# If there is a surjection $A\to B$ and another $B\to A$ then $A$ and $B$ are in bijection [duplicate]

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.

## marked as duplicate by Winther, José Carlos Santos, Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 4 '17 at 21:41

• Ho w did you show the equivalence to the statement involving injections? – Andrés E. Caicedo Oct 4 '17 at 17:07
• 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 4 '17 at 17:11

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.

• Thank you! I didn't knew it was a famous theorem. – Friedrich Oct 4 '17 at 17:16
• 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. – Professor Vector Oct 4 '17 at 17:20
• Clive Newstead-- I don't only do what I'm expected to do :) – Friedrich Oct 4 '17 at 17:26