# 'Categorical' characterization of invertible elements of monoid

Is it possible to give a 'categorical' criterion for when an arbitrary element $m$ of a monoid $M$ is invertible, by which I mean, a criterion that ideally only refers to monoid homomorphisms and does not explicitly refer to the monoid structure?

For example, it is (obviously) true that an element $m$ of a monoid $M$ is invertible iff its image under any monoid homomorphism is invertible. But here the right-hand criterion still involves the notion of invertibility in a monoid. Is there a right-hand criterion that does not make use of the explicit monoid structure? I'm guessing the answer will be 'no', but I just wanted to be sure.

UPDATE: The application I have in mind is the following: I want to determine, for an arbitrary algebraic/equational theory $T$, whether I can create a suitable definition of an 'invertible element' of a model $M$ of $T$. So if I could determine a criterion for when an element of a monoid is invertible, which could be stated in a 'categorical' or universal algebraic way that could be applied to any algebraic theory, then this would likely give me what I want.

• Regarding your application, consider the theory of semirings (like rings but additive inverses need not exist). Do you want your definition to pick out the elements of semirings which are invertible with respect to addition or multiplication? – Qiaochu Yuan Nov 17 '17 at 6:07
• Good point. I would say that I want it to pick out the elements that are invertible with respect to multiplication. But now I see that there may be many different notions of invertible elements for a single algebraic theory. – User7819 Nov 17 '17 at 19:30

Without having your actual application in mind, my first reaction is as follows.

"Categorically", we tend to identify elements $m \in M$ with homomorphism $\mathbb{N} \to M$. And we can identify invertible elements of $M$ with homomorphisms $\mathbb{Z} \to M$.

So, an element $m : \mathbb{N} \to M$ is invertible if and only if it can be written as a composite of the form $\mathbb{N} \to \mathbb{Z} \to M$, where the first map is the canonical inclusion.

My next reaction, again still not knowing if its suitable for your application, is to pick out the functor $\mathrm{Core}$ (the subgroup of invertible units) in an abstract way.

If $U : \mathbf{Groups} \to \mathbf{Monoids}$ is the forgetful functor, then I believe there is an adjunction $\hom_{\mathbf{Monoids}}(U(G), M) \cong \hom_{\mathbf{Groups}}(G, \mathrm{Core}(M))$.

Thus, you can identify the functor $\mathrm{Core}$ as being the right adjoint to $U$, and the counit $U \mathrm{Core}(M) \to M$ picks out its submonoid of invertible elements.

And, of course, an element $m \in M$ is invertible if and only if it lies in this submonoid.

• There's an additional question here, though, which is whether you can identify the functor $U$ given only the abstract structure of the category of monoids (possibly together with the free monoid $\mathbb{N}$ so you can talk about elements). For example, it might be the case that $\text{Mon}$ has an automorphism sending $U$ to a different functor. – Qiaochu Yuan Nov 16 '17 at 19:38
• Thanks for your response, it definitely gives me some ideas. I'll edit the original question to describe the application I have in mind. – User7819 Nov 16 '17 at 19:51