# Why is the forgetful functor representable?

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, $$U : \mathbf{Mon} \to \mathbf{Sets}$$, 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 $$\mathrm{Hom}(A_0, -)$$ for some monoid $$A_0$$, and some googling led me to believe it would be a good idea to set $$A_0 = \mathbb{N}$$, but this doesn't seem to work for the following reason:

Fix the monoid $$A = (\mathbb{Z}_2, +)$$, 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 $$\mathrm{Hom}(A_0, A)$$ is not equal to $$\mathrm{Hom}(A_0, B)$$, 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"?

• Could you explain in more detail what you mean by "the $0$ and $1$ are swapped"? – Ruben Jul 21 '19 at 16:43
• @Ruben $B$ is the monoid with set ${0, 1}$, with the table: $1 + x = x + 1 = x$, $0 + 0 = 1$ – Robly18 Jul 21 '19 at 16:52

If representable meant $$F$$ is equal to $$\mathrm{Hom}(R, -)$$, 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 $$\mathrm{Hom}(R, -)$$ 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 $$\mathrm{Hom}(ℕ, A)$$ and $$\mathrm{Hom}(ℕ, B)$$ are isomorphic as well, so there's no contradiction with representability of $$U$$.

• As of now this is the only answer that actually adresses the question (and gives the correct answer !), +1 – Maxime Ramzi Jul 21 '19 at 17:16
• Well, while this addresses the immediate confusion, the other explain why $ℕ$ is supposed to be the representing object. – user54748 Jul 21 '19 at 17:33

When you have an adjunction $$\mathsf{F} \dashv \mathsf{U}$$ $$\mathsf{F}: \mathsf{Set} \leftrightarrows \mathsf{K}: \mathsf{U},$$ where $$\mathsf{U}$$ is a forgetful like functor, $$\mathsf{U}$$ 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 $$\mathcal{V}$$ in $$\mathcal{V}$$-$$\mathsf{Cat}$$ when $$\mathcal{V}$$ is monoidal closed). In fact $$\mathsf{U}(\_) \cong \mathsf{Set}(1, \mathsf{U}(\_)) \cong \mathsf{K}(\mathsf{F}1, (\_)),$$ and thus the forgetful functor is representable and is represented by the free algebra over the one element set. This is evident when $$\mathsf{K}$$ is $$\mathsf{R}$$-$$\mathsf{Mod}$$, 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 $$n\mapsto a^n$$ for some $$a\in M$$, 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.

• That does not address my question. Consider the monoids $A$ and $B$ I describe in my post. We have $U(A) = U(B)$, yet the set of morphisms from $\mathbb{N}$ to $A$ is not the same as the set of morphisms from $\mathbb{N}$ to $B$. – Robly18 Jul 21 '19 at 16:51
• $A$ and $B$ are isomorphic monoids, and $U(A)$ are $U(B)$ are "isomorphic" sets (isomorphism in the category of sets means equal cardinality). @Robly18 – Angina Seng Jul 21 '19 at 17:02