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?



Is the category of rings co-well-powered?
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?](https://mathoverflow.net/questions/56453/epimorphisms-have-dense-range-in-tophausgrp))


