Is multiplication by a cancellative element a bijection between divisors? Let $M$ be a commutative monoid and $m,x \in M$. Let $m^{-1}x$ be the set of elements $t$ such that $mt=x$ and suppose it not empty. Let finally $c$ be a cancellative element.
Multiplication by $c$ defines a map:
$$f_c:m^{-1}x\rightarrow m^{-1}(xc).$$
Such map is injective by definition. I can't find though any $M$ such that it is not surjective.
Could someone help me?
 A: We need to come up with some case where $mt = xc$ but $c$ is not a factor of $t$.
Consider the case of $\mathbb{Z}[\sqrt{5}]$, which is an integral domain but is not a UFD. We take it to be a monoid under multiplication. Note that this ring can be expressed as $\{a + b \sqrt{5} \mid a, b \in \mathbb{Z}\}$.
Then let $c = 2$, $m = \sqrt{5} - 1$, and $x = 2$. Note that both $m$ and $c$ are irreducible, but neither $m$ nor $c$ is prime.
Then note that $m \cdot (\sqrt{5} + 1) = 4$, so $\sqrt{5} + 1 \in m^{-1}(4)$.
But clearly, we see that 2 is not a factor of $\sqrt{5} + 1$.
Edit: as stated, this doesn't work because $m^{-1}(2)$ is empty and OP edited the question to specify that it be nonempty.  I will see if I can modify the example to make it work.
A: Found it. Let $M$ be the monoid of natural numbers with one extra element, say $\epsilon$, such that $\epsilon$ is an absorbing element. Now, every number is a cancellative element. Let $x=m=\epsilon$. Then $m^{-1}x=M$, and clearly, if $c\neq 0$, $f_c$ is not surjective, since $1$ doesn't belong to its image.
Very well.
It can be interesting now to find an example of $M$ where $xc \neq x$. I think the example above can be manipulated someway to get this other fact...
