# The image of a mono (or epi) under forgetful functors

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)?

• See Example 5.1.30 in Leinster. – 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.