let $C$ be the category of $\tau$-algebras for some type $\tau$. consider the statements:

  1. every monomorphism is regular.
  2. every epimorphism in C is surjective.

it is easy to see that 1. implies 2. what about the converse?


Update: the following exchange appeared on the categories mailing list several years ago: http://article.gmane.org/gmane.science.mathematics.categories/3094. Walter Tholen's response strongly suggests that the answer to Martin's question is that the converse does not hold, although I don't have access to the four-author article he cites as reference. It's probably worth a look though, and if I learn anything more I'll post another update.

Second update: My surmise was correct. Walter Tholen kindly emailed to me the relevant two pages (pp. 88-89) of the four-author paper

  • E.W. Kiss, L. Marki, P. Prohle, W. Tholen: Studia Sci. Math. Hungaricum 18 (1983) 79-141.

where the following example is given on page 89: in the category of semigroups with zero such that all 4-fold products are zero, all epimorphisms are surjective but not all monos are regular [a specific nonregular mono is described]. (I can forward this email if you write to me at topological dot musings at gmail dot com.)

Assuming amalgamated products exist (as they do in categories of algebras of a Lawvere theory), a mono i: A >--> B is regular if it is the equalizer of the pair of canonical maps from B to the amalgamated product B *_A B (i.e., the coprojections of the pushout of i with itself, aka the cokernel pair of i). The equalizer of the cokernel pair defines a closure operator on the lattice of subalgebras Sub(B), called the dominion operator Dom_B. So to prove a subalgebra is not regular is to show that it is not Dom-closed. The key technical result needed to prove the claim above is Isbell's Zig-Zag theorem (given in his paper Epimorphisms and Dominions in the 1965 La Jolla conference proceedings on categorical algebra), as recalled here, which gives a precise and useful criterion for an element to belong to the dominion (= Dom-closure) of a subalgebra.

Hope this helps. I am voting up your question, Martin, since it's rather nontrivial!


Certainly not. Consider the category of groups: every epimorphism is surjective (see, e.g., Categories for the Working Mathematician, p. 21 exercise 5) but not every mono is a kernel.

  • $\begingroup$ I was about to give the same answer, but after some googling I discovered that actually every monomorphism in Grp is regular: for an example of the inclusion of a non-normal subgroup as a regular monomorphism, consider the equalizer of the identity map of G and the map given by conjugation by g. $\endgroup$ – Reid Barton Jan 1 '10 at 14:16
  • $\begingroup$ actually the category of groups motivated my question! todd, a monomorphism is regular if it is a equalizer of two morphisms. besides, kernels are not defined in the general setting above. $\endgroup$ – Martin Brandenburg Jan 1 '10 at 14:36
  • $\begingroup$ Whoops! Sorry, you guys are right. I hastily identified equalizers in Grp with kernels (I know btw that kernels are not defined generally). I'll think harder then. $\endgroup$ – Todd Trimble Jan 1 '10 at 15:03

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