What do the multiplication and unit mean in a monoid in a monoidal category?

Wikipedia says

In category theory, a monoid (or monoid object) $$(M, μ, η)$$ in a monoidal category $$(C, ⊗, I)$$ is an object $$M$$ together with two morphisms

• $$μ: M ⊗ M → M$$ called multiplication,
• $$η: I → M$$ called unit,

Is it correct that the monoidal category $$(C, ⊗, I)$$ is irrelevant to the definitions of $$\mu$$ and $$\eta$$? In particular, is the definition of $$\mu$$ independent of $$⊗$$, and the definition of $$\eta$$ independent of $$I$$?

$$\mu$$ represents the binary operation on the monoid $$M$$, just like the binary operation on a monoid in the usual sense (in $$Set$$). Then what does $$\eta$$ mean here? For example, what is $$\eta$$ in each of the following examples?

A monoid object in $$Set$$, the category of sets (with the monoidal structure induced by the Cartesian product), is a monoid in the usual sense.`

and

For any category $$C$$, the category $$[C,C]$$ of its endofunctors has a monoidal structure induced by the composition and the identity functor $$I_C$$. A monoid object in $$[C,C]$$ is a monad on $$C$$.

• Example one: $\mu$ is a binary operation on a set which makes it a monoid, and $\eta$ maps $\ast$ from $\{\ast\}$ (the one-element set) to the identity element (with respect to $\mu$). Jul 11 '19 at 7:31
• Thanks. (1) If the purpose of $\eta$ is to specify the identity element in the monoid, why not directly specify the identity element in the monoid? Is it because in the general definition of a monoid in a monoidal category, a monoid might not be a set and therefore might not have elements? (2) In the second example, what is the identity "element" (wrt $\mu$) and what is $\eta$?
– Tim
Jul 11 '19 at 7:48
• $\mu$ and $\eta$ are just two arrows that meet special conditions. CWM (great book!) is in your possesion right? Have a good look there. I do not understand what you mean with "independent" here. Jul 11 '19 at 7:51
• (1) Yes. One of category theory's fundamental mantra is replace elements with arrows to generalize. (2) Dunno. Also, I'm with drhab in not understanding what you mean by independent. Jul 11 '19 at 7:55
• @drhab by "independent" I mean "irrelevant". see my update. Yes, I am reading CWM besides Wikipedia. I cited WIkipedia only because it is easier to copy. Do you know what is the identity "element" (wrt $\mu$) and what is $\eta$ in the category of endofunctors (the second example)?
– Tim
Jul 11 '19 at 8:55

I am not entirely sure I understand your question, but it appears to me you are trying to understand the relation between a monad and a monoid of endofunctors, specifically in your last point:

For any category C, the category [C,C] of its endofunctors has a monoidal structure induced by the composition and the identity functor IC. A monoid object in [C,C] is a monad on C.

Maybe I can illustrate this relation in more detail.

Let's first ask what a monoid object in a category $$\mathcal C$$ is. We would like to say it is a triple $$(M,\cdot,1)$$ with $$\cdot:M\times M\to M$$ and $$1:*\to M$$ satisfying the usual identity and associativity, where $$*$$ is the terminal object. This definition requires two things, namely that $$\mathcal C$$ has products and a terminal object.

We would like to define such an object in our category of endofunctors $$\mathcal D=[\mathcal C,\mathcal C]$$. However we do not in general have a product and terminal object. But the category $$\mathcal D$$ has a weaker structure, which is the structure of a (strict) monoidal category. Roughly, this means we have a tensor product $$S\otimes T \equiv S\circ T$$ for any two objects (note this is not commutative), and a unit object $${\bf 1} \equiv \text{Id}_{\mathcal C}$$ such that $$(S\otimes T)\otimes U = S\otimes (T\otimes U)$$ and $$S\otimes {\bf 1} = S = {\bf 1}\otimes S.$$ (Note: the equalities here come from the "strict" monoidal category. Otherwise we would only have isomorphisms)

We can still define a monoid-like object in such a category. It is given by a triple $$(M,\mu,\eta)$$ with $$\mu:M\otimes M\to M$$, and $$\eta:{\bf 1}\to M$$ with the usual associativity and identity conditions $$\mu\circ (\mu\otimes\text{Id}_M) = \mu\circ(\text{Id}_M\otimes \mu)$$ and $$\mu\circ (\text{Id}_M\otimes\eta) = \text{Id}_{M\otimes I} = \text{Id}_M=\text{Id}_{I\otimes M} =\mu\circ (\eta\otimes\text{Id}_M).$$

Now if we consider such a monoid-like object in the endofunctor category $$\mathcal D$$ with the given monoidal structure, then this is a triple $$(M,\mu,\eta)$$ with $$\mu:M^2\to M$$, $$\eta:{\text{Id}_{\mathcal C}}\to M$$ satisfying $$\mu\circ (\mu M) = \mu\circ (M\mu)$$ and $$\mu\circ (M\eta) = \text{Id}_M = \mu\circ (\eta M).$$

These are precisely the conditions for a monad on the endofunctor $$M$$, where $$\mu M$$, $$M\mu$$, $$\eta M$$, and $$M \eta$$ are the whiskering operations.

Now you may be wondering how I got from $$\epsilon\otimes\text{Id}_M$$ to $$\epsilon M$$ and $$\text{Id}_M\otimes\epsilon$$ to $$M\epsilon$$. The trouble is that we need to define the tensor $$\delta\otimes \epsilon$$ of two maps $$\delta:S\to U$$ and $$\epsilon:T\to V$$ inside of $$\mathcal D$$. This is not easy to define in this context, but to give an idea, this can be defined pointwise by $$(\delta\otimes \epsilon)_X \equiv U(\epsilon_X)\circ\delta_{T(X)} = \delta_{V(X)}\circ S(\epsilon_X).$$ One can check that when $$\delta$$ or $$\epsilon$$ is the identity natural transformation, then this corresponds to a whiskering operation.

If I recall correctly this (roughly) corresponds to defining composition in the category of monads (I also think this is a good exercise), and there are actually multiple ways of doing this.

• $\otimes$'s action on arrows in this context is just horizontal composition of natural transformations. Jul 11 '19 at 18:16
• @DerekElkins Thanks for pointing that out. Do you know if this is the unique such product? Jul 11 '19 at 19:38