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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 $C$.

Tim
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  • 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 $\mu$). – anon Jul 11 '19 at 07:31
  • 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 07:48
  • $\mu$ 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 07:51
  • (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. – anon Jul 11 '19 at 07:55
  • @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 08:55
  • @runway44 see above. – Tim Jul 11 '19 at 08:55
  • In usual adjunction (with $G$ as right-adjoint functor with left-adjoint $F$) and $T=GF$ as endofunctor we have natural transformation $\eta:\mathsf{id}_X\to T=GF$ as unit and $\mu:=G\varepsilon F:T^2\to T$. And every monad $(T,\eta,\mu)$ can be represented like that. You can al find this in CWM. – drhab Jul 11 '19 at 10:01

1 Answers1

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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.

Couchy
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