Mfcttrf

I'm reading Adowey's Category Theory, and I'm struggling with the last exercise of the second chapter, which is to show that the forgetful functor for monoids, $U : mathbfMon to mathbfSets$, is representable.

Well. My first reaction upon looking at the question was thinking it was wrong, because a representable functor takes objects to sets of functions, while the underlying set of a monoid has no obligation to be of that sort. I then came to the conclusion that what was actually necessary was to show that the forgetful functor was, in some way, isomorphic to a functor of the sort $mathrmHom(A_0, -)$ for some monoid $A_0$, and some googling led me to believe it would be a good idea to set $A_0 = mathbbN$, but this doesn't seem to work for the following reason:

Fix the monoid $A = (mathbbZ_2, +)$, and the monoid $B$ which is exactly the same but the $0$ and $1$ are swapped. These two monoids have the same underlying set, that is, it should definitely be true that $U(A) = U(B)$. However, notice that $mathrmHom(A_0, A)$ is not equal to $mathrmHom(A_0, B)$, for any monoid $A_0$ in fact: simply consider that the function constant equal to $0$ is an element of the former but not the latter.

This leads me to believe that there is no way that the forgetful functor can be thought of as a representable functor, because, in some sense, homomorphisms from a monoid $A_0$ to $A$ "care" about which element of $A$ is the unit, while the underlying set of $A$ does not.

Where is the error in my thinking, and how should I interpret the statement "The forgetful functor is representable"?

If representable meant $F$ is equal to $mathrmHom(R, -)$, then, as you can see from your (correct) observations, the notion would be of practically no use. Representability means instead that the functor is naturally isomorphic to $mathrmHom(R, -)$ for some $R$.

Loosely speaking, it is a general principle of Category theory that you shouldn't ask whether two things are equal, but only whether they're isomorphic instead. In particular your monoids $A$ and $B$ are isomorphic, which means that the sets $mathrmHom(ℕ, A)$ and $mathrmHom(ℕ, B)$ are isomorphic as well, so there's no contradiction with representability of $U$.

When you have an adjunction $mathsfF dashv mathsfU$ $$mathsfF: mathsfSet leftrightarrows mathsfK: mathsfU,$$ where $mathsfU$ is a forgetful like functor, $mathsfU$ is always representable. This is due to a very specific entanglement that is characteristic of the category of sets (and in general will kinda apply for $mathcalV$ in $mathcalV$-$mathsfCat$ when $mathcalV$ is monoidal closed). In fact $$mathsfU(_) cong mathsfSet(1, mathsfU(_)) cong mathsfK(mathsfF1, (_)),$$ and thus the forgetful functor is representable and is represented by the free algebra over the one element set. This is evident when $mathsfK$ is $mathsfR$-$mathsfMod$, groups, monoids and algebraic structures in general.

Notice that in the proof I used isomorphic (!!!) functors!

The "free" monoid on one element is $Bbb N_0=0,1,2,,ldots$
with addition as operation. Each monoid morphism from $Bbb N_0$
to a monoid $M$ is $nmapsto a^n$ for some $ain M$, so the monoid maps
from $Bbb N_0$ to $M$ correspond to the elements of $M$.

In short, $Bbb N_0$ represents the forgetful functor.