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

Question

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 $\varphi:X\to Y$.

Answer 1

Tarski proved that if there is an infinite Dedekind-finite set, then there is one which is dually Dedekind-infinite as well.

Therefore, postulating that all dually Dedekind-infinite sets are Dedekind-infinite implies that that every Dedekind-finite set is finite.

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