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This question is really about proof writing. I'm new and having trouble grasping it.

If I have two sets of the same cardinality. Say, $|A| = |B| = n$ and a function $A \to B$.

If the function is injective, how do I write a proof to show that it is also surjective? I really need to just see a template.

I completely understand the logic, I understand in this particular case why it's surjective if it's injective, I'm just very unsure on how to approach writing a proof and want to understand.

Gary
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DanielMack
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  • As both the sets have the same cardinality and A is injective;it means that all the images would have unique pre- images and thus all the n elements of B would be used up,thus proving it to be surjective(/codomain=range). –  Mar 11 '22 at 10:55
  • @RamanujanXV Thank you, I completely understand that logic. Is that how I'd write the proof for it though? – DanielMack Mar 11 '22 at 11:03
  • No,you would have to be a bit more formal involving the use of sets,etc. –  Mar 11 '22 at 11:04
  • @RamanujanXV The formal part is what I'm having trouble with. Could you show me how it's done? – DanielMack Mar 11 '22 at 11:09
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    See https://math.stackexchange.com/a/989059/589 – lhf Mar 11 '22 at 11:11

1 Answers1

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Let $A=\{a_1, \dots ,a_n\}$ and $B = \{b_1, \dots ,b_n\}$ two finite sets and $f:A\rightarrow B$ a injective function.

Then, $f$ maps every $a_i$ into some $b_j$, for $i=1,\dots ,n$ and $j=1, \dots , n$, i.e., given $b_k \in B$ exists a $a_{i_{k}}$ such that $f(a_{i_k}) = b_k$.

Then $f$ is surjective.

This is a "light" proof of your problem, but I think this would help you to understand some approach to "how write a proof".

Bruno
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