Injective but not surjective on an infinite set left inverse not unique. $f : \mathbb {N} \to \mathbb {N}$ be a function such that $ \forall n \in \mathbb{N} $ 
$f(n) =n+1$
i)Prove that f has at least two different left inverses.
ii)Find at least 2 of them.
I made this question/function up to help me understand the concept behind injectivity and surjectivity. I was trying to make a function that acted on a set back to that set that was injective but not surjective and find at least 2 left inverses.
 A: I'll assume $\Bbb N = \{1, 2, 3, \ldots\}$, that is, that $\Bbb N$ starts with $1$ and not $0$; everything is easily modified if your $\Bbb N$ starts with $0$.
The key here is to identify what is forced to happen for any $g : \Bbb N \to \Bbb N$ such that $g \circ f = \operatorname{id}_\Bbb N$, and what leeway such a function $g$ has.
Picking $n = 5$ as an arbitrary example, then because $g$ is a left-inverse of $f$, we know that 
$$5 = (g \circ f)(5) = g(f(5)) = g(6)$$
and so $g(6)$ cannot be anything but $5$: because $6$ is in the image of $f$, the value of $g(6)$ is completely determined.
More generally, 


*

*$g$ is forced to take particular values on all of $\operatorname{im}(f) = \{2, 3, 4, \ldots\}$. Consequently, 

*we are free to define the value of $g$ on $\Bbb N \setminus \operatorname{im}(f) = \{1\}$ in any way we'd like. 
This is where our leeway is to be found.
A: How about the function $g(n) = n-1$ if $n=1,2,3,\ldots$ and $g(0) = 10$. And then $h(n) = n-1$ for $n=1,2,3,\ldots$ and $h(0) = 20$? These are both left inverses.
