Is it true that any morphism in any category can be written as a combination of monomorphism and epimorphism? In SET and categories where monomorphism is an injective function and epimorphism is a surjective function is it true. But I am interested about the issue in any category. How to prove that statement or there is a counterexample?


In general there is no reason for this to be true. A simple counterexample is the category with one object $x$ and exactly one non-identity arrow $f:x\to x$, where $f\circ f=f$. In this category $f$ is neither a monomorphism nor an epimorphism, so the only monomorphism/epimorphism is the identity, and $f$ is not a composition of the identity with itself. More generally, if $\mathcal{C}$ is a category where every epimorphism or monomorphism is an isomorphism, then clearly an arrow that is not an isomorphism cannot be decomposed like that (though I can't think of any other example of this kind).

It is true, however, for most concrete categories that one usually considers; in particular, a category where every morphism factors as a strong epimorphism followed by a monomorphism and strong epimorphisms are stable under pullbacks is called a regular category, and any quasivariety of universal algebra, any abelian category or any quasitopos is regular. Note that the category of topological spaces is not regular, but it still has the property that every arrow factors as an epi followed by a mono (in fact its dual is regular).

  • $\begingroup$ Is having zero object, kernels and cokernels enough? $\endgroup$ – Vladislav Oct 2 '19 at 18:09
  • $\begingroup$ @Vladislav It's certainly not enough to be regular or coregular, and I'm pretty sure it is not enough to have epi-mono factorizations. I think one can make a variation of my counterexample (to a different question) in this answer by adding a collection of endomaps $\alpha_n:A\to A$ and a kernel of $f$, in such a way that $f$ would not have any factorization. $\endgroup$ – Arnaud D. Oct 3 '19 at 9:52

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