# Grp and Ab are not equivalent as categories

I am trying to understand an example leading to the notion of adjoint.

Let $$\mathbf{Grp}$$ and $$\textbf{Ab}$$ denote the category of groups and abelian groups, respectively. Further, suppose that $$U: \textbf{Ab} \to \textbf{Grp}$$ denotes the forgetful functor and $$F: \textbf{Grp} \to \textbf{Ab}$$ be the abelization functor, i.e. for $$G \in \text{Ob}(\textbf{Grp})$$, we set $$F(G) := G/G'$$ with $$G' = [G,G]$$ being the commutator subgroup of G.

The claim is that (1) $$F \circ U \cong \text{id}_{\textbf{Ab}}$$, but (2) $$U \circ F \not \cong \text{id}_{\textbf{Grp}}$$.

Can you explain why (1) and (2) holds?

• Is $U$ the inclusion functor? The forgetful functor takes us to the category of sets, right? Commented Nov 6, 2020 at 20:01
• @JohnDouma : the inclusion is also a forgetful functor (it forgets that the group is abelian) Commented Nov 6, 2020 at 20:04
• @MaximeRamzi If I include the integers in the category of groups it doesn't cease to be abelian. Commented Nov 6, 2020 at 21:02
• @JohnDouma : I know. That's not the point - you can still forget that it was abelian and treat it as a group (which happens to be abelian). That's a forgetful functor which forgets a property - the forgetful functor to sets forgets structure, but forgetting a property is still forgetting something. See e.g. ncatlab.org/nlab/show/stuff%2C+structure%2C+property#examples Commented Nov 6, 2020 at 21:07

Let $$A$$ be abelian group. Apply $$U$$ to get $$U(A)=A$$. Now apply $$F$$ to get $$FU(A) = A/A'$$. Now, since $$A$$ is abelian, $$A' = \{e\}$$ is trivial. Thus, $$FU(A)=A/A' \cong A$$.
Now, choose $$G$$ nonabelian. Apply $$F$$ to get $$F(G) = G/G'$$. Now apply $$U$$ to get $$UF(G) = G/G'$$. Since $$G$$ was nonabelian, $$G' \neq \{e\}$$ and therefore $$UF(G) = G/G' \not\cong G$$.
Claim 1 holds, because for every group you can find a (natural) isomorphism between $$FU(G)$$ (the abelianization of $$G$$, that was already abelian and has just been regarded as a mere group, forgetting abelianity: the commutator subgroup $$[G,G]$$ is trivial if $$G$$ is abelian; actually, if and only if). If you're new to the concept of "natural isomorphism", this is a good first instance of how they work.