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An epimorphism $f$ is said to be extremal, if for any decomposition $f=i\circ p$ with $i$ a monomorphism, the morphism $i$ is automatically an isomorphism. (This is from the textbook by F.Borceux.)

Let us say that $f$ is weakly extremal, if for any decomposition $f=i\circ p$ with $i$ a monomorphism and $p$ an epimorphism, the morphism $i$ is automatically an isomorphism.

Are these definitions equivalent?

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Which categories have you tried so far? – Martin Brandenburg Nov 22 '11 at 21:33
Of course, this must be the same in some categories. For example, in Abelian categories, or, more generally, in categories where every morphism $f$ can be represented as a composition $f=m\circ g\circ e$, with $m$ a strong monomorphism, $e$ a strong epimorphism, and $g$ a bimorphism (see details in, Theorem 1.3; unfortunately, in Russian). But I do not know counterexamples at all. Perhaps, these definitions are equivalent in any category? – Sergei Akbarov Nov 22 '11 at 22:37
Thanks for the background. I strongly believe that there are counterexamples, but I haven't found one. After all, it may happen that every epi is an iso? What about considering categories given by generators and relations, perhaps even finite ones? – Martin Brandenburg Nov 23 '11 at 8:46
up vote 3 down vote accepted

A counterexample:

Consider the monoid $\langle a,b,c\mid ac=bc\rangle$ as a category with one object. Then $a,b$ are monics, $bc$ is an epic and $c$ is not an epic. So $bc$ is not an extremal epic, but it is easily to see weakly extremal.

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Any extremal epi is weakly extremal. Do you mean the opposite? – Benjamin Steinberg Nov 23 '11 at 17:51
I don't see why this works. Clearly bc is not an extremal epi since b is not an iso (only 1 is an iso). Why is it weakly extremal? – Benjamin Steinberg Nov 23 '11 at 17:53
Sorry, I meant a is not an iso. – Benjamin Steinberg Nov 23 '11 at 17:55
Oh, I see it is weakly extremal because it cannot be written as a monic times an epi. Very nice. – Benjamin Steinberg Nov 23 '11 at 17:57
@ Benjamin: Thank you very much. I mixed and correct the answer now. – Boris Novikov Nov 23 '11 at 19:33

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