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I am trying to prove this (it looks true to me) :

Let $A,B $ be two sets. If there is a surjection $A\to B $ and a surjection $B\to A$ then $A $ and $B $ are in bijection.

I showed that is it equivalent to the following statement :

If there is an injection $A\to B $ and an injection $B\to A$ then $A $ and $B $ are in bijection.

But I am stuck, I don't see how to prove either.

Andrés E. Caicedo
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Friedrich
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  • Ho w did you show the equivalence to the statement involving injections? – Andrés E. Caicedo Oct 04 '17 at 17:07
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    If we have a surjection $f:A\to B $ then, for each $b \in B $ we can (via the axiom of choice) choose an element $a\in f^{-1}(b) $. Then we have an injection $g:B\to A $ by defining $g (a)=b $ (we do that for every $b $). – Friedrich Oct 04 '17 at 17:11

1 Answers1

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The assertion that if there are injections $A \leftrightarrows B$ then there is a bijection $A \to B$ is the Cantor–Schröder–Bernstein and has a reasonably involved proof—involved enough that I doubt you'd reasonably be expected to prove it yourself.

Every surjection $f : A \to B$ has a right inverse $r : B \to A$, i.e. a function such that $f(r(b))=b$ for all $b \in B$, and this right inverse is injective. You have (in the comments to your question) correctly constructed such a right inverse and identified that it is injective, and so the result follows from the Cantor–Schröder–Bernstein theorem... that is, if you're allowed to assume it.

Clive Newstead
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    Thank you! I didn't knew it was a famous theorem. – Friedrich Oct 04 '17 at 17:16
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    It should be noted that the existence of the right inverse is equivalent to the axiom of choice, while the Cantor–Schröder–Bernstein theorem can be proved without it. –  Oct 04 '17 at 17:20
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    Clive Newstead-- I don't only do what I'm expected to do :) – Friedrich Oct 04 '17 at 17:26