We say an infinite set $X$ is *splittable* if there are $X\_1, X\_2\subseteq X$ with $X\_1\cap X\_2 = \emptyset$, $X\_1\cup X\_2 = X$ and there are bijections $\varphi:X\_1\to X\_2$ and $\psi:X\_1\to X$.


Does the statement "Every infinite set is splittable" imply $\mathsf{AC}$?



Does "Every infinite set is splittable" imply $\mathsf{AC}$?
The answer is no and it follows from the following: 



> 
> It is consistent that $AC$ fails but for all infinite cardinals $\kappa, 2 \cdot \kappa=\kappa.$
> 
> 
> 


The above result is proved by Sageev:


Sageev, Gershon [An independence result concerning the axiom of choice](http://www.sciencedirect.com/science/article/pii/0003484375900029). Ann. Math. Logic 8 (1975), 1–184.


In a model as above, every infinite set is splittable but $AC$ fails in it.


