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Consider the function $m: \mathbb{N}^2 \to \mathbb{N}$ given by $m((a,b))=ab$. I think the function has an inverse only when $a=b=ab=1$. Am I right? Here's my reasoning.

For the function to have an inverse somewhere, it must be injective there, i.e. only $1$ element in the preimage must target a given element in the image.

i) For any $(a,b)$ s.t. $a \neq 1$ and $b \neq 1$ (i.e. $ab$ composite), we have $m((a,b))=m((1,ab))$, so there are multiple elements in the preimage mapping to $ab$.

ii) For any $(a,b)$ s.t. $a \neq b$, we have $m((a,b))=m((b,a))$, so again there are multiple elements in the preimage mapping to $ab$.

The only $(a,b) \in \mathbb{N}^2$ satisfying both these requirements is $(1,1) \; \square$.

Merry Christmas.

mjc
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    A single function either has inverse or it doesn't, there is no 'where'. You could be trying to see if some restriction of the function to some subset of its codomain or its domain results in a (new) function that has an inverse. In this case there are many more restrictions of the multiplication that have inverses. For example, all restrictions $m|_{{(a,b)}}:{(a,b)}\to {ab}$ have an inverse. – MoonLightSyzygy Dec 25 '19 at 18:12
  • A question for which what you wrote is the answer could be: Which restriction of the domain of the inverse relation is a function? – MoonLightSyzygy Dec 25 '19 at 18:14

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Your proof is correct. Here’s a slightly more direct approach.

For every $d\mid n$, $$m\left(\left(d,\frac nd\right)\right)=n,$$ and viceversa, so that every $n$ has $\sigma_0(n)$ inverses (where $\sigma_0$ denotes the divisor function). The only number with one unique inverse will be the number with only one positive divisor: $1$.

ViHdzP
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