[Math] Show that for $\mathbb{N} = \{ 1, 2, 3, \ldots \}$, $s: \mathbb{N} \to \mathbb{N}$ where $s(n) = n + 1$ has infinitely many left inverses.

discrete mathematicsinverse

The exact textbook question is:

Let $\mathbb{N}$ denote the set $\{1,2,3,\ldots,\}$ of natural numbers, and let $s:N \to N$ be the shift map,
defined by $s(n) = n + 1$. Prove that $s$ has no right inverse, but that it has infinitely many
left inverses.

The only piece of this throwing me off is the infinitely many left inverses part.

The map $s$ is injective and it clearly has a valid left inverse function:

$l(n) = n – 1$.

$l \circ s = \text{id}_\mathbb{N}$

However, I don't see how $s$ has infinitely many left inverses. It seems to have just one left inverse.

Best Answer

Hint: Your definition of $l(n)$ (which is otherwise correct) fails for $n=1$, so you need to define that case separately.

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