Denurable - onto function $\mathbb N \to A$ then $A$ is denurable

Question

I know that if there is a bijective function from $\mathbb N$ to $A$ then $A$ has the same cardinality of $\mathbb N$, therefore, $A$ is denurable. However, I was reading one answer that $A$ is denurable if there is a onto function from $\mathbb N$ to $A$. How can I prove this?

Answer 1

Hint:

If $A$ is finite, then it is denumerable.

If $A$ is infinite, then there is an injection $\mathbb N \hookrightarrow A$, which coupled with the assumed surjection $\mathbb N \to A$ shows that a bijection exists. Do you see how to use, say, the Cantor-Bernstein-Schroder theorem to show this?

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I hope this helps ^_^