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Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

Suppose that $f$ is a mapping from a set $A$ to a finite set $B$. If $A$ is finite and $|A| > |B|$, then $f$ is not injective. If $A$ is infinite, then there exists $b \in B$ such that $f^{−1}[\{b\}]$ is infinite.

  1. If $A$ is finite and $|A| > |B|$, then $f$ is not injective

Lemma: Suppose that $B$ is finite and that $g:B\to A$ is surjective. Then $A$ is finite and $|A|\le|B|$. (I presented a proof here)

Assume the contrary that $f$ is injective. We define a mapping $g$ from $B$ to $A$ by $g(b)=f^{-1}(b)$ for all $b\in\operatorname{ran}f$ and $g(b)=\bar a$ for all $b \in B\setminus \operatorname{ran}f$ for some $\bar a \in A$. We have $f^{-1}[\operatorname{ran}f]= \operatorname{dom}f$ $=A$. Thus $g[\operatorname{ran}f]=A$. Hence $g:B\to A$ is surjective. Moreover, $B$ is finite, then $A$ is finite and $|A|\le|B|$ by Lemma. This contradicts the fact that $|A|>|B|$. Hence $f$ is not injective.

  1. If $A$ is infinite, then there exists $b \in B$ such that $f^{−1}[\{b\}]$ is infinite

Lemma: Finite union of finite sets is finite

Assume the contrary that $f^{−1}[\{b\}]$ is finite for all $b\in B$. It's clear that $\bigcup\limits_{b\in B}f^{−1}[\{b\}]=A$. Furthermore, $\bigcup\limits_{b\in B}f^{−1}[\{b\}]$ is a finite union of finite sets. It follows from the Lemma that $A=\bigcup\limits_{b\in B}f^{−1}[\{b\}]$ is finite. This contradicts the fact that $A$ is infinite. Hence there exists $b\in B$ such that $f^{−1}[\{b\}]$ is infinite.

Akira
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    Both seems good, but let me suggest a different way to 1, $|A|>|B|\implies |A|\ge |B|$ and if $f$ is injective then $|A|\le |B|$ by Schröder–Bernstein theorem $|A|=|B|$, contradiction. – ℋolo Sep 22 '18 at 11:11
  • Thank you @Holo! It seems to me that ($f:A\to B$ is injective and $A,B$ are finite $\implies |A|\le |B|$) requires a proof. Coincidentally I presented a proof at https://math.stackexchange.com/questions/2785884/let-fi-m-to-i-n-be-injective-then-m-leq-n :) – Akira Sep 22 '18 at 12:49
  • Hi @Holo Many of my recent proofs depend heavily on the lemma above. If you don't mind, please have have a check on it at https://math.stackexchange.com/a/2926113/368425! – Akira Sep 22 '18 at 13:26
  • Hi, Let's move to chat, because, although the general approach looks okay I have a few things to point and it will be easier there $\Large{}^{\frown}_{~\ddot{}}$ – ℋolo Sep 22 '18 at 13:50

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