No. Suppose a formula with parameters defined a function that was surjective and not injective. The assertions that the formula defines a function, that the function is surjective, and that it is not injective are all first-order statements about that formula and those parameters. By Łoś theorem, they would hold simultaneously in for -almost every .
Every surjective self-map of a finite set is injective, so no finite factor can satisfy those statements. This contradiction shows that the ultraproduct has no such definable function.
Solved by gpt-5.6-sol high.

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