Left and Right Inverses with semigroups

Question

Having the semigroup $(F,\circ)$ where $F=\{f: f: \mathbb{N}\to \mathbb{N}, \mathrm{Dom}(f) = \mathbb{N}\}$.

The identity $e∈F$ is the function $e(n) = n$, define the function

$g(n) = m$ if $5(m-1)<n\le 5m$,

How do I show that $g$ is onto $\mathbb{N}$ and that $g$ has multiple right inverses?

Answer 1

Translate $g$ into some more familiar language about division to see that it's onto. A right inverse $f$ is just a choice for every $m$ of an $n$ with $g(n)=m$; the fact that the defining inequality for $g$ has an interval of length $5$ in it shows that you have five different choices for $f(m)$.

EDIT: Here's some more detail.

• $g$ is onto. $g$ is simply division by $5$, rounding up remainders. So for instance $g(5n)=n$ for any $n$, thus $n$ is in the image of $g$ and $g$ is onto.
• $g$ has multiple right inverses. The most natural one is $f(n)=5n$; from the previous argument we've seen $g\circ f(n)=g(5n)=n$, as is required for a right inverse. But we can have $f$ send $n$ a bit lower as well. For instance, let $h$ be equal to $f$ except at $n=7$, and $h(7)=34$. Now $g(34)=7$, so $h$ is another right inverse of $g$ and $g$ has multiple right inverses (indeed, infinitely many.)