The dual version of trichotomy of cardinality without AC Let $A$, $B$ be sets and denote by $A\leq^* B$ an assertion that if $A$ is non-empty, then there's a surjection from $B$ onto $A$.
This might be understood as a dual definition of cardinality. (Typically we use injections to capture the notion of A not being bigger than B. The $\leq^*$ notation is to be read as B not being smaller than A.)
Is this relation trichotomous without using the axiom of choice? Meaning for every two sets $A$ and $B$, it holds $A\leq^* B$ or $B\leq^* A$. (I would be surprised). Would trichotomy imply some interesting extra-ZF axiom?
More broadly, is there anything to be said about $\leq^*$? For example how does it relate to the usual comparisons of cardinality by injections?
I suspect there's no clear answer, since it is somehow reminiscent of Partition Principle ($A\leq^* B \implies A\leq B$) and Weak Partition Principle ($A\leq^* B \implies B\not< A $) which both are independent of ZF.
 A: Not even a little bit.
If $\leq^*$ satisfies the trichotomy, then the axiom of choice holds.

Theorem. (Lindenbaum) If $X$ is a set, then there is an ordinal $\alpha$ such that $\alpha\nleq^*X$.

(Proof. If $\alpha\leq^*X$, then $\alpha$ injects into $\mathcal P(X)$; apply Hartogs' theorem.)
Let us denote the least such (non-zero) ordinal as $\aleph^*(X)$ and name it the Lindenbaum number of $X$.

Theorem. If $X\leq^*\alpha$ for an ordinal $\alpha$, then $X$ can be well-ordered.

(Proof. If $X$ is empty, we're done; otherwise fix a surjection from $\alpha$ onto $X$, and for every $x\in X$ let $\beta_x$ be the least $\beta<\alpha$ mapped to $x$. Then $x\prec y\iff\beta_x<\beta_y$ is a well-ordering of $X$.)

Theorem. If $\leq^*$ satisfies trichotomy, then the axiom of choice holds.

Proof. Compare $X$ and $\aleph^*(X)$, it has to be the case that $X<^*\aleph^*(X)$, and therefore $X$ can be well-ordered.

Now, what can we say about $\leq^*$ in general? Not too much. 
It does not have to be anti-symmetric; but if it is anti-symmetric then there are no Dedekind-finite sets.
