Construction of $g:\mathbb{N}\rightarrow\mathbb{N}$ surjective, such that $g^{-1}(n)=\infty,\;\forall n\in\mathbb{N}$

Question

The question is as the title describes. I found this exercise where it was asked to prove the existence of a subjective function $g:\mathbb{N}\rightarrow\mathbb{N}$, such that $g^{-1}(n)=\infty,\;\forall n\in\mathbb{N}$.

I thought in expressions like $g(n)=\left[\frac{n^2}{|\mathbb{N}|}\right]$, or $g(n)=|\mathbb{N}|-n$, but I don't fill very comfortable with them. Any suggestions or explanations for the possible expressions that I thought for $g$? :(

PS: $|\mathbb{N}|$ stands for the cardinality of $\mathbb{N}$, and $[x]$ is the integer part of $x$.

Answer 1

Consider a bijection $f : \mathbb{N} \rightarrow \mathbb{N}^2$, and consider the function $h : \mathbb{N}^2 \rightarrow \mathbb{N}$ defined for every $(x,y) \in \mathbb{N}^2$ by $$h(x,y)=x$$

Then, $g : =h \circ f$ should do the job.

Answer 2

I shall take $\mathbb{N}$ starting with $1$ so that $0\notin\mathbb{N}.$

If you want a surjective function with the preimage to have infinite cardinality for all $n\in\mathbb{N}$, you could try the following:

$f:\mathbb{N}\to \mathbb{N}$ such that $f$ sends perfect squares to $1$ and any non-squares to the smartest divisor (that is not $1$).

Explicitly, let $m\in\mathbb{N}$ and by unique factorisation to primes, we can write it as $p_1^{l_1}\cdots p_k^{l_k}$. WLOG, rearrange so that $p_1<p_2<\cdots<p_k. $

If $k=1$ and $l_1=1$ then $f(m)=1$. If $k=1$ and $l_1> 1$ then $f(m)=p_1$. If $k>1$ then $f(m)=p_1l_1$. (And $f(1)=1$).

It is surjective:

First $f(1)=1$ so $1$ is in the image. For $n\neq1$ and $n$ a composite number, we could write $n=pq$ and $p$ is prime. Then we can map $p^q\alpha$ for some $\alpha >p$ and $\alpha$ is prime to $p$. For instace, let's consider $16=2\times 8$, then $2^8\times 3$ will be mapped to $16$. $f$ is thus surjective. For $n\neq1$ and prime then any of its powers would be mapped to this prime. So surjective!

$f^{-1}(n)$ has infinite cardinality:

$f^{-1}(1)$ has infinite cardinality since there are infinitely many primes. So let us consider $n>1$. Again, say $n=pq$, $p$ prime. Notice since there are infinitely many primes, there are infinitely many choices of $\alpha$ where $\alpha$ is given as above. Lastly, if $n>1$ and prime then any of its powers work. Thus it has infinite cardinality as required.