Monomorphisms of monoids

Question

Notation Let $X$ be a set. $M(X)$ will denote the free monoid on $X$. $U$ will be the forgetful functor from the category of monoids to the category of sets. Then $M(X)$ has the universal property

• for every monoid $N$ and for every function $f:X\to UN$, there exists a unique monoid homomorphism $\hat{f}:M(X)\to N$ such that $U(\hat{f})\circ i_X=f$

where $i_X:X\to UM(X)$ is the "inclusion" of $X$ in $M(X)$. I am trying to prove that:

• If $h:M\to N$ is a monomorphism in the category of monoid and monoid homomorphisms, then $U(h)$ is an injective function.

So let $h:M\to N$ be monic. Let $x,y\in M$ be such that $h(x)=h(y)$. I want to prove that $x=y$. I can consider $x$ and $y$ as functions $1\to M$ form the one-element set $1$ to $M$. So we have $h\circ x=h\circ y$ by assumption. I can't cancel $h$ on the left because $x$ and $y$ are not monoid homomorphisms, just functions. So I use the free monoid on $1$, denoted $M(1)$. I know there are monoid homomorphisms $\hat{x},\hat{y}:M(1)\to M$ such that $U(\hat{x})\circ i=x$ and $U(\hat{y})\circ i=y$. I would like to deduce that $\hat{x}=\hat{y}$, hence that $x=y$ by universality of $M(1)$. To do that, I need $h\circ \hat{x}=h\circ\hat{y}$, which I don't know how to prove:

May you give some help please?

Answer 1

You do know that $$U(h\hat{x})i = U(h)U(\hat{x})i = U(h)x = U(h)y = U(h\hat{y})i$$ though. The universal property of $M(1)$ hence yields (apply it to the function $f=U(h)x=U(h)y$) that $h\hat{x}=h\hat{y}$ and as $h$ is mono we find $\hat{x}=\hat{y}$, hence $x=\hat{x}i=\hat{y}i=y$.

Answer 2

Monomorphisms in $\textbf{Set}$ are injective functions. What you're asking for is a reflection of the fact that right adjoints preserve monomorphisms, and the forgetful functor $U:\textbf{Mon}\to\textbf{Set}$ is a right adjoint (to the free functor).

Proposition. Right adjoints preserve monomorphisms.

Proof:

An adjunction $F:C\rightleftarrows D:G$ is characterized by natural isomorphisms $$ \alpha:C(c,Gx) \xrightarrow{\sim} D(Fc,x). $$

Let $f:c\to d$ be a monomorphism in $D$, a diagram in $C$ such as $$ x\rightrightarrows Fc \xrightarrow{Ff} Fd. $$ can be transposed via $\alpha$ to a diagram in $D$ $$ Gx\rightrightarrows c \xrightarrow{f} d. $$ Since $f$ is a monomorphism, both morphisms on the left are equal, and applying $\alpha^{-1}$ to here we find out that the original pair in $C$ had to be equal i.e. $Ff$ is a monomorphism.