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This question already has an answer here:

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.

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marked as duplicate by TheSilverDoe, Community Mar 17 at 17:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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 $n$.

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