On pseudofiniteness and injective-surjective

Question

In many literatures it is noted that “let M be pseudofinite and f a definable function, then f is injective if and only if it is surjective.” Let's break it down into parts, Let A --- M be a pseudofinite and φ definable function, and B --- be f injective if and only if it is surjective. $A \implies B$, but $B \implies A$ is not always true. This is noted in the work "Lou van den Dries and Vinicius Cifú Lopes (2010). Division rings whose vector spaces are pseudonite. The Journal of Symbolic Logic, 75, pp 1087-1090 doi:10.2178/jsl/1278682217".

Now the question: In many literatures on pseudofinite models, the proof of pseudofiniteness comes by proving property B. For example, the additive аbelian group Z and an algebraically closed field. How correct is this? Is it possible to use other methods?

Answer 1

You are correct that $A$ implies $B$, but $B$ does not imply $A$ in general.

You wrote:

In many literatures on pseudofinite models, the proof of pseudofiniteness comes by proving property B. For example, the additive аbelian group Z and an algebraically closed field.

It would be helpful to give some references to the many places you've seen such arguments. But let's be clear about something: the additive abelian group $\mathbb{Z}$ is not pseudofinite, and algebraically closed fields are not pseudofinite. And both have obvious definable functions which violate $B$. The map $x\mapsto 2x$ in $\mathbb{Z}$ is injective but not surjective, and the map $x\mapsto x^2$ in an algebraically closed field (of characteristic $\neq 2$) is surjective but not injective.

So it seems most likely that in sources you were reading that the authors proved these structures are not pseudofinite by showing that they violate $B$. This is a valid argument, of course! If $A\implies B$, then the contrapositive $\lnot B\implies \lnot A$ is true.

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In the comments, you wrote:

In the work "Division rings whose vector spaces are pseudofinite" is noted that $B\implies \lnot A$.

This is just wrong. There is a big difference between $B \;\not\!\!\!\implies A$ ($B$ does not imply $A$) and $B\implies \lnot A$ ($B$ implies not $A$).