Unit of a monoid object in a monoidal category

Question

Let $(M, \mu, \eta)$ be a monoid object in a monoidal category $(C, \otimes, I)$, with multiplication $\mu:M\otimes M\rightarrow M$ and unit morphism $\eta:I\rightarrow M$. I'm trying to understand what $\eta$ does. It'd help if someone could illustrate through specific examples.

If $C= $Vect$_k$, then a monoid object would be a $k$-algebra $M$. Now what is the map $\eta:k\rightarrow M$? Where does it map each element of $k$ to?

Likewise, if $C$ is the category of abelian groups with tensor product, then a monoidal object is a ring $M$. What would be the map $\eta:\mathbb{Z}\rightarrow M$? Is it just any map that embeds a copy of $\eta$ in $M$? Or does it not even have to be injective? I'd appreciate some concrete example. Thanks in advance.

Answer 1

As the name suggests, the unit map $\eta$ picks out the unit of the monoid $M$, so in the first example the map $k\to M$ is determined by sending $1\in k$ to the multiplicative unit of the $k$-algebra $M$, and likewise for the second example.

Answer 2

As is often the case it might be a good approach to first consider a monoid object in the cartesian monoidal category $\mathsf{Set}$. Here the unit object of the monoidal structure is given by the singleton set $\{*\}$. Hence, given a monoid object $(M,m,u)$ in $\mathsf{Set}$, the unit $u:\{*\}\rightarrow M$ just picks the unit element of the multiplication operation.

Another example is given by the monoidal categories $\mathsf{Mod}_R$ of modules over some commutative ring $R$ and tensor products over $R$ (a module essentially an vector space over some ring). This is the right setting of both of your examples, since $\mathsf{Ab}=\mathsf{Mod}_\mathbb{Z}$ and $\mathsf{Vect}_K = \mathsf{Mod}_K$. Here the unit object is not a singleton set anymore but the ring $R$ considered as $R$-module over itself. While this might potentially have infinitely many elements, the ring structure on $R$ yields a uniquely determined element $1_R$ and the axioms for a morphism of modules force that a morphism of modules $R \rightarrow M$ is uniquely determined by its image of $1_R$. In this sense the unit morphism $u:R \rightarrow A$ of a monoid object $(A,m,u)$ in $\mathsf{Mod}_R$ determines a unit element of the $R$-algebra multiplication $m:A \otimes_R A \rightarrow A$.

Answer 3

There are already good answers to the specific question asked, but I think it might help to add a generality on how to think of maps $I\to X$ for $I$ the unit object in a monoidal category, and $X$ any object of the monoidal category.

Notation

Let's set up notation. I'm going to slightly modify it for my convenience (I like lower case objects, so I'll change the identity of the category to be called 1 so it doesn't look weird). Let $(C,\otimes,1)$ be a monoidal category, objects in $C$ will be lower case letters, $x,y,z,w,\cdots$. The homs in $C$ from $x$ to $y$ are denoted $C(x,y)$.

How do we think of $C(1,x)$?

We should think of $C(1,x)$ as being "the underlying set" of $x$. Why does this make sense? Well, firstly, $C(1,-)$ is a functor from $C$ to $\newcommand\Set{\mathbf{Set}}\Set$.

Moreover, this functor respects the monoidal structure (it is a lax monoidal functor), in that we have canonical maps
$$C(1,x)\times C(1,y) \to C(1,x\otimes y)$$ and $$\{*\}\to C(1,1),$$ where the first is given by $(f,g)\mapsto (f\otimes g) \circ \mu^{-1}$, where $\mu :1\otimes 1\to 1$ is the unit isomorphism, and the second is given by $*\mapsto \mathrm{id}_1$.

In addition, these maps satisfy certain associativity and unitality conditions that you can find on the linked nlab page.

In other words, the "take the underlying set" operation has good properties, and it actually comes up a lot in the theory of enriched categories, but I'll leave it there for now.

Why should we think of this as the underlying set, rather than some other associated set?

Well, the short answer is that in a lot of examples, $C(1,x)$ does return the actual underlying set of $x$, where this makes sense. Here are some examples:

$R$-modules:

If $R$ is a commutative (unital) ring (like $\Bbb{Z}$ or a field), then the category of $R$-modules comes equipped with a tensor product that makes it a monoidal category whose unit is $R$ regarded as an $R$-module.

Then we have a well known natural isomorphism $$\operatorname{Hom}_R(R,M)\simeq M$$ for all $R$-modules $M$ induced by the correspondence $$\phi \mapsto \phi(1),$$ $$(a\mapsto a\cdot m) \longleftarrow{\raise{.4pt}{\hspace{-5pt}\shortmid}} m$$

Sets, topological spaces, etc:

For sets and topological spaces, we give the category the cartesian monoidal structure, and the terminal object is the point, $\{*\}$. Morphisms from the point to a set or topological space correspond bijectively to the underlying set of points of our set or space.

Sheaves/Presheaves

Similarly, for (pre)sheaves of sets, we also use the cartesian monoidal structure, and morphisms from the terminal object to a (pre)sheaf correspond to global sections of the (pre)sheaf.

Comment It might be also be helpful to keep this perspective of $C(1,-)$ being the global sections functor in mind as well.

Relating this back to the unit morphism

(This has already been explained in other answers, so I'll keep it brief)

The way you should think about the map $\eta : 1\to m$ is as picking out the unit element of $m$. Just like with a monoid in $\Set$, we need to know what the unit of a monoid in $C$ is, and $\eta$ tells us.