In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in <a href="http://mathoverflow.net/questions/82177/a-profinite-group-which-is-not-its-own-profinite-completion">this question</a>. However, all three constructions given use consequences of the axiom of choice: ultrafilters in the first answer, and "every vector space has a basis", in Milne's notes as referenced in the second answer, and used to compute the number of finite-index subgroups in the third answer.

Is it possible to prove the existence of a discontinuous homomorphism from a profinite group to a finite group without the axiom of choice? Instead is it consistent with ZF that there is none?