# What is surjective comparability equivalent to? [duplicate]

By surjective comparability I mean the following principle:

"For any nonempty sets $$A,B$$: there exists a surjective function $$f$$ from $$A$$ to $$B$$ or there exists a surjective function $$g$$ from $$B$$ to $$A$$."

1. Is this statement provable in $$ZF$$?

2. If not then what exactly that principle is equivalent to (over $$ZF$$)?

This principle (for nonempty sets, to avoid a trivial counterexample) is equivalent to the axiom of choice. One direction is in a previous answer. For the converse, let $$X$$ be an arbitrary set (which I want to well-order). Apply Hartogs's theorem to get an ordinal $$\alpha$$ so large that it cannot be mapped one-to-one into the power set $$\mathcal P(X)$$. Then $$X$$ cannot be mapped onto $$\alpha$$, because if $$f:X\to\alpha$$ were surjective then $$\alpha\to \mathcal P(X):\beta\mapsto f^{-1}(\{\beta\})$$ would map $$\alpha$$ one-to-one into $$\mathcal P(X)$$. So, by your principle, there is a surjection $$g:\alpha\to X$$. Then $$X$$ can be well-ordered by setting $$x\prec y$$ iff the first $$\beta$$ with $$g(\beta)=x$$ is smaller than the first $$\beta$$ with $$g(\beta)=y$$.
• possibly you meant: $\alpha \to \mathcal P(X): \beta \mapsto \{x \in X| f(x)=\beta\}$, this would be a one-to-one map from $\alpha$ to $\mathcal P(X)$ and the proof would work through. – Zuhair Oct 16 '19 at 11:56
• @Zuhair What you wrote in your comment seems to be just a long way to say the same thing that I wrote. Remember that, when $Z$ is a subset of the domain of a function $f$, the notation $f^{-1}(Z)$ means $\{x:f(x)\in Z\}$. In the case at hand, $Z$ is $\{\beta\}$. – Andreas Blass Oct 16 '19 at 12:46
• Yes, but that should have been written as: $\alpha\to \mathcal P(X):\beta\mapsto f^{-1}(\{\beta\})$ – Zuhair Oct 16 '19 at 13:43
• @Zuhair Thanks for pointing out the missing $\mathcal P$. I'll correct it. – Andreas Blass Oct 16 '19 at 16:50