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To prove two sets are isomorphic is it sufficient to find 2 injections $f:A\to B$ and $g:B\to A$ or 2 surjections $h: A\to B$, $j:B\to A$?

YoTengoUnLCD
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    Yes it is true. –  Aug 06 '15 at 20:52
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    If by isomorphic, you mean have the same cardinality, yes. Usually isomorphic refers to having the same structure in my experience, which has something to do with more than just size – Alan Aug 06 '15 at 20:56
  • The word for two sets with the same cardinality is "equipollent", but (to quote Michael Caine) not a lot of people know that, so "having the same cardinality" may be preferable. "Isomorphic" is misleading as Alan's comment points out. – Rob Arthan Aug 06 '15 at 21:27
  • Hmm I thought it was appropriate in the usual sense, as sets don't have any additional structure (as say, a group), but I get it, I'll edit my post. – YoTengoUnLCD Aug 06 '15 at 21:31

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If you want to prove that there are a bijective function between the sets, finding two injections is sufficient. If the set are finite, this is automatic since an injection leads to an inequality on the cardinal of the sets. If the set have inifite cardinals, this is still true thanks to the Schrer-Bernstein (or Cantor-Bernstein) theorem: https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem

The proof is a bit involved so I would recommand to read the wikipedia page directly. It is worth noticing, that even with if the sets are not finite, the proof does not require the axiom of choice.

Tantto
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Assuming choice, yes. Without the axiom of choice, there is a difference, a big difference, between injections and surjections.

The Cantor-Bernstein theorem says that if there are injections between $A$ and $B$, then there is a bijection between $A$ and $B$. It is provable without the axiom of choice, since we essentially define the bijection.

The dual-Cantor-Bernstein theorem says that if there are surjectiosn between $A$ and $B$, then there is a bijection between $A$ and $B$. With the axiom of choice, the surjections can be reversed to injections and the usual Cantor-Bernstein theorem, but without the axiom of choice this is not necessarily true. For example, if the dual-Cantor-Bernstein theorem holds, then every infinite set has a countably infinite subset. Such statement is not provable without at least some weak form of choice. So the theorem itself is not provable without choice.

Asaf Karagila
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