I've been solving this problem:

In $\mathbf{Ab}$, show that all monics are regular.

Suppose $m:G\to H$ is a monic. Note that the diagram $G\to H\to H/m(G)$, where the arrow $G\to H$ is $m$ and the arrows $H\to H/m(G)$ are $\pi$ (the quotient map) and $0$, is a fork. To conclude the proof, let $F:A\to H$ be an arrow with $\pi\circ F=0$. We have that for all $a\in A$, $\pi(F(a))=0$, which is equivalent to $F(a)\in m(G)$, so $F(a)=m(g_a)$ for some $g_a\in G$. Define $\bar F:A\to G$ by $\bar F(a)=g_a$. To show that this map is well-defined, we need to show that if $m(g_a)=m(g_b)$, then $g_a=g_b$. That is, we need to show (at least as far as I understand) that the image $U(m)$ of $m$ under the forgetful functor $U:\mathbf {Ab}\to \mathbf {Set}$ is an injective function. I know how to prove that in $\mathbf {Set}$ an arrow is monic iff injective, but why is the image of a monic under the forgetful functor a monic?

If $m:G\to H$ is a monic in $\mathbf {Ab}$, then $mh=mh'\implies h=h'$ for any abelian group $X$ and any group homomorphisms $h,h':X\to G$. Now the diagram $X\to G\to H$ gives rise to the diagram $U(X)\to U(G)\to U(H)$. We need to show that for all sets $S$ and all functions $k,k':S\to U(G)$, $U(m)k=U(m)k'\implies k=k'$. I don't see how to prove this and how this is related to $U(h)$ or $U(h')$.

Also, more generally, is it always true that the image of a mono (or epi) under forgetful functors from a category of abstract algebra is again a mono (or epi)?

  • $\begingroup$ See Example 5.1.30 in Leinster. $\endgroup$ – user634426 Jul 26 '19 at 3:29

For any variety of universal algebras, the forgetful functor is representable (in fact, has a left adjoint), and hence preserves monomorphisms. In the case of $\mathbf{Ab}$, or even $\mathbf{Grp}$, the representing object is the (abelian) group $\mathbf{Z}$.

But in general, the forgetful functor need not preserve epimorphisms, as the inclusion $\mathbf{Z} \hookrightarrow \mathbf{Q}$ in $\mathbf{Ring}$ shows. The same example also shows that a variety of universal algebras need not, in general, be a balanced category.

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  • $\begingroup$ I was hoping that there is an easy direct argument proving the claim about monics. The fact that functors that have a left adjoint preserve monomorphisms is not familiar to me, so I'll probably have to ask another question about it. $\endgroup$ – user634426 Jul 26 '19 at 1:09
  • $\begingroup$ @user634426: you can prove it directly by using Hom(F1, X) ≅ UX, ie. that the morphisms from the free algebra on one generator correspond to the elements of X, so monomorphisms have to distinguish them $\endgroup$ – user54748 Jul 26 '19 at 1:30

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