Is there a homomorphism of finitely presented groups $f:G\to H$ and an element $h\in H$ such that the statement "$f^{-1}(h)$ is empty" is independent of ZFC?
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$\begingroup$ It seems to me that any homomorphism between finitely presented groups is computable, whatever it means. Do I miss something? $\endgroup$– YCorCommented Apr 21, 2021 at 17:48
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$\begingroup$ We can take $f$ a morphism taking everything to identity. Then for appropriate $H$ we have that ZFC can't tell whether $H$ is nontrivial, so it can't check the nontriviality on all generators. $\endgroup$– WojowuCommented Apr 21, 2021 at 17:50
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$\begingroup$ @Wojowu isn't the preimage of the identity always non-empty? $\endgroup$– user178109Commented Apr 21, 2021 at 17:51
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$\begingroup$ @Oniqa But ZFC won't be able to prove that these elements are or aren't identity. $\endgroup$– WojowuCommented Apr 21, 2021 at 17:53
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$\begingroup$ @Wojowu but the identity is certainly getting sent to the identity $\endgroup$– user178109Commented Apr 21, 2021 at 17:54
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1 Answer
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The answer is yes, as a consequence of my answer to your other question.
Namely, in that answer, we have a finite group presentation $H$ and a word $h$ such that the question $h=1$ in $H$ is independent of ZFC. So if we take $G$ to be trivial and $f:G\to H$ the unique homomorphism, we have the statement "$f^{-1}(h)$ is nonempty" being independent of ZFC.
The general lesson of that answer supplies also an answer to Benjamin Steinberg's comment here concerning a confusion between computable undecidability and ZFC or logical undecidability. The general lesson, which I argue on the other post, is that every computably undecidable enumerable decision problem is saturated with logical undecidability. So the two notions of undecidability are actually intimately connected.
answered Apr 21, 2021 at 20:22
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$\begingroup$ But the independence from ZFC thing seems to make some assumptions about cardinals $\endgroup$ Commented Apr 22, 2021 at 2:11
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$\begingroup$ @BenjaminSteinberg The $\Sigma_1$-correctness was only for the very general fact. Meanwhile, the example at the end of my other answer produces a ZFC independent instance assuming only Con(ZFC), which is required in order to have anything independent of ZFC. $\endgroup$ Commented Apr 22, 2021 at 8:35