This question seems elementary and there's probably a really obvious argument for this that I'm not thinking of but I can't find it anywhere.
For equinumerous $A$, $B$, must every surjection $f:A \twoheadrightarrow B$ have a right inverse? What about if just $A = B$? If for some reason this doesn't apply in $ZF$ does it hold with something weaker than full choice like the partition principle or countable choice?
No. In fact even taking $A=B$ this is equivalent to the axiom of choice.
This is an observation made by Jech and myself a while ago when we talked about this sort of propositions.
The point is to take a function which has no injective inverse and use the domain and range to define a set and a self-surjection which has no inverse.
You can find the details here.