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"?

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

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

  • 1
    $\begingroup$ 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 $B$. $\endgroup$ – Robly18 Jul 21 '19 at 16:51
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    $\begingroup$ $A$ and $B$ are isomorphic monoids, and $U(A)$ are $U(B)$ are "isomorphic" sets (isomorphism in the category of sets means equal cardinality). @Robly18 $\endgroup$ – Angina Seng Jul 21 '19 at 17:02

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