This is a slight variation of [this recommended blog post](https://karagila.org/2022/dual-df-maps/) by [Asaf Karagila](https://mathoverflow.net/users/7206/asaf-karagila). Let $A$ be a set. Then:


1. $A$ is said to be *Dedekind-finite* if every injective map $f:A\to A$ is also surjective.
2. $A$ is said to be *dually Dedekind-finite* if every surjective map $f:A\to A$ is also injective.
3. A map $g:A\to A$ is said to be *(dually) Dedekind-finite-to-one* if for all $b\in A$ the preimage $f^{-1}(\{b\})$ is (dually) Dedekind-finite.


**Question.** In ${\sf (ZF)}$, suppose that the set $A$ has the property, that every surjective map is (dually) Dedekind-finite-to-one. Does this imply that $A$ is (dually) Dedekind-finite?



Dedekind-finite-to-one vs Dedekind-finite
The answer is Yes.


For the first question, suppose that $A$ is Dedekind infinite and let $f$ be an injection from $\omega$ into $A$. Then the function that maps $f(2n)$ to $f(n)$ for each $n\in\omega$, maps $f(2n+1)$ to $f(0)$ for each $n\in\omega$, and leaves all other elements of $A$ fixed, is a surjection from $A$ onto $A$ which is not Dedekind-finite-to-one.


For the second question, suppose that $A$ is dually Dedekind infinite and let $f$ be a surjection from $A$ onto $A\cup\{b\}$, where $b\notin A$. For each $n\in\omega$, let $A\_n=(f^{n+1})^{-1}[\{b\}]$. Clearly, for all $m,n\in\omega$ with $m<n$, $f^{n-m}|A\_n$ is a surjection from $A\_n$ onto $A\_m$. Then the function $g$ that maps each $x\in A\_{2n}$ to $f^n(x)$ for all $n\in\omega$, maps each $x\in A\_{2n+1}$ to a fixed $a\in A$ for all $n\in\omega$, and leaves all other elements of $A$ fixed, is a surjection from $A$ onto $A$ which is not dually-Dedekind-finite-to-one. $g$ is not dually-Dedekind-finite-to-one, because the inverse image of $a$ under $g$ includes $B=\bigcup\_{n\in\omega}A\_{2n+1}$, and $B$ is dually Dedekind infinite since $f^2|B$ is a surjection from $B$ onto $B\cup\{b\}$.


