# 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 $\mu$). – runway44 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 $\eta$? – StackExchange for All 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. – drhab 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. – runway44 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)? – StackExchange for All 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. – Derek Elkins left SE Jul 11 '19 at 18:16
• @DerekElkins Thanks for pointing that out. Do you know if this is the unique such product? – Couchy311 Jul 11 '19 at 19:38