# Abstract Nonsense

## Category Theory: Monoids (Part I)

Point of Post: To give a perfunctory background into monoid theory, in preparation for their appearance as both an object of study, and method to study categories.

What is a monoid?

In essence a monoid is a group without inverses. Said differently, a monoid is just a set with an associative binary operation and an identity element. In fact, as we’ll see in the examples monoids come up quite often.

If you remember back to your days of group theory, you were often consternated by sets which just barely didn’t qualify as groups. More often than not, it was because they lacked inverses. $\mathbb{N}$ with addition, $\mathbb{Z}$ with multiplication, $\mathbb{R}$ with multiplication,  etc. In all these cases (and much more) the structure is actually a monoid (a commutative monoid). Also, categories can be seen as somewhat of a generalization of monoids, in some sense. In fact, not only is every monoid a category (as was commented on in the last post) but there is a natural (not in the categorical sense) to turn a category into a monoid. We’ll discuss this later. So, let us give the definition

Definition

A monoid $\left(M,\mu\right)$ is a set $M$, and a binary operation

$\mu:M\times M\to M$

such that:

a) $\mu$ is associative, in the sense that

$\mu\left(x,\mu(y,z)\right)=\mu\left(\mu(x,y),z\right)$

for all $x,y,z\in M$.

b) There exists some element $e\in M$ such that

$\mu_{\mid{\{e\}\times M}}=\mu_{\mid{M\times\{e\}}}=\text{id}_M$

Remark: It is common practice to drop the function notation $\mu$ in lieu of concatenation. Meaning, instead of having to write $\mu(x,y)$ all the time, we simply let $xy$ to mean $\mu(x,y)$. With this in mind the, admittedly menacing, axioms above become the friendly lookin

a) For all $x,y,z\in M$ it’s true that $x(yz)=(xy)z$

b) There exists some $e\in M$ such that $ex=xe=x$ for all $x\in M$

So, we now state some examples so that the reader can see the diversity in which monoids appear

Examples

Along with the previous mentioned examples, and all groups, the following are monoids:

Example: Let $X$ be a set and let $M=\left\{f:X\to X\mid f\text{ is right invertible}\right\}$. We claim that $\left(M,\circ\right)$ where $\circ$ is normal function composition forms a monoid.

To see this we first must show that $\circ$ is really a well-defined map $M\times M\to M$, but this is equivalent to showing that if $f,g:X\to X$ are right invertible, then so is $g\circ f$. But, this is clear since there exists some $\tilde{f}$ and some $\tilde{g}$ which map $X\to X$ such that $f\tilde{f}=g\tilde{g}=\text{id}_X$. So then, $\tilde{f}\circ\tilde{g}:X\to X$ and

$\left(g\circ f\right)\circ\left(\tilde{f}\circ\tilde{g}\right)=g\circ\left(f\circ\tilde{f}\right)\circ\tilde{g}=g\circ\text{id}_X\circ\tilde{g}=g\circ\tilde{g}=\text{id}_X$

so that $g\circ f$ is indeed right invertible

Next, we need to check that $\circ$ is associative, but this is true since it’s just plain function composition. Thus, all that remains is to find an identity element, but since the identity map $\text{id}:X\to X$ is right invertible we have that $\text{id}_X\in M$ and evidently $f\circ\text{id}_X=\text{id}_X\circ f=f$ for all $f\in M$ from where the conclusion follows.

Example: Let $\mathfrak{M}_2=\left\{\begin{bmatrix}a & b \\ c & d\end{bmatrix}:a,b,c,d\in F\right\}$ for any field $F$. Then, $\left(\mathfrak{M}_2,\cdot\right)$ where $\cdot$ is the normal matrix multiplication is a monoid with identity element $I$ (the identity matrix)

Example: Let $M$ be a monoid, then $\mathcal{P}\left(M\right)$ with $XY=\left\{xy:x\in X\text{ and }y\in Y\right\}$ is a monoid.

Remark: This maybe isn’t so obvious since it’s clearly not true for groups. So, we’ll prove it.

We first note that clearly if $X,Y\subseteq M$ then $XY=\left\{xy:x\in X\text{ and }y\in Y\right\}\subseteq M$ since we assumed that $M$ was closed under multiplication. So $(X,Y)\mapsto XY$ is indeed a binary operation.

Let $X,Y,Z\in\mathcal{P}\left(M\right)$ and let $\alpha\in X(YZ)$ then $\alpha=x(yz)$ for some $x\in X, y\in Y$ and $z\in Z$. But, since we assumed that $M$ was a monoid it follows that $\alpha=x(yz)=(xy)z$ and so $\alpha\in (XY)Z$. The reverse is done similarly: let $\alpha\in (XY)Z$, then $\alpha=(xy)z$ for some $x\in X,y\in Y$ and $z\in Z$. But, since monoid multiplication is always associative we see that $\alpha=(xy)z=x(yz)$ so that $\alpha\in X(YZ)$. It follows that $X(YZ)=(XY)Z$ so that the prescribed multiplication on $\mathcal{P}\left(M\right)$ is associative.

Furthermore, let $e$ be the identity element of $M$, then $\{e\}\subseteq \mathcal{P}\left(M\right)$ and

$\{e\}X=\left\{ex:x\in X\right\}=\left\{x:x\in X\right\}=X=\left\{xe:x\in X\right\}=X\{e\}$

The following theorem for monoids will prove useful, namely

Theorem: Let $M$  be a monoid, then the identity element $e$ of $M$ is unique

Proof: Let $e'\in M$ be another such element with the property that $xe'=e'x=x$ for all $x\in X$. Then, combining this with the same property for $e$ we get

$e=ee'=e'$

from where the conclusion follows. $\blacksquare$

October 8, 2010 -

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