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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?

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    $\begingroup$ Is $U$ the inclusion functor? The forgetful functor takes us to the category of sets, right? $\endgroup$
    – John Douma
    Commented Nov 6, 2020 at 20:01
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    $\begingroup$ @JohnDouma : the inclusion is also a forgetful functor (it forgets that the group is abelian) $\endgroup$ Commented Nov 6, 2020 at 20:04
  • $\begingroup$ @MaximeRamzi If I include the integers in the category of groups it doesn't cease to be abelian. $\endgroup$
    – John Douma
    Commented Nov 6, 2020 at 21:02
  • $\begingroup$ @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 $\endgroup$ Commented Nov 6, 2020 at 21:07

2 Answers 2

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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 $UF(G) = G/G' \not\cong G$.

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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.

So, (somewhat sloppily, but this is the main idea) claim 2 holds by virtue of the fact that there exists at least one group that is not abelian. For example, the free group of two elements.

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