What is the inverse of $g$?

Question

Define $A := \{(a,b) \in \mathbb{N}^2 \, \mid \, \gcd(a,b) = 1 \text{ and } 1 \le a < b\}$.

Does a bijection exist between $A$ and $\mathbb{N}$?

I tried the following map:

Let $g: \mathbb{N} \to A$ such that $$\begin{equation*} g(n) =\left\{ \begin{array}{ll} (1,n), \quad \text{iff } n = p^k, \text{ for some } 2 \le k \in \mathbb{Z}^{\ge 0} \text{ and } p \in \mathbb{N}\\ (a,b), \quad \text{iff } n \ne p^k \text{ for any } k \in \mathbb{Z}^{\ge 0} \text{ and } p \in \mathbb{N}, n = ab \text{ with largest factor } b>a, \text{ and } \gcd(a,b) = 1 \end{array} \right. \end{equation*} $$

For example, $g(28) = (4,7)$. Also, $g(100) = (1,100)$ because $10^2 = 100$.

Questions:

What is the explicit inverse of $g$?
Furthermore, can someone simplify the definition of $g$?
Is $g$ a bijective map? I proved it, but I want to ensure that my map is "ok." If anyone wants to see the proof, let me know.

Why am I interested in this?

I want to make $A$ into a group with a well-defined operation $\star$. I used this post for reference. However, it is not important to make $A$ into a group, but to make a bijection from $A$ to $\mathbb{N}$, because it's the only thing I am confused about from curiosity.

Edit

Using the comments below, how about changing $g$ to

$$\begin{equation*} g(n) =\left\{ \begin{array}{ll} (1,n), \quad \text{iff } n = p^k, \text{ for some } 2 \le k \in \mathbb{Z}^{\ge 0} \text{ and } p \in \mathbb{N}\\ (a, \frac{n}{a}), \quad \text{iff } n \ne p^k \text{ for any } k \in \mathbb{Z}^{\ge 0} \text{ and } p \in \mathbb{N}, \gcd(a,\frac{a}{n}) = 1, \text{ and } a \in \min_{x \in \mathbb{N}} \{x \text { divides } n \text{ and } x \ge 2\} \end{array} \right. \end{equation*} $$

There's something wrong with your definition of $g$, unless you imply $p$ to be prime (which is not stated) or $k \ge 2$, since $n = p^1$ trivially for all $p \in \mathbb N$. — heropup, Jul 25 '22 at 05:03
You still have another problem with $g$: for example, if $n = 3 \cdot 5 \cdot 7 = 105$, then we have three factorizations $n = 3 \cdot 35 = 7 \cdot 15 = 5 \cdot 21$, but $g$ cannot possibly map to all three of these. — heropup, Jul 25 '22 at 05:15
But notice it says largest factor in the following sense: largest factor $d$ of $n$ such that $d < n$ means that if other $c \mid n$, then $c \le d$. — Yoyos Tutoring, Jul 25 '22 at 05:19
The question becomes, which of these three pairs $(3,35), (7,15), (5,21)$, all of which are obviously relatively prime, maps to $n = 105$? There are no other possible choices for $n$. — heropup, Jul 25 '22 at 05:22
A general observation is that a bijection $g$ cannot have the property that $n=ab$ whenever $g(n)=(a,b)$. This is because, using @heropup's example, $(3,35)$, $(7,15)$ and $(5,21)$ are all elements of $A$, so they cannot all be mapped to $105$ by the inverse function. — Jyrki Lahtonen, Jul 25 '22 at 05:45
My previous comment applies to the edited version of $g$ as is. It is impossible to find a bijection $g:\Bbb{N}\to A$ such that $g(n)=(a,b)$ implies $n=ab$. — Jyrki Lahtonen, Jul 25 '22 at 05:48
I linked to this in chat but as it essentially answers the question, it bears repeating here: https://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals — heropup, Jul 25 '22 at 05:54

What is the inverse of $g$?

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