**The Generalized Continuum Hypothesis** can be stated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. We know that GCH implies AC (Jech, *The Axiom of Choice*, Theorem 9.1 p.133).

In fact, a relatively weak formulation: $|X|\le|Y|< 2^X\implies |X|=|Y|$ would already imply the axiom of choice, although in this case the proof is slightly longer. [Note: in Herrlich's book he refers to **GCH** stated above as "The Aleph Hypothesis" and the weak formulation is called GCH]

GCH itself is independent of the axiom of choice, we can have the axiom of choice and power sets can grow wildly, or just "a little bit". We can have the continuum function to be injective, but the continuum hypothesis can fail on a proper class of cardinals (for example $2^\kappa=\kappa^{++}$ for regular cardinals).

Let **ICF** (Injective Continuum Function) be the assertion: $$2^X=2^Y\implies |X|=|Y|.$$

**Question:** Assuming ZF+ICF, can we deduce AC?

(I looked around Equivalents of the Axiom of Choice, but couldn't find much. It is possible that I missed this, though.)

**Edits:**

In an exercise in Jech he states that if there exists an infinite Dedekind-finite cardinal, then ICF does not hold. From the assumption that it holds we can deduce that there are no infinite D-finite sets.

Note that if $f\colon X\to Y$ is a surjection then $A\subseteq Y\mapsto f^{-1}(A)$ is an injection from $P(Y)$ into $P(X)$. This means that

**ICF**implies the Dual Cantor-Schroeder-Bernstein theorem:Assume that $X$ and $Y$ have surjections from one onto the other, then there are injections between their power-sets therefore $2^X=2^Y$ and thus $|X|=|Y|$.

We shall abbreviate Goldstern's variant of ICF as

**Homomorphic Continuum Function**, or**HCF**: $$2^X\leq 2^Y\implies |X|\leq|Y|$$ One major observation is that HCF implies*The Partition Principle*(PP), which states that $A$ can be mapped onto $B$ (or $B$ is empty) if and only if $B$ can be injected into $A$. This principle is quite an open choice principle, and it is unknown whether or not it implies AC in ZF.To see that HCF implies PP, we observe the following: if $f\colon A\to B$ is surjective then the preimage map is an injection from $2^B$ into $2^A$, i.e. $2^B\leq 2^A$, from HCF it follows that $B\leq A$, i.e. there is $g\colon B\to A$ injective.

veryhard but indeed less trivial than I'd expected. – Asaf Karagila Nov 18 '11 at 10:15