We usually say that a monoid is just a category with only one object. Thus, for every monoid $M$, viewed in as a category, there corresponds a monoid $M$ in the category of monoids and vice-versa. But this a little bit vague, isn't? What machinery do we need to make this more elegant?

Is there a way to I establish that $M \in \bf{\text{Mon}}$ and $M \in \bf{\text{Cat}}$ are equivalent? My guesses:

($\Rightarrow$) I think we can define a functor $F : \bf{\text{Mon}} \to \bf{\text{Cat}}$ that takes objects to themselves and each monoid homomorphisms to a functor between those monoids viewed as categories.

($\Leftarrow$) Really have no idea. What domain would this functor have?

Also, since $\bf{\text{Mon}}$ and $\bf{\text{Cat}}$ are different categories, is an isomorphism even possible?


  • $\begingroup$ Your functor $F : \mathbf{Mon} \to \mathbf{Cat}$ has a left adjoint. This is a natural candidate. $\endgroup$ – Zhen Lin Nov 4 '15 at 15:16
  • $\begingroup$ @ZhenLin So, my functor $F$ is well-defined? Sorry, I am a beginner in Category Theory, I am not very familiar with the concept of an adjoint yet. Would you please elaborate it a little bit more? $\endgroup$ – StudentType Nov 4 '15 at 15:25
  • $\begingroup$ @StudentType Yes, $F$ is well-defined. $\endgroup$ – Patrick Stevens Nov 4 '15 at 15:33
  • $\begingroup$ I'll lift this out to a comment, because my answer was so very wrong: Mon and Cat are not isomorphic. Indeed, the initial object is the same as the terminal object, in Mon: they're both the trivial monoid. However, in Cat, the initial and terminal objects are the empty category and the one-object one-arrow category respectively, and they're not isomorphic. $\endgroup$ – Patrick Stevens Nov 4 '15 at 16:45

You want to establish something like an equivalence of categories between monoids and the full subcategory of categories on the categories with a single object. Given a monoid $M$, there is a category $FM$, but you shouldn't say this is "the object $M$ itself" before you've proved your proposition! $FM$ is instead simply defined to be the single-object category whose morphisms $FM(*,*)$ are given by $M$. Then $F$ is a functor-just check explicitly that the monoid homomorphism axioms are the same as the axioms for a functor between one-object categories.

The inverse is similarly simple: send a one-object category to the endomorphisms of its single object. This is a monoid, and you'll check that these two functors are inverse to each other.

  • $\begingroup$ I should note, that what you describe provides an equivalence between the category of monoids and the category of one-object categories, not an isomorphism of categories. $\endgroup$ – Tashi Walde Jul 11 '18 at 15:07
  • $\begingroup$ @TasiWalde Yes, that's right, thanks, although it's an isomorphism with the one-object categories whose only object is $*$. $\endgroup$ – Kevin Carlson Jul 11 '18 at 15:51
  • $\begingroup$ @KevinCarlson: There are other fine details about how these categories are defined that would make them not obviously isomorphic. For example, the category of monoids defined as sets with a binary function and chosen element is not (obviously) isomorphic to the category of monoids defined to be the models of a finite limit sketch; e.g. because each model gets to pick its own singleton set to use in the definition of the identity element, or its own product diagram to use in the definition of multiplication. Because of things like this, I like to avoid ever mentioning isomorphisms of categories $\endgroup$ – Hurkyl Jul 12 '18 at 7:41

The more precise and invariant statement is that the category of monoids is equivalent to the category of pointed connected categories; "connected" means there is a single isomorphism class of object (as opposed to just a single object; that's an evil condition), and "pointed" means you've distinguished such an object.

If you remove "pointed," then connected categories naturally form a 2-category, not a category, and this 2-category is a bit more complicated than the 1-category of monoids.

  • $\begingroup$ The point (no pun) being that choosing a point trivialises the 2-dimensional structure in this case. $\endgroup$ – Zhen Lin Nov 4 '15 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.