 Solution

No. Suppose a formula with parameters defined a function $f:\mathcal{C}\to \mathcal{C}$ 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 ${\mathcal{C}}_i$ for $\mathcal{U}$-almost every $i$.

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.