Cancellation property for commutative monoid

Question

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 $n x=e$.

Answer 1

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.

Answer 2

Consider the monoid $M=\mathbb{N}\times \{0,1\}$ where $$(n,a)*(m,b):=(n+m, a\cdot b).$$ The unit element is $e:=(0,1)$. Note that this monoid is torsion free. Now consider the maps $$g:(M,*,e)\rightarrow (\mathbb{N}, +,0), g(n,a)=n$$ and $f: (\mathbb{N}, +,0) \rightarrow (M,*,e)$ such that $f(0)=e$ and $f(n)=(n,0)$. Then we have $g\circ f=id$, but $f(1)=(1,0)$ is not cancellative as $$ (1,0)*(0,0)=(1,0)=(1,0)*(0,1).$$