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If $|A| \geq |B|$, then there exists an onto function $f: A \rightarrow B$.

If $|B| \leq |A|$, then there exists a one-to-one function $f: B \rightarrow A$.

My issue is that I don't think that $|A| \geq |B|$ implies $|B| \leq |A|$. If there is an onto function from A to B, there isn't necessarily a one-to-one function from B to A.

Asaf Karagila
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    This statement is precisely equivalent to the Axiom of Choice, so it cannot be proved in some sense. However, if you assume that $A,B$ are both finite sets, then this is easy to be verified. – Crostul Feb 03 '16 at 18:08
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    @Crostul: That is incorrect. The axiom of choice is equivalent to "every surjection has an inverse". – Asaf Karagila Feb 03 '16 at 18:18
  • By the way, using $\geq$ for the existence of a surjection is godawful and bordering on wrong. The standard use for almost any relation symbol written in reverse is just stating the reverse relation to hold. Using injections and surjections to order the cardinals is using two different definitions and you need an actual theorem to prove their equivalence. – Asaf Karagila Feb 03 '16 at 18:22

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If $f$ is a surjection from $A$ to $B$, then you can define an injection from $B$ to $A$ using the Axiom of Choice: $$ g:B\to A, g(b)=a,\text{ where }f(a)=b. $$ Essentially $g(b)$ just picks out one element from the preimage of $b$ under $f$.

TomGrubb
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