# The equivalence of Dedekind-infinite and dually Dedekind-infinite as a weak form of (AC)

A set $$X$$ is Dedekind-infinite if there is an injective map $$f: X\to X$$ that is not surjective.

A set $$X$$ is dually Dedekind-infinite if there is a surjective map $$f: X\to X$$ that is not injective.

In $${\sf (ZF)}$$, it is easy to see that any Dedekind-infinite set is dually Dedekind-infinite: Let $$f:X\to X$$ be a non-surjective injection. Pick $$x_0\in X$$ and let $$g: X\to X$$ be defined by $$\big\{\big(f(y), y\big): y\in f(X)\big\}\,\cup\,\big\{(z, x_0):z\in X\setminus f(X)\big\}.$$ Then $$g$$ is a non-injective surjection.

Consider the statement

(DD) Every dually Dedekind-infinite set is Dedekind-infinite.

It is not hard to see that (AC) implies (DD). Consider the partition principle:

(PP) If $$X,Y$$ are sets and there is a surjection $$f:X\to Y$$, then there is an injection $$g:Y\to X$$.

Question. In $${\sf (ZF)}$$, are there any implications between (DD) and (PP)?

Note. It would also be interesting to see whether there is any implication between (DD) and the Dual Cantor Bernstein statement (CB)*, which is implied by (PP) in $${\sf (ZF)}$$:

(CB)* If $$X,Y$$ are sets and $$f:X\to Y$$ and $$g:Y\to X$$ are surjections, then there is a bijection $$\varphi:X\to Y$$.

This is a consequence of $$\sf DC$$, of course, which is a consequence of $$\sf PP$$.