monoids.md

We spent quite a while talking about whether a monoid M (thought of as a one object category) could be equivalent to its opposite.

First, I want to point out that this is not what Leinster was actually asking in 1.2.23(b) --- he asked if it could be isomorphic to its opposite, contrary to my insistence that you shouldn't even talk about isomorphism of categories, only equivalence.

Lemma. A functor between monoids is exactly a monoid homomorphism. Lemma. A natural transformation between two monoid homomorphisms F,G : M → N is an element s : N showing that F and G are conjugate, i.e. s G(r) = F(r) s for all r : M. Lemma. A natural transformation is a natural isomorphism if the element s : N is invertible.

(These are all straightforward exercises.)

Thus an equivalence between monoids M, N consists of a pair of monoid homomorphisms F : M → N and G : N → M, and invertible elements m : M, n : N, such that m F(G(r)) = r m for all r : M, and n G(F(s)) = s n for all s : N. It is an isomorphism only if m and n are the identities.

Let's first try to find a monoid M not isomorphic to its opposite.

• The free monoid we discussed in class was not useful --- the map which reverses words is a perfectly good functor.
• No commutative monoid can work, since then the identity map will be a functor, and easily enough an isomorphism.
• There are two good solutions on Stack Exchange:
  • End(X) for any set X with at least two elements, because any constant function is left-absorbing (i.e. c ∘ f = c for any f : End(X)), but there are no right-absorbing functions, and an anti-isomorphism End(X) → End(X) would have to take left-absorbing elements to right-absorbing elements.
  • The monoid with elements { 1, a, b } satisfying a a = a, a b = a, b a = b, b b = b easily gives another counterexample.

Are these counterexamples also not equivalent to their opposites?

• The second example is easy: the only invertible element is 1, so equivalence immediately reduces to isomorphism. (That is, this monoid is not even equivalent to its opposite.)
• The endomorphism example is a bit trickier... using the characterisation of equivalences as functors which are fully faithful and essentially surjective, we can see that the homomorphism F must be a bijection, and from this we can see that the image of a left-absorbing elements must be right-absorbing, giving the same contradiction as above. Is there an easier way, that doesn't go via that characterisation?