The $2$-category of monoids People sometimes say that monoids are "categories with one object". In fact people sometimes suggest that this is the natural definition of a monoid (and likewise "groupoid with one object" as the definition of a group).
But categories naturally form a $2$-category $\mathbf{Cat}$. So if we took the above definition seriously then we would view monoids as forming a $2$-category $\mathbf{Mon}$. The objects would be monoids and the morphisms would be monoid homomorphisms, but there would also be $2$-morphisms between homomorphisms. A $2$-morphism between $f,g:M\to N$ is an $n\in N$ such that $nf(m)=g(m)n$ for all $m\in M$.
If one takes the principle of equivalence seriously then this poses a problem because we lose the ability to talk about the "underlying set" of a monoid. There's no $2$-functor $U:\mathbf{Mon}\to\mathbf{Set}$ (treating $\mathbf{Set}$ as a $2$-category with no nontrivial $2$-morphisms) that sends each monoid to its underlying set and each homomorphism to its underlying function. In the $1$-category of monoids this would be given by applying the functor $\mathrm{Hom}(\Bbb N,-)$. But in the $2$-category $\mathbf{Mon}$ two homomorphisms $f,g:\Bbb N\to M$ are isomorphic whenever $f(1)=mg(1)m^{-1}$ for some $m\in M$, so this construction only gives us the set of conjugacy classes of $M$ rather than its set of elements.
Clearly this poses a problem if we want to work with monoids and groups. In particular proofs involving finite groups often require the ability to count the number of elements in some subset of a group. It becomes impossible to state Lagrange's Theorem. We also lose the ability to talk about the free group on a set, since we can't construct the adjoint to the nonexistent functor $U$.
In light of this, I want to know if it's actually possible to take "category with one object" as our definition of monoid, and still be able to prove things in a practical way. I can see two ways to do this:
1) Recover the $1$-category of monoids from $\mathbf{Mon}$ in some natural way
or
2) Show that we can reconstruct group theory in a way that never uses concepts like "order of a group" or "free group on a set"
Does anybody know a way to do either of these?
 A: Kevin Carlson answered the question, but I thought I'd add my own answer based on his, with some more details.
The ($1$-)category $\mathbf{Set}$ lives inside the $2$-category $\mathbf{Cat}$, as the full sub-$2$-category on the discrete categories. The inclusion $F:\mathbf{Set}\to\mathbf{Cat}$ has a right adjoint $U:\mathbf{Cat}\to\mathbf{Set}$ that sends a category to its set of isomorphism classes.
So a "category with one object" (or, to better respect the principle of equivalence, a "category with one isomorphism class of objects") is precisely a category $\mathcal{M}$ such that there's a bijection $1\to U\mathcal M$. Since there's at most one such bijection we could equally well say that it's a category equipped with a bijection $a:1\to U\mathcal M$. But as I said in the question, this gives a $2$-category with unwanted $2$-morphisms.
Instead, the correct definition is to look at categories equipped with a particular object to which every other object is isomorphic. An object is precisely a functor from the terminal category, and the terminal category is equivalent to $F1$. So we define a monoid to be a category $\mathcal M$ equipped with a functor $a:F1\to\mathcal M$ which corresponds to a bijection $1\to U\mathcal M$ under the isomorphism $\mathrm{Hom}(F1,\mathcal M)\cong\mathrm{Hom}(1,U\mathcal M)$ given by the adjunction.
Based on this definition, it makes sense to say that a morphism between monoids $(\mathcal M,a)\to(\mathcal N,b)$ is a functor $f:\mathcal M\to\mathcal N$ such that $f\circ a\simeq b$, and that a 2-morphism $f\to g$ is given by a natural transformation $\alpha:f\to g$ such that $(\alpha\circ f)_\bullet = g(\mathrm{id}_\bullet)$ (where $\bullet$ is the object of $1$). Of course there is in fact only one such natural transformation, so this version of $\mathbf{Mon}$ is indeed a $1$-category.
The sort of definition we gave above is actually quite common in mathematics. Two similar definitions arise from the usual adjunction between $\mathbf{Set}$ and $\mathbf{Vect}$. A basis $S$ of a vector space $V$ is precisely a function $f:S\to UV$ that the corresponding function $FS\to V$ is an isomorphism. Dually a vector space structure $V$ on a set $S$ is a function $FS\to V$ such that the corresponding function $S\to UV$ is a bijection.
By analogy, we could say that a monoid is not a "category with one object" but rather a "category structure on the set with one element". This gives some intuition for why monoids only form a $1$-category. Categories naturally form $2$-categories, but structures based on sets are only sophisticated enough to form $1$-categories.
In fact, I believe that if we allow any set $S$ in place of $1$ in the above definition of a monoid then we get a definition of the $1$-category of categories. So we could also define monoids by first passing to this $1$-category, and then looking at the "categories with one object" within it.
A: First of all, even having one object is not invariant under equivalence. So perhaps a monoid is actually a category with a unique isomorphism class of objects. Then the way this issue is handled in topology is to consider a different 2-category: that of pointed categories, that is, categories with a distinguished object, functors preserving that object, and natural transformations which are the identity on that object. This fixes your problem: the category of pointed functors between two pointed monoids is discrete.
