Problem: Show the collection of all sets with the collection of all monoid homomorphisms is a category.
By definition, to satisfy a category, I need to prove two axioms 1): for every morphism $f:X \rightarrow Y$, the composites $1_Y f$ and $f1_X$ are equal to $f$ (here, X,Y are objects). and 2): Every composable triple of morphisms $h(gf)$ and $(hg)f$ are equal and denoted by $hgf$. That means $f:X \rightarrow Y, g:Y \rightarrow Z, h:Z \rightarrow W \Rightarrow hgh:X \rightarrow W$.
From Wiki, monoid has an identity element and is associative. And monoid homomorphisms have the properties such that $f(x*y)=f(x)*f(y)$ for all $x,y$ in $M$ and $f(e_M)=e_N$ where M,N are monoids.
So given a collection of monoidss M and a collection of monoid homomorphisms M. For the first axioms, to show this. Given a homomorphism $n \in N$ and $(m,*),(m',*) \in M$, I need to show $1_{m'}*N=N*1_m$. How to argue this since monoid does not contain commutative property?
For the second one,I use the associative property of homomorphism, that means we can construct three morphisms and the result is straightforward.
I need help with the first one. Thanks!
After some discussion, I think I get the idea.
Call the category $\mathbb{M}$. Objects are the class of all monoids (denoted M,N,P and so on) and morphisms here are monoid homomorphisms (denoted by f,g,h and so on).
Claim: identity is a monoid homomorphism.
Proof of claim: denote the function $id:(M,*) \rightarrow (M,*)$ such that $id(m)=m$ for all $m \in M$. Then $id(x*y)=x*y=id(x)*id(y)$ and $id(e)=e$ holds by definition. $\square$
Define $1_M$ for any $M$, which is the corresponding $id$ function. We need to show $f \circ 1_M=f=1_N \circ f$. Take any $m \in M$, then $f(1_M(m))=f(m) \in N=1_N(f(m)).$ The first axiom is shown.
Now define $f:(M,*) \rightarrow (N,*), g:(P,*) \rightarrow (M,*),$ and $h:(N,*)\rightarrow (P,*)$. To show $fg$ is defined we need to show $fg$ is a monoid homomorphism.
$fg(x*y)=f(g(x)*g(y))=fg(x)*fg(y)$ and $fg(e_P)=f(e_M)=e_N$
By similar method, $gh$ is defined. We arrive at $f(gh)=(fg)h$.
Hence $\mathbb{M}$ is indeed a category. $\square$