Reflective subcategories of monoids

Question

An exercise in The Joy of Cats, p. 59, is as follows:

Show that no finite monoid, considered as a category, has a proper reflective subcategory.

The obvious idea is to let $r : \cdot \to \cdot$ be a reflector. Then by assumption every arrow $f$ factors as $f' \circ r$, where $f'$ is in the subcategory. Now if we can show that $r$ is itself in the subcategory, then we win... But this is giving me some trouble.

The theorem is false for infinite monoids (and the second part of this problem, which I've done, gives a counterexample), but I'm not sure how to leverage finiteness without knowing my monoid is cancellative. We can conclude lots of things by pigeonhole, but I'm not sure how to apply them.

Any help is appreciated ^_^

Answer 1

Let $M$ be a finite monoid and let $N$ be a proper submonoid of $M$. For each $r \in M$, one has $|Nr| \leqslant |N| < |M|$. Therefore, for each $r \in M$, there exists $f \in M$ such that for all $f' \in N$, $f \not= f'r$.

Answer 2

I wrote about Monads on Monoids and Adjunctions between Monoids on my blog.

If you write out the definition of an adjunction between one-object-categories in terms of the corresponding monoids then you get the following:

An adjuction from a monoid $M$ to a monoid $N$ consists of homomorphisms $\lambda:M\to N$ and $\rho:N\to M$ along with elements $h\in M$ and $e\in N$ such that

\begin{align} &&\rho(e)h&=1_M,\tag{1}\\ &&e\lambda(h)&=1_N,\tag{2}\\ &\forall x\in M:&\rho(\lambda(x))h&= hx,\tag{3}\\ &\forall y\in N:& e\lambda(\rho(y))&= ye.\tag{4} \end{align}

Combining (1) and (3) gives $\rho(e)\rho(\lambda(x))h= x$, so $\lambda$ is injective since it has left inverse $y\mapsto \rho(e)\rho(y)h$. Likewise (2) and (4) imply $\rho$ is injective. So $\rho\circ\lambda$ and $\lambda\circ\rho$ are injections $M\to M$ and $N\to N$. If $M$ and $N$ are finite this means they must also be bijections, and hence $\lambda$ and $\rho$ are also bijections. So if $\lambda$ or $\rho$ is an inclusion map then it must in fact be the identity.