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An injective map between two sets of the same cardinality is bijective if at least one of the sets is finite. This is not true if we drop the assumption that at least one of the sets is finite. Is there a strengthening of the notion of injectivity such that the statement continues to hold?

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closed as off-topic by YCor, Yoav Kallus, Jeremy Rickard, LSpice, Yemon Choi Jul 3 at 1:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – YCor, Yoav Kallus, Jeremy Rickard
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ A linear linjection of vector spaces is bijective if at least one of them is finite-dimensional. $\endgroup$ – Gerald Edgar Jun 1 at 11:54
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    $\begingroup$ you probably mean something different, consider the inclusion of the zero into the line. Either that their dimensions are equal as cardinals, or your coefficient field is finite (so you can compare the cardinalities of the vector spaces themselves). $\endgroup$ – user140765 Jun 1 at 12:23
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    $\begingroup$ Every infinite structure has a proper elementary extension. So there is no first-order structure-dependent notion which strengthens injectivity and entails surjectivity for infinite sets. $\endgroup$ – Monroe Eskew Jun 1 at 12:43
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    $\begingroup$ For maps preserving some particular structure there are plenty of examples, see "co-hopfian". $\endgroup$ – YCor Jun 1 at 16:48
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    $\begingroup$ Since on the one hand the question is little focussed and in some more precise forms has already been discussed, and on the other hand the OP has closed their account, I'm voting to closed the question. $\endgroup$ – YCor Jul 1 at 17:33
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If $A$ and $B$ are countable and equipped with probability measures $\mu_A$, $\mu_B$ that give each element positive probability then any measure-preserving map is a bijection. (I guess per @MonroeEskew's comment this is not first order).

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