After reading Proving existence of a surjection $2^{\aleph_0} \to \aleph_1$ without AC I became curious if there is a generalization to arbitrary cardinals. That is, if $\frak m<n$, does it follow that $\frak m^+\le^*n$? Or in terms of sets:
Suppose $A,B$ are sets, and there is an injection from $A$ to $B$ but not from $B$ to $A$ (so $|A|<|B|$). Does there exist a surjection from $B$ to $\aleph(A)$, where $\aleph(A)$ is the least ordinal that does not inject into $A$?
Of course all these questions are trivial under choice, where $\le^*$ coincides with $\le$ and $\frak m<n$ coincides with $\frak m^+\le n$ by trichotomy.
The proof methods given in the linked problem generalize in the following manner in the case $|B|\ge^*|{\cal P}(A\times A)|$ (for convenience I assume that $B={\cal P}(A\times A)$ but it is easy to compose with a surjection). For each $\alpha\in\aleph(A)$ we know that there is an injection from $\alpha$ to $A$, which induces a well-order of a subset of $A$. Thus for each $\alpha\in\aleph(A)$ there is an $R\in{\cal P}(A\times A)$ such that $R$ is a well-order of type $A$ (note that by using reflexive well-orders we can keep track of the domain of the well-order), so we can define our surjection ${\cal P}(A\times A)\to\aleph(A)$ by sending each well-order of type $\alpha$ to $\alpha$ and everything else to $0$.
In the language of cardinals, this means that ${\frak n}\le^* 2^{\frak m\times m}\to{\frak m}^+\le^*{\frak n}$, and if $\frak m\times m=m$ as is the case with alephs we get ${\frak m}^+\le^*2^{\frak m}$, which gives the original question for ${\frak m}=\aleph_0$.
