/M factorization in the category of algebras. For example, by taking epi/regular mono in the category of monoids, you get that groups are an example. And indeed groups are closed in monoids under regular subobjects. You don't get every full subcategory, since you have rational vector spaces are a full subcategory of abelian groups, but it is possible to show that if F^T --> F^S is epic, then the you always get a full subcategory. Michael 10-Oct-2002 16:53:11 -0300,1216;000000000000-00000000
Well, it may be that completeness is not required, but I am not now interested in hypothesis chopping. Maybe later. As for whether it is necessary that E consist of epis, if it doesn't, then not every regular monic is in M and then the category of S-algebras is not even closed under equalizers, therefore not reflective. While this case, might have some interest, I am trying to get a clean statement with a direct proof. I have come to the conclusion that a better theorem might just use an E/M factorization in the category of algebras. For example, by taking epi/regular mono in the category of monoids, you get that groups are an example. And indeed groups are closed in monoids under regular subobjects. You don't get every full subcategory, since you have rational vector spaces are a full subcategory of abelian groups, but it is possible to show that if F^T --> F^S is epic, then the you always get a full subcategory. Michael 13-Oct-2002 19:55:44 -0300,1490;000000000000-00000000
From (2) you necessarily have that M contains the isos and is contained in
Just to echo some of Michael's points, here is how I would look at his problem. Call (as I have done since the seventies) a category C, for any class M of morphisms, M-complete if (1) a pullback of an M along any morphism exists and is in M, (2) an intersection (=multiple pullback) of any-size family of M's with common codomain exists and (its diagonal) is in M. (Of course, "small" suffices when you assume C to be M-wellpowered.) the class of monos; the latter fact comes from the observation that when you take a multiple pullback of as many copies of a given morphism as the size of the category, that morphism got to be monic: see my paper with Reinhard Boerger in Math. Z. 160 (1978) 135-138. Now, add to (1) and (2) the condition that (3) M is closed under composition, then M is the right factor of a factorization system in C, and except for the existence assumptions in (1), (2), all conditions are necessary. Actually, you not only have orthogonal factorizations of morphisms, but of any-size families of morphisms with common codomain, and that statement is fully equivalent to (1)-(3). From this angle you see that you also get the existence of equalizers from (1) and (2), which is of interest when you want to prove that the left companion E of M is a class of epimorphisms iff M contains all regular monos. In Michael's situation where you have a monadic (or tripleable, if you insist) category A over C, with its limit-creating forgetful functor U, if C satisfies (1)-(3) for M, A will satisfy the same conditions for U^{-1}M, and you are in business. All this is of course old stuff of more than 20 years ago (see, for example JPAA 15 (1979) 53-73), but the correspondence Michael indicates I haven't seen before in this form at all. However, I am wondering how his correspondence relates to the almost equally old Cassidy-Hebert-Kelly paper and its predecessors (Gabriel, Ringel)? Regards, Walter.
Further to my previous posting, it occured to me that the second sentence of the following paragraph wasn't formulated correctly:
Actually, you not only have orthogonal factorizations of morphisms, but of any-size families of morphisms with common codomain, and that statement is fully equivalent to (1)-(3). From this angle you see that you also get the existence of equalizers from (1) and (2), which is of interest when you want to prove that the left companion E of M is a class of epimorphisms iff M contains all regular monomorphisms.
It should read: Having (1)-(3), ie an M-complete category C with M closed under composition, then the companion E of M is a class of epimorphisms iff C has equalizers and M contains all regular monomorphisms. Walter. 15-Oct-2002 15:26:37 -0300,2406;000000000001-00000000
participants (4)
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Bill Rowan -
Jiri Adamek -
Michael Barr -
Walter Tholen