Consider the following statements in $\sf ZF$:



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> (S) If $A, B$ are nonempty sets, then there is a surjection $s:A \to B$, or there is a surjection $t:B\to A$.
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> (I) If $A, B$ are sets, then there is an injection $i:A\to B$, or there is an injection $j:B\to A$.
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Note that (I) implies (S). Assuming $\sf AC$, both statements are true.


**Question.** Is there a model of $\sf ZF$ in which (S) holds, but not (I)?



Existence of surjection vs injection over $\sf ZF$
The answer is no. Both (I) and (S) are equivalent to AC over ZF. Indeed, for any set $S$ the class of ordinals $\alpha$ such that $\alpha$ injects into $S$ (resp. $S$ surjects onto $\alpha$) is a set, so there is some ordinal $\beta$ outside this set. Assuming (I) (resp. (S)), there is an injection $S\to\beta$ (resp. surjection $\beta\to S$). In the latter case, we also get an injection $S\to\beta$, by taking any element $S$ to the least element in its preimage. In either case, we see $S$ is well-orderable. Thus both (I) and (S) imply well-ordering theorem and hence AC.


Since both (I) and (S) are equivalent to AC, they are also equivalent to each other.


