Showing MON and CAT are equivalent categories

I am struggling with an equivalence of categories.
Let $\mathbf{Mon}$ be the category of monoids, and let $\mathbf{Cat}$ be the category whose objects are all categories with exactly one object. The morphisms from object $A$ to $B$ in $\mathbf{Cat}$ are all functors from $A$ to $B.$ I want to show that $\mathbf{Mon}$ and $\mathbf{Cat}$ are equivalent categories. Please help. Thanks in advance.

• Do you see how a monoid is a category with only one element?
– M.B.
Nov 7 '13 at 21:08
• yes. take a monoid and identity map Nov 7 '13 at 21:22
• I can construct 1 natural functor from MON to CAT, but I cannot construct the inverse Nov 7 '13 at 21:23
• If CAT is the category of locally small categories with only one element, then define a monoid from an object of CAT by letting the Hom-set be the elements and the binary product the composition.
– M.B.
Nov 7 '13 at 21:41
• They are clearly isomorphic. Nov 9 '13 at 21:24

A monoid $(M,*,1)$ can be seen as a category with only one object $\bullet$, the same way a group is a groupoid with a single object. The composition of morphisms corresponds to the multiplication $"*"$ of elements and the rule $f\circ(g∘h)=(f∘g)∘h$ means that multiplication is associative, while the single identity arrow $1_\bullet$ is the unit of $*$ since $1∘f=f∘1=f$ for any $f\in M$.
A functor $F:M\to N$ between two categories with a single object is uniquely determined by its arrow function, and since there is only one hom-set, any function is allowed, as long as it respect the composition of arrows and the identity, i.e. $F(1_∙)=1_∙$ and $F(g∘f)=F(g)∘F(f)$. But these are just the properties that we require from a homomorphism between monoids. Hence, a functor between two monoids, considered as categories with a single object, is basically the same as a homomorphism, and vice versa.