4

I know why an object G gives you a group when you take all its automorphisms.

But how does an object G give you a monoid when you take all its endomorphisms? What does it mean to compose two endomorphisms?

John Smith
  • 1,385
  • 8
  • 28

3 Answers3

3

Any category, by definition, contains the information of what composition of morphisms means. Specifically, a category $\mathcal{C}$ consists of

  • a collection of objects $\mathrm{ob}(\mathcal{C})$

  • for any objects $X,Y\in\mathrm{ob}(\mathcal{C})$, a set of morphisms $\mathrm{Hom}_\mathcal{C}(X,Y)$

  • for any objects $X,Y,Z\in\mathrm{ob}(\mathcal{C})$, a function (which by abuse of notation is always named $\circ$) $$\circ:\mathrm{Hom}_\mathcal{C}(X,Y)\times\mathrm{Hom}_\mathcal{C}(Y,Z)\to\mathrm{Hom}_\mathcal{C}(X,Z)$$

(which must also satisfy certain axioms). But regardless, the point is that for any object $X$ of any category $\mathcal{C}$, the composition function $$\circ:\mathrm{Hom}_\mathcal{C}(X,X)\times\mathrm{Hom}_\mathcal{C}(X,X)\to\mathrm{Hom}_\mathcal{C}(X,X)$$ is built into the structure of the category itself. This is the binary operation on $\mathrm{Hom}_\mathcal{C}(X,X)$ that makes it into a monoid.

Zev Chonoles
  • 129,973
  • Nice explanation! I didn't think that Monoid could have multiple identities, which is the case with endomorphism one. – Hi-Angel Apr 01 '16 at 07:42
  • 1
    @Hi-Angel: Thanks for the kind words! However, it is not possible for any monoid to have multiple identities. The endomorphism monoid $\mathrm{Hom}_{\mathcal{C}}(X,X)$ of any object $X$ in any category $\mathcal{C}$ has exactly one identity, namely the identity morphism $\mathrm{id}_X:X\to X$. – Zev Chonoles Apr 01 '16 at 18:41
1

It is just function composition. Say $\phi,\psi:G\rightarrow G$ are functions, if we were in the category of sets, (or endomorphisms assuming they are both endomorphisms in an arbitrary category) then $\phi\circ \psi:G\rightarrow G$ is a function, defined by $x\mapsto \phi(\psi(x))$, and thus an endomorphism, similarly for $\psi \circ \phi$. We can compose any way that we want but, we can't necessarily take inverses (as in the category of sets we can look at when an inverse to a function exists with domain $G$ i.e. injective and surjective). There is an identity so this is a monoid.

Eoin
  • 5,809
  • 2
    The most intuitive place to look is in the category of sets, since this is just composition of functions who has domain and codomain $G$. In other categories it is just composition and we do not have a notion of elements necessarily but, this is more abstract. It is merely formal. – Eoin Jun 16 '15 at 21:10
1

A category comes with a composition law; it's part of how we define categories. Generally, given any morphisms $f: A \to B$ and $g: B \to C$, we can compose them to get a morphism $g \circ f: A \to C$. We had to say what $g \circ f$ means when we defined the category that they live in.

Now if you take two morphisms $f, g: G \to G$, then since they have the same domain and codomain, you can always compose them to get a new morphism $f \circ g : G \to G$. This is a binary operation on the set $\text{End}(G)$. It's an associative operation because that's one of the conditions of what it means for something to be a category. It also has an identity element $id_G \in \text{End}(G)$, which by definition satisfies $id_g \circ f = f \circ id_G = f$ for all $f \in \text{End}(G)$. Thus $\text{End}(G)$ is indeed a monoid under this operation.

Alex G.
  • 8,848