Suppose there exist a surjection from set A to set B and there also exist a surjection from set B to set A . Does that imply there should exist a bijection from set A to set B?
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1Yes, if you assume the axiom of choice. In the absence of choice there need not be one. – Brian M. Scott Aug 22 '16 at 21:06
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their inverses are injections and then you apply Cantor Schroeder – Asinomás Aug 22 '16 at 21:06
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1@CarryonSmiling: Without $\mathsf{AC}$ they may not have (right) inverses. – Brian M. Scott Aug 22 '16 at 21:07
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can you provide a proof for that ? – rathod dinesh Aug 22 '16 at 21:09
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@BrianM.Scott I see, thanks. I was aware of that though. – Asinomás Aug 22 '16 at 21:13