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A,B nonempty finite sets f is a bijection from A to B. Prove if $g:A\rightarrow B$ is injective then it's surjective. Is this true for infinite sets?

If f is a bijection from A to B all we can say is that A and B have the same cardinality because bijection means every element in A has a unique element in B and all elements in both are mapped.

If g is injective then that means every element in A has a unique element in B that it's mapped to, but because A and B are the same cardinality that means that every element in B must have an element it's mapped to in A, therefore it is surjective.

Let me know if any of the above proof is incorrect.

I don't know if this holds true for infinite sets though. I want to say no? Because if they're both infinite sets does that mean they have the same cardinality?

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  • $\begingroup$ Well in this case it must it be, because A and B have the same cardinality? If something is injective then all elements in A have a unique element in B so if it's injective it must be surjective no? $\endgroup$ – Dylan Y Nov 28 '19 at 11:45
  • $\begingroup$ O sorry I edited it. Yes g is from A to B $\endgroup$ – Dylan Y Nov 28 '19 at 11:49
  • $\begingroup$ There is bijection $\mathbb N\to\mathbb N$ (e.g. the identity). Let $g$ be defined as $n\mapsto2n$. Is it injective? Is it surjective? $\endgroup$ – drhab Nov 28 '19 at 11:53
  • $\begingroup$ It's injective but it's not surjective because 1 in the range isn't mapped to anything in the domain $\endgroup$ – Dylan Y Nov 28 '19 at 11:54
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    $\begingroup$ You better say: no element is mapped to $1$ (so not surjective). So this makes clear that the statement is not true for infinite sets, doesn't it? $\endgroup$ – drhab Nov 28 '19 at 11:56
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Your reasoning for $g$ being injective then it's surjective is intutiively fine, but formally it's a bit lacking, since the (essential) assumption of finiteness isn't used anywhere in the proof. This is a red flag, because you need $A$ and $B$ to be finite for this to be true!

You can piece together a more rigorous proof using the following two lemmas:

Lemma 1. Let $A$ and $B$ be sets and let $g : A \to B$ be an injection. Then $|A| = |g[A]|$, where $g[A]$ is the image of $g$.

Lemma 2. Let $B$ be a finite set and let $V \subseteq B$. If $|V| = |B|$, then $V=B$.

These lemmas together imply that $g$ is a bijection: Lemma 1 implies that $|g[A]| = |B|$ since $|A| = |g[A]|$ and there is a bijection $A \to B$. Lemma 2 is where the assumption of finiteness is used: setting $V = g[A]$ tells you that $g[A]=B$, which is equivalent to saying that $g$ is surjective.

Now, of course, you need to worry about whether you can use Lemmas 1 and 2 in your proof. Lemma 1 has an easy proof, but Lemma 2 is more fiddly: you can prove it by induction on $|B|$.


P.S. As drhab mentioned in the comments, this does indeed fail when $A$ and $B$ are not assumed to be (Dedekind-)finite.

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