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Consider the following statement in $\sf ZF$:

(I) Whenever $X$ is a set with more than $1$ element, there is an injective map $\iota: X\to X$ such that $\iota(x) \neq x$ for all $x\in X$.

The Axiom of Choice (AC) implies (I) -- but does (I) imply (AC)?

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It is shown in

Tachtsis, E.
On the existence of permutations of infinite sets without fixed points in set theory without choice.
Acta Math. Hungar. 157 (2019), no. 2, 281-300.

that ZF+(every infinite set supports a permutation with no fixed points) does not imply AC. It is easy to see in ZF that a finite set supports a cyclic permutation, so that should be sufficient for your question.

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  • $\begingroup$ I mean, yes. But it's an easy consequence of Sageev's work from the late 60s/early 70s that is mentioned in the comments. $\endgroup$
    – Asaf Karagila
    Jan 4, 2022 at 11:50
  • $\begingroup$ @AsafKaragila: My original answer gave the bib info for Sageev's paper, but I edited my answer to point to the paper that answers the question asked on this page. $\endgroup$ Jan 4, 2022 at 12:03
  • $\begingroup$ My point is that saying "it was shown in..." and referring to anything but Sageev's paper is a bit odd. It's as though you'd refer to a paper from 2022 proving that the axiom of choice is equivalent to "in a partial order where every chain has a lower bound, below every element there is a minimal element". $\endgroup$
    – Asaf Karagila
    Jan 4, 2022 at 12:15

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