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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1$\begingroup$ A linear linjection of vector spaces is bijective if at least one of them is finite-dimensional. $\endgroup$– Gerald EdgarCommented Jun 1, 2019 at 11:54
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1$\begingroup$ 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). $\endgroup$– user140765Commented Jun 1, 2019 at 12:23
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6$\begingroup$ 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 EskewCommented Jun 1, 2019 at 12:43
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2$\begingroup$ For maps preserving some particular structure there are plenty of examples, see "co-hopfian". $\endgroup$– YCorCommented Jun 1, 2019 at 16:48
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1$\begingroup$ 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. $\endgroup$– YCorCommented Jul 1, 2019 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).
answered Jun 1, 2019 at 15:51