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Can anybody explain me if the category of (associative) rings is co-well-powered? (This is the MacLane definition, in Russian literature this is called "locally small from the right side".) I mean, it is well-powered, of course, since for any ring A one can easily find a skeleton in the category Mono(A) (of all monomorphisms with values in A) and this will be a set (the set of all subrings in A). But is it true, that it is co-well-powered, i.e. for any ring A there exists a skeleton in the category Epi(A) (of all epimorphisms from A into other rings) and is this skeleton again a set?

Sergei Akbarov
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  • I think that the right word is cowellpowred – Buschi Sergio Nov 01 '11 at 16:19
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    Or well-copowered? – Guillaume Brunerie Nov 01 '11 at 16:24
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    MacLane writes "co-well-powered"... – Sergei Akbarov Nov 01 '11 at 17:25
  • There are at least two different meaninigs of locally small category in mathematics. One is that a category has small hom-sets and other is that set of subobjects is small for all objects. His question is if ring has small set of quotient objects. – Anonymous Nov 01 '11 at 13:23
  • For me, if every object has a (small) set of subobjects, the category is called well-powered. – Guillaume Brunerie Nov 01 '11 at 14:29
  • This terminology also occurs and is probably more popular. In "CFWM" Maclane suggested not to use "locally small", because it may lead to confusion(as we saw here he predicted well). I was only trying to explain what Sergei Akbarov meant. – Anonymous Nov 01 '11 at 15:47
  • I think I should apologize here also. The problem was that initially I wrote "locally small", and only later I edited the text. (Actually it was a surprise for me that the terminology in English is not the same than in Russian.) So explanation given by Anonymous was absolutely correct. – Sergei Akbarov Nov 01 '11 at 19:23

3 Answers3

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Yes, rings as any algebraic theory make a locally presentable category, and this is well- copowred (Adámek and Rosicky, Locally Presentable and Accessible Categories Cambridge University Press, Cambridge, (1994))

About topological case, in the MAria Clementino article "Categorical and topological aspects of semi-abelian theories" (http://www.math.yorku.ca/~tholen/HB07BournClementino.pdf) ther is a call from a Wyler article (see 10.1 in Clementino article), from this follow that: the forgetful funtor from topological rings category to rings category preserve colimits (and these categories have colimits), the it preserve epimorphism, then from the well copoweredness of rings category follow the well cowellpowerness of the topological rings category

I seems (and hope) that this work well...

(Please, excuse my porr English).

About ALgebraic theory modeled on $T_2$ spaces (Hausdorff topological spaces), for example Hausdorff topological rings, or vector (on real or complex numbers) hausdorff topological spaces (with infinite unary operation, one for any number) we have that epimorhpisms in $T_2$ are maps by dense images. Now the $T_2$ topological spaces (and continuous maps) is a cowellpowred category, this follow from the fact that the cardinaliy of a $T_2$ space $Y$ by a dense subset X is bounded by $2^{2^X}$ (take two different point $a, b\in Y$ consider the class of neighborhoods of these points, and their intersection by $X$).

THen if the epimorphism in vectorial hausdorf spaces are continuous maps with dense image (of course a such maps is a Ephimorphism) then VEctor topological housdorff spaces are cowellpowred.

but I seems that the answere to this question is yes: if a morphism $f: X\to Y$ on linear $T_2$ spaces has a nodense image, let $N\subset Y$ the closure of image of $f$, then we have a non null ($T_2$) quotient $Y/N$ and the two morphism $0, \pi: Y\to Y/N$ that have the some composition (it is $0$) with f.

I see that for hausdorff topological groups the answere to the question "has a epimorphism dense image?" is not (Epimorphisms have dense range in TopHausGrp?)

Community
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Buschi Sergio
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  • This turns out to be unexpectedly (and amazingly) deep... OK, thank you, Sergio, I'll try to understand what they write in this book. – Sergei Akbarov Nov 01 '11 at 19:34
  • Just in case, I am curious: nothing similar is known about topological algebras, isn't it? – Sergei Akbarov Nov 02 '11 at 10:02
  • I Improved my answere considering the topological case. – Buschi Sergio Nov 02 '11 at 14:05
  • Sergio, apparently, I do not understand something... You write: "the forgetful funtor from topological rings category to rings category preserve colimits (and these categories have colimits), the it preserve epimorphism" Does this mean that the forgetful functor preserves epimorphisms? I thought, this is not true, because, for example, in the theory of topological vector spaces the forgetful functor does not preserve epimorphisms. How can this effect diappear in topological algebras? – Sergei Akbarov Nov 02 '11 at 18:17
  • As in the my answere Email to you, topological (real) vector space aren't make by an a (finitary) algebraic theory (in universal algebra way) on topological spaces. But rings are an a algebraic theory. – Buschi Sergio Nov 02 '11 at 20:23
  • I see that also for (Hausdorff) topological goup the question about characterization of epimorphisms is hard (open question still on 1992). see: http://www.sciencedirect.com/science/article/pii/0166864194900558

    for C^star algebra the answere is $yes$ just because epimorphisms are surjective maps http://arxiv.org/abs/funct-an/9405003.

    ABout Banach spaces (with contractive maps) and Hilbert spaces see the Adamek book pag. 70 and pag. 100 (from it follow that the first is well-copowered, and the second too if it has pullshout's).

     – Buschi Sergio Nov 02 '11 at 20:58
  • Still about your question "Sergio, apparently, I do not understand something".

    for a comparationbetwen the theorem 10.1 in the Clementino article (from it follow that the forgetful funcor preserve Epi), and the fact that "Epimorphism" of topological linear space arent epimorphism on the base spaces...I guess you mean the they arent surjective (but have dense image), but these are epimorphism on Hausdorff topological spaces, and the topological vector spaces (as infinitary algebraic theory ) are modelled on Housdorff topological spaces, non on general topological spaces.

     – Buschi Sergio Nov 02 '11 at 21:29
  • Ah, yes... So they (I mean, Bourn and Clementino) have in mind a category of not necessarily Hausdorff topological algebras, do they? It is strange for me that there are people who are interested in this... And I suppose, for Hausdorff topological algebras the question is open? – Sergei Akbarov Nov 03 '11 at 16:43
  • About Hausdorff topological vector spaces: of course, this is a co-wellpowered category, this follows from the description of epimorphisms there. But, as far as I understand, there is no such a description for Hausdorff topological algebras... Or I am wrong? Is is true that a morphism of Hausdorff topological algebras f:A->B is an epimorphism only if it can be represented as a composition of an epimorphism of (non-topological) algebras, and a continuous injection with a dense image? – Sergei Akbarov Nov 03 '11 at 22:18
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Sergio Buschi already gave a general answer, and here is a pedestrian way of thinking about it. First, apparently if $e : R \to S$ is epi, then the cardinality $|S|$ of $S$ cannot exceed the cardinality $|R|$ of $R$. This is not obvious, as $e$ need not be surjective. I am no expert on rings, so I simply Googled the fact and found this. For each cardinality below $|R|$, there are can be only set-many non-isomorphic rings. So we have a set of sets of candidates, which is again a set. That's co-well-poweredness.

Andrej Bauer
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  • It would be great to understand, why this is true: "if $e : R \to S$ is epi, then the cardinality $|S|$ of $S$ cannot exceed the cardinality $|R|$ of $R$". – Sergei Akbarov Nov 01 '11 at 19:40
  • Indeed, it would. I sense another Mathoverflow question... – Andrej Bauer Nov 01 '11 at 20:16
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    Arturo Magidin gave an answer on math.SE http://math.stackexchange.com/q/26934/ from which it follows that for an epimorphisms $e: R \to S$ of rings you can express every $s \in S$ as arising from certain triples $(CD,D,DE)$ of matrices with coefficients in $R$ (whose cardinality doesn't exceed the cardinality of $R$ if $R$ is infintite). This yields that the set of these triples are partitioned into equivalence classes from which you can describe $S$ entirely. Details are given in the article by Mazet linked to in the blog post mentioned in your answer (p. 2-07). – Theo Buehler Nov 04 '11 at 06:18
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A category is locally small if for all objects $A$ and $B$, $\mathrm{Hom}(A,B)$ is a set. In the category of rings, $\mathrm{Hom}(A,B)$ is a subset of the set of all functions between the underlying sets of $A$ and $B$, so it's a set.

I don't understand what you are trying to do with your categories of monomorphisms and epimorphisms (perhaps you meant something else than "locally small"?)

Guillaume Brunerie
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