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My book says that:

A monoid is a set M equipped with a binary operation $.:M\times M\to M$ and a distinguished unit element $u\in M$ such that for all $x,y,z\in M$:

$$x\cdot(y\cdot z) = (x\cdot y)\cdot z$$ and

$$u\cdot x = x = x\cdot u$$

Equivalently, a monoid is a category with just one object. The arrows of the category are the elements of the monoid. In particular, the identity arrows is the unit element $u$

So, a monoid is a category with just one object. What it means exactly? $M$ has more than one object. Later the book cites $\mathbb{N}$ as being a monoid. Why?

Poperton
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  • Like your book says, the elements of the monoid correspond to the arrows of the category; the composition function for the category is the binary map $\cdot$ on the monoid, and this is why the identity arrow corresponds to the unit element in the monoid. Associativity of the monoid operation is then exactly the same as associativity of the composition of arrows. – Hayden Mar 03 '18 at 07:22

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The monoid $M$ may have more than one element, but the elements of the monoid are not objects of the category in question.

The category in question is defined as the category whose sole object is $M$, and whose morphisms are the elements of $M$, where composition is defined as element-wise multiplication in the monoid.

The key is to then verify that the axioms for a category are satisfied.

quasi
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  • Ok, I'm trying to verify. Each arrow is an element of $M$. For example, $x\in M$. The first thing about arrows is that they must have a domain and codomain. I think the arrow $x$ has domain $M$ and codomain $M$. I can pick two arrows, $x,y$ and compose them giving me a third arrow that also goes from $M$ to $M$. Now I must show that for each element $x$, there is an identity arrow that goes from $M$ to $M$. Every element of $M$ will be an identity, right? Because every goes from $M$ to $M$. They're associative by definition. And the unit property also follows. – Poperton Mar 03 '18 at 07:38
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    So, every element is an identity? What's the usefullness of this definition? – Poperton Mar 03 '18 at 07:39
  • Actually I think not every arrow can be the identity. The definition from awodey's book says "THE identity", implicit saying that it's unique for each category element – Poperton Mar 03 '18 at 07:45
  • @Guerlando OCs: All the arrows go from $M$ to $M$, but there is only one arrow that qualifies as an identity. Since the composition is defined as monoid multiplication, the identity morphism must be the identity of the monoid (which is unique). – quasi Mar 03 '18 at 07:55
  • @Guerlando OCs: And by the identity of the monoid, of course I mean the identity element of the monoid. – quasi Mar 03 '18 at 08:24
  • by this reasoning I can understand, but I still cannot see in a pure categoric way why the unit is unique. All elements of $M$ go from $M$ to itself, so any element should be the identity. What is missing? – Poperton Mar 03 '18 at 08:40
  • @Guerlando OCs: Sure, all the elements of $M$, when regarded as arrows, are arrrow from the object $M$ to itself. Just because the arrows have domain $M$ and codomain $M$ doesn't make them identity arrows. A necessary condition for an arrow to be an identity arrow is for it to have its domain the same as its codomain. But that's not sufficient. To be an identity morphism, it has to act like an identity with respect to compostion, which in this case, is the same as saying that the corresponding monoid element acts like an identity with respect to mutliplication. – quasi Mar 03 '18 at 08:53
  • @Guerlando OCs: Consider the category of sets, whose objects are nonempty sets, and whose morphisms, from set $A$ to set $B$ are functions with domain $A$, and codomain $B$. Let $S={1,2,3}$. There are $3^2=9$ morphisms which have domain $S$ and codomain $S$, but only one of those is the identity morphism from $S$ to $S$. – quasi Mar 03 '18 at 08:57
  • @Guerlando OCs: As regards uniqeness, if $e,f$ are multiplicative identitities of the monoid $M$, then $ef=e$, since $f$ is an identity, and also, $ef=f$, since $e$ is an identity. Thus $e=ef=f$, so $e=f$. Thus, there can'r be two distinct identities for the same monoid. – quasi Mar 03 '18 at 09:06
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If $C$ is a category with a single object, then the morphisms of that category, that is, the morphisms from that single object into itself, from a monoid (the identity morphism is the unit element of that monoid).

And if $M$ is a monoid, you can create from it a category $C$ with a single object in which the morphisms of that category (which are, again, the morphisms from that single object to itself) are the elemnts of the monoid.

José Carlos Santos
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"$M$ has more than one object..."

In most cases a monoid has indeed more than one element, but the elements are not objects in this context, but are arrows/morphisms.

The composition of arrows is then the binary operation on them.

The unique object serves as domain as well as codomain for these arrows so that for every pair of arrows $f,g$ composition $f\circ g$ is defined (this in contrast with the situation in which there is more than one object).

Composition in categories is associative and the unit serves as identity.

You might wonder: "then what is that unique object?"

Actually that does not matter so much (categories can be defined without the use of the concept "object"), and you make your own choice. It is not common, but it won't hurt if for instance the identity arrow also gets the predicate "object".

drhab
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