Let $(M,+,e)$ be a commutative monoid with unit $e$. An element $a\in M$ is called cancellative element if


for any $b,c \in M$ such that $a+b=a+c$ implies that $b=c$.


Let $(\mathbf{N},+,0)$ the commutative monoid of natural numbers.


suppose that


1. we have two morphisms of monoids $f:(\mathbf{N},+,0)\rightarrow (M,+,e)$ and $g:(M,+,e)\rightarrow (\mathbf{N},+,0) $ such that $g\circ f= id$.
2. The monoid $(M,+,e)$ is torsion-free.


My question is the following: is the element $a=f(1)$ automatically a cancellative element in $(M,+,e)$ ?


Edit: By torsion-free I do mean that there does not exist a natural number $n>0$ and some element $x\in M-\{e\}$ such that $n x=e$.



Cancellation property for commutative monoid
The answer is no. Let $U=\{0,1\}$ under multiplication. Let $P$ be the semigroup of positive integers under $+$. Consider $S=P\times U$, the direct product and let $M=S\cup \{I\}$ where $I$ is an adjoined identity. Then $M$ is torsion-free, there is a homomorphism $f\colon \mathbf N\to M$ given by $f(0)=I$ and $f(n)=(n,0)$ for $n>0$ and $g\colon M\to \mathbf N$ with $g(I)=0$ and $g(n,x)=n$ for $n>0$ and $x\in \{0,1\}$ and clearly $gf=1$ but $(1,0)$ is not cancellable.


