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Dear colleagues,

Can anybody explain me if a 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?

Thank you in advance, Sergei Akbarov

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I think that the right word is cowellpowred –  Buschi Sergio Nov 1 '11 at 16:19
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Or well-copowered? –  Guillaume Brunerie Nov 1 '11 at 16:24
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MacLane writes "co-well-powered"... –  Sergei Akbarov Nov 1 '11 at 17:25

4 Answers 4

up vote 9 down vote accepted

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?)

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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 1 '11 at 19:34
    
Just in case, I am curious: nothing similar is known about topological algebras, isn't it? –  Sergei Akbarov Nov 2 '11 at 10:02
    
I Improved my answere considering the topological case. –  Buschi Sergio Nov 2 '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 2 '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 2 '11 at 20:23

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.

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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 1 '11 at 19:40
    
Indeed, it would. I sense another Mathoverflow question... –  Andrej Bauer Nov 1 '11 at 20:16
    
Arturo Magidin gave an answer on math.SE 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 4 '11 at 6:18

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"?)

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There are locally small categories which are not well-powered, or co-well-powered. Local smallness has little to do with the question. –  Andrej Bauer Nov 1 '11 at 18:47
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The original question was using the "locally small" term instead of "co-well-powered". –  Guillaume Brunerie Nov 1 '11 at 19:01
    
That was my fault -- initially I wrote "locally small", and only after Guillaume's suggestion I made a correction: "co-well-powered". –  Sergei Akbarov Nov 1 '11 at 19:09

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.

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For me, if every object has a (small) set of subobjects, the category is called well-powered. –  Guillaume Brunerie Nov 1 '11 at 14:29
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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 1 '11 at 15:47
    
This should have been a comment. Why is it an "answer"? –  Andrej Bauer Nov 1 '11 at 18:48
    
@Andrej Because Anonymous does not have enough reputation to leave comments. –  Guillaume Brunerie Nov 1 '11 at 19:17
    
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 1 '11 at 19:23

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