$f,g: \mathbb{N} \rightarrow \mathbb{N} $, where $f(n)=g(2n)$. Prove that if $f$ is surjective, $g$ is not injective.

Question

Trying this question for a while.

$f,g: \mathbb{N} \rightarrow \mathbb{N} $

$\forall n \in \mathbb{N}: f(n)=g(2n)$

I need to prove that if $f$ is surjective $g$ is not injective.

Now I see that $g$ is all the even numbers, and $f$ includes both odd and even, but I cant seem to find a way to use that $f$ is surjective to prove $g$ is not injective.

Answer 1

If $f$ is surjective, then $\{g(2n):n\in\Bbb N\}=\Bbb N$.
In particular the value $g(1)\,\in\Bbb N$ is also obtained as $g(2n)$ for some $n$.