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I struggle real had trying to understand the definition of monoid in Category theory. At the first glance, the definition of monoid seems nothing but the definition in abstract algebra, but when I try to understand it further by looking at some examples, nothing really make sense to me. I've looked some related post on the site (like this and this and related post inside) and on wiki, nLab etc. None of these really help. The following is the definition given in wikipedia:

In category theory, a branch of mathematics, a monoid (or monoid object, or internal monoid, or algebra) $(M, \mu, \eta)$ in a monoidal category $({\bf C}, \otimes, I)$ is an object $M$ together with two morphisms

$\bullet$ $\mu: M \otimes M \rightarrow M$ called multiplication,
$\bullet$ $\eta: I \rightarrow M$ called unit,
such that the pentagon diagram....

follow with diagrams saying the associativity law and identity law hold.

The definition itself looks very straight forward and intuitive as the categorical version of monoid.

As the first example, we have the monoid in $({\bf Set}, \times, I)$, category of set, just as the monoid with underlying set $C$. The multiplication is the binary operation of the monoid and the unit gives the identity of the monoid.

The second example, the one that confusion starts, is monoid in $({\bf Ab}, \otimes_\mathbb{Z}, \mathbb{Z})$, category of Abelian group, which recover (non-commutative) rings. The underlying object is just the abelian group. The multiplication is a bilinear map from the tensor product of abelian group with itself to itself associatively. Finally the unit map $\eta$ from $\mathbb{Z}$ to the object is suppose to give the identity of multiplication, but neither does ring always have the unity element, nor can I see how should it be a function from $\mathbb{Z}$. I feel like I'm missing something really obvious but I cannot get around it. It's very possible for my lack of understanding in tensor product though. I also had a look at the tensor product of abelian group in nLab, but that didn't help either.

Next example I tried to look at is monoid in ${\bf Top}$, the category of topological spaces, which gives the Topological monoid. It sounds like something induced from the categorical monoid and I have no idea what this is as well.

Other examples on the wiki are just totally out of my reach.

Thanks in advance!

Edit:

To make my confusion clearer, let me put down details of what I think it's happening in these examples.

The monoid in $({\bf Set}, \times, \{*\})$:
$\bullet$ An object $C$ is a set.
$\bullet$ The (associative) multiplication morphism $\mu: C\times C \rightarrow C$ defines the distributive, associative binary operation on the set $C$.
$\bullet$ The unit morphism $\eta : \{*\} \rightarrow C$ gives the identity element of the binary operation.

Together we get the monoid in abstract algebra.

The monoid in $({\bf Ab}, \otimes_\mathbb{Z}, \mathbb{Z})$:
$\bullet$ An object $A$ is an abelian group.
$\bullet$ The multiplication morphism $\mu : A\otimes A \rightarrow A$ again gives an associative, distributive binary operation on $A$, which is not the same as addition in $A$. Call it "multiplication".
$\bullet$ The unit morphism $\eta : \mathbb{Z} \rightarrow A$ gives the identity element of "multiplication"

Together we get a unital ring, except we didn't? If we only need "the identity" of multiplication, why do we need $\mathbb{Z}$ as the domain? In PrudiiArca's answer they says the case $I=\{*\}=0$ the trivial group has nothing interesting to discuss, but I can't really understand why.

Monoid in $({\bf Top}, \times, \{*\})$:
Thank to Qi Zhu's comment I think I now understand this example. Somehow in my head this wasn't a topological space with monoid structure but something unimaginable.
$\bullet$ An object $(X, \tau)$ is a topological space.
$\bullet$ I think multiplication and unit morphism are defined similarly to the morphisms in ${\bf Set}$.

and we get a topological space with monoid operation, the topological monoid.

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    $\begingroup$ A monoid in $\mathbf{Ab}$ is a unital ring. $\endgroup$ Nov 13, 2021 at 20:42
  • $\begingroup$ @GeoffreyTrang Ah that makes alot of sense. It is unital by construction! $\endgroup$
    – cct
    Nov 13, 2021 at 20:52
  • $\begingroup$ @GeoffreyTrang And in general, if $R$ is a unital ring then a monoid in $R-Mod$ is a unital $R$-algebra. $\endgroup$ Nov 13, 2021 at 23:14
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    $\begingroup$ A topological monoid is really just a monoid with a topology such that the monoid operation is continuous. (Maybe you're familiar with topological groups? That would be a group object in $\mathbf{Top}$.) $\endgroup$
    – Qi Zhu
    Nov 14, 2021 at 7:14
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    $\begingroup$ @cct Not group structure, but monoid structure. Yes, $(\mathbb R, +)$ is a topological group (and thus a topological monoid). So yes, a monoid object in any category is some object along with an additional monoid structure and similarly for group objects. Hence, typical examples like groups, topological groups, Lie groups, group schemes, etc. $\endgroup$
    – Qi Zhu
    Nov 14, 2021 at 14:51

1 Answer 1

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Since it is not perfectly clear to me, what is unintuitive to you, let me make some remarks.

While people often think category theory is about doing things without elements, it rather is about working with morphisms between objects. It is very often the case that we can express elements as morphisms, hence we can still work with them.

In the case of sets and topological spaces we can express elements as functions / continuous maps $* \rightarrow X$, where $*=\{*\}$ is the terminal object given by the singleton set/space. Hence given an element $(x,y): * \rightarrow X \times X$ the composite $m \circ (x,y) : * \rightarrow X$ defines an element of $X$. Try to convince yourself that in this light the axioms stating a monoid object precisely capture the usual axioms, when we make the notational convention $m(x,y) := x\cdot y$. Note further that in the case of topological spaces the fact that we work in the category of spaces enforces the multiplication map to be a continuous map. For monoids this is quite lame, but you can imagine that it is much easier to say „a topological ring is a ring object in $\mathsf{Top}$“ than to say „a topological ring is a space $R$ together with continuous maps … which satisfy …“

In more general categories elements can have more general forms. In the case of abelian groups we consider group homomorphisms and the terminal object is the trivial group $*=\{0\}$. Since group homomorphisms send 0 to 0, elements in the sense of morphisms from the terminal object always pick the zero element, which becomes really uninteresting really really fast. The „right“ notion of an element of a group (in categorical language) is that of a map $\Bbb Z \rightarrow A$. This is by the universal property of $\Bbb Z$ being the free abelian group on one generator, stating that given any element $a\in A$ of the set $A$ there is a unique group homomorphism $\Bbb Z \rightarrow A$ sending 1 to $a$. Hence again we can pick elements of our groups just talking about morphisms in the category at hand (by which I mean without referencing to things like underlying sets or such), but we need a different object than the terminal one to get them.

There are two natural „products“ in the category of abelian groups, the biproduct and the tensor product. I don’t actually know what happens when considering monoid objects with respect to the biproduct, but I would guess that it turns out to be always the group structure itself. Anyway, monoids become interesting, when we use the tensor product. A map $m: A \otimes A \rightarrow A$ classifies a bilinear map $A \times A \rightarrow A$ and with our discussion of elements above it is not hard to see that this bilinearity precisely captures a distributivity law of our multiplication $m$ with the additive structure of our abelian group $A$. You are perfectly right in that the dedicated homomorphism $\Bbb Z \rightarrow A$ makes this monoid thing into a unital ring.

What is the advantage of this categorical perspective? Well, there are instances, in which the category we want to consider does not admit a good definition of elements. For example we can consider for any category $\mathcal C$ its endofunctor category $\operatorname{End}(\mathcal C)$. Besides the pointwise cartesian product (for which $\mathcal C$ has to be good enough) it admits a natural (in general not commutative!) product given by composition of endofunctors. The monoid objects with respect to this product structure are called monads and give rise to a very nice theory, with which many effects in algebra and category theory can be explained.

So at the end it is not important to abandon elements per se, but writing things without explicitly mentioning elements helps to discover very useful tools…

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  • $\begingroup$ Thank you very much for you answer. I still can't fully digest what you wrote but let me elaborate what confuses me. I can understand the group homomorphism $\mathbb{Z} \rightarrow A$ exists naturally and defining the bilinear "product" map on the tensor product gives us distributivity for the "mutiplication". What's stuck in my head is that, the "unit" morphism should give the identity element of multiplication of the ring, i.e. $\eta$ maps $\mathbb{Z}$ to the identity of multiplication. However from you answer, it looks more like we are considering $\mathbb{Z}$ because it's free on $\{*\}$? $\endgroup$
    – cct
    Nov 14, 2021 at 14:33
  • $\begingroup$ I added more details to the question! $\endgroup$
    – cct
    Nov 14, 2021 at 15:28
  • $\begingroup$ I edited my answer to be more precise. You are absolutely correct that we consider $\Bbb Z$, because it is free on on generator. You may think of this like that: the terminal object in groups is at the same time the initial object. But this means there are no interesting maps out of it. So we need a tiny bit more structure to probe elements. But as soon as we have two elements we can double this element etc. Hence we need the free abelian group on one generator. $\endgroup$ Nov 14, 2021 at 16:47
  • $\begingroup$ @PrudiiArac Thank you for the extra details. I think I'm almost there. You are absolutely right about the misconception that I'm thinking the category only as the objects forgetting detail. I'm missing out that we can just recover the elements of the underlying set in most category by the morphism from a singleton set to the object. In the case of ${\bf Set}$ and ${\bf Top}$, an element in the set (object) $C$ is just a morphism $\{*\} \rightarrow C$, but in ${\bf Ab}$, we cannot recover an element of the group from $\{*\} \rightarrow A$, since $\{*\}=0$ is initial in ${\bf Ab}$. $\endgroup$
    – cct
    Nov 14, 2021 at 18:41
  • $\begingroup$ Hence to find a suitable object in ${\bf Ab}$ which have morphisms to the target object $A$ that represent the elements of $A$, we utilize the property that $\mathbb{Z}$ is free on the singleton set in ${\bf Ab}$. Although the function sending $*$ to an element of $A$ is not a morphism in ${\bf Ab}$, this function induces a morphism $\mathbb{Z} \rightarrow A$ "uniquely" and hence can be used to recover the elements of $A$! Since $\{*\}$ is actually free on itself is it safe to say we are taking $I$ in the definition that is free on $\{*\}$ in the category? $\endgroup$
    – cct
    Nov 14, 2021 at 18:41

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