# Are two forms of the Dual Schroeder-Bernstein property equivalent?

We know the Shroeder-Bernstein (SB) theorem can be proved in ZF, while the Dual Schroeder-Bernstein (DSB) can be proved in ZF+AC but not in ZF. Define as ISB the property that whenever there are both an injection and a surjection $X \to Y$, then there must be a bijection too. Trivially ZF+DSB $\implies$ ZF+ISB (because an injection $X\to Y$ allows to define a surjection $Y\to X$, without using AC). Is the converse true? I'm assuming here that ZF $\nRightarrow$ ISB, but I don't know a proof of that either (UPDATE: this is the case, as pointed out by Asaf Karagila in a comment).

• What do you mean "the converse"? – Asaf Karagila Jan 14 '15 at 21:44
• As for your last remark, this thread shows that $\sf ISB$ is not provable in $\sf ZF$. – Asaf Karagila Jan 14 '15 at 21:45
• Does ZF+ISB $\implies$ DSB? Should I edit? – Yaakov Baruch Jan 14 '15 at 21:46
• This is a difficult question, since we don't know any explicit models where $\sf ZF+\lnot AC+DSB$ hold, or $\sf ISB$ for that matter. At best we have "local examples" (where the general principle fails, but nontrivially holds for some set). My guess is that the implication is false. – Asaf Karagila Jan 14 '15 at 21:51

What you call $\sf ISB$, is better known as $\sf WPP$ (Weak Partition Principle) and can be formulated as "If there is a surjection from $X$ onto $Y$, then $X$ cannot have a strictly smaller cardinality than $Y$" (alternatively, $|X|\leq^*|Y|\leq|X|\rightarrow |X|=|Y|$ or $|X|\leq^*|X|\rightarrow |X|\nless|Y|$).