TITLE: On surjections, idempotence and axiom of choice
QUESTION [10 upvotes]: The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF?
(I couldn't find it in the Consequences site so far.)

If $A$ is an infinite set such that $A$ can be mapped onto $A\times 2$ then $|A\times 2|=|A|$

The problem is that we cannot necessarily choose from every fiber of $f$, so we cannot construct an injection from $A$ to $f^{-1}(A\times\lbrace 0\rbrace)$, which will prove the assertion.
While I'm on the topic, is it possible for a D-finite set to have such property? It is possible for a D-finite set to be surjected onto a larger set than itself, but what about that large?

REPLY [2 votes]: A simpler version (using the length of a sequence for coding, rather than its entries):  Let $X$ be infinite but Dedekind finite.   Let $S$ be the set of all finite 1-1 sequences from $X$.  Then $S$ is Dedekind finite. 
Now define a surjective map $f$ from $S$ onto $S\times \{0,1\}$ as follows.

If $s\in S$ has length $2k+i$ with $i\in \{0,1\}$, map $s$ to $(s\mathord\restriction k, i)$ where $s\mathord\restriction k$ is the initial segment of $s$ of length $k$.


To get a surjective  map from  $S$ onto $S^{<\omega}$, we  fix a surjective function $$H:\omega\to (\omega^{<\omega})^{<\omega}$$
with the property that all numbers appearing in $H(n)$ will be $< n$. 
Now define $f:S\to S^{<\omega}$ as follows:  Given a sequence $s\in S$ of length 
$n$, write $H(n)$ as $ (h_0,\ldots, h_{k-1})$, and let $F(s) := (s\circ h_0, \ldots, s\circ h_{k-1})$. (If some $h_i$ is not 1-1, let $F(s):=()$; alternatively, modify the function $H$ so that its range is the set of all sequences of 1-1 sequences.)
It is clear that $f$ is onto: given any sequence $t=(t_0,\ldots, t_{k-1})$ of elements of $S$, first find a 1-1 sequence $s$ enumerating  the union of the ranges of the $t_i$.  So there 1-1 sequences $h_i$ such that t $t_i = s\circ h_i$ for all $i$.  Now extend $s$ (in a 1-1 way) to a sequence $r$ whose length codes the family of the $h_i$'s. We will have $f(r)=t$.