Consider the following two very similar statements in ${\sf ZF}$:
(Mat_1) There is a set $A$ a map $\alpha: \omega \to {\cal P}(A)$ such that for all $n\in \omega$ we have $\alpha(n+1) \subseteq \alpha(n)$, and there is no injective map from $\alpha(n)$ into $\alpha(n+1)$.
and
(Mat_2) There is a set $A$ a map $\alpha: \omega \to {\cal P}(A)$ such that for all $n\in \omega$ we have $\alpha(n+1) \subseteq \alpha(n)$, and there is no surjective map from $\alpha(n+1)$ onto $\alpha(n)$.
In both settings we have a sequence of subsets of a ground set $A$ getting smaller and smaller - reminscent of Matryoshka dolls.
Note that neither statement is compatible with the Axiom of Choice ${\sf (AC})$ since the $\{|\alpha(n)|: n\in \omega\}$ would be an infinite strictly descending sequence in the cardinal $|A|$, contradicting the fact that $|A|$ is well-ordered.
Note the following:
Fact. In ${\sf ZF}$, if $A, B$ are non-empty sets and $f:A\to B$ is an injection, then there is a surjection $g:B\to A$ such that $g\circ f$ is the identity on $A$.
From this we get that (Mat_2) implies (Mat_1).
Are (Mat_1) and (Mat_2) equivalent in ${\sf ZF}$? Does one of $\neg$(Mat_1) and $\neg$(Mat_2) imply ${\sf (AC)}$?
