# Is the following equivalent to the axiom of choice?

For sets $A,B$, let $|A|\leq^*|B|$ say that there exists an onto map $f:B\rightarrow A$ or $A=\emptyset$.

My question is, is $$\forall A,B(|A|\leq^*|B|\longrightarrow |A|\leq|B|)$$ equivalent to the axiom of choice?

Thanks

• @AsafKaragila Well, not really. If we $|\emptyset|\le^*|B|$ is false by the simplified definition, then the case $A=\emptyset$ is trivially true, and that does not hurt. All this $\le, \le^*$ is just fancy writing for "If there exists a surjection $B\to A$, there exists an injection $A\to B$". – Hagen von Eitzen Sep 11 '13 at 21:28
• @Hagen: There are no surjections from a non-empty set onto the empty set. And no, $|A|\leq^*|B|$ need not imply, generally, the existence of an injection from $A$ into $B$. – Asaf Karagila Sep 11 '13 at 21:35
• @Hagen: I realized your comment is the result of a misunderstanding in the question. The definition of $\leq^*$ is not the statement given in the displayed equation. It is given on the top part of the text. The question asks whether or not the displayed statement is equivalent to the axiom of choice (because it trivially follows from it). – Asaf Karagila Sep 13 '13 at 14:50

This is known as the partition principle (note that the other implication is trivial in $\sf ZF$). The problem whether or not this is equivalent to the axiom of choice has been open for over a century now.
There isn't much to say about it, really. We know it implies some basic choice principles such as "Every infinite set is Dedekind infinite", but we don't know a lot more. Every time I think about the problem I run into the same problem, we don't have enough tools to manage - or even understand - the structures of cardinals in arbitrary models of $\sf ZF$. Not even under $\leq$ and let alone under $\leq^*$.