Why a monoidal category with only one object is a monoid in the category of monoids?

Question

I'm reading Chapter 4 in Steve Awodey's Category Theory and it mentions

A discrete monoidal category, that is, one with a discrete underlying category, is obviously just a regular monoid (in $Sets$), while a monoidal category with only one object is a monoidal monoid, and thus exactly a commutative monoid, by the foregoing remark 4.7.

The remark says

Note that we did not really need the full group structure in this argument. Indeed, the same result holds for monoids in the category of monoids: these are exactly the commutative monoids.

Maybe it's obvious to a category theory expert that monoidal monoid and the monoid in the category of monoids have the same meaning but I can't make the connection. My understanding is the category formed by morphisms in the monoidal monoid is indeed a monoid $(M,id, \circ)$ and because it is a monoidal category $(M,\otimes,I)$ we can define the homomorphisms $\mu = \otimes$ and $\eta = I$ to makes it a monoid in the category of monoids. Is that right?

And I was also wondering why the same does not apply to the discrete monoidal category. I am grateful for explanations.

Answer 1

The only object in a monoidal category with one object is the unit $I$. Thus the only morphisms are those from $I$ to itself. These include the identity morphism and are closed under the associative composition, so form a monoid by definition. Now the monoidal structure necessitates that $I\otimes I$ is the same as $I$. Consequently, given two morphisms $f,g\colon I\to I$, you can interpret $f\otimes g\colon I\otimes I\to I\otimes I$ as a morphism $I\to I$. Thus, $\otimes$ gives a second way of combining morphisms $I\to I$. Moreover, the properties of $\otimes$ require that identity morphisms are a unit for this way of combining as well, and that it distributes over the first way: $ff'\otimes gg'=(f\otimes g)(f'\otimes g')$. This is the "monoid in the category of monoids" that is meant.