Surjective order-preserving map $f:{\cal P}(X)\to \text{Part}(X)$ Let $X$ be a set, and let $\text{Part}(X)$ denote the collection of all partitions of $X$. For $A, B\in \text{Part}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. This relation defines a partial order on $\text{Part}(X)$.
If $X$ is an infinite set, is there a surjective order-preserving map $f:{\cal P}(X)\to \text{Part}(X)$ (where ${\cal P}(X)$ denotes the power-set of $X$, ordered by $\subseteq$)?
 A: The positive solution uses an equivalent of the Axiom of Choice:
for every infinite set $A$ there is a bijection $f:A\to A\times A$.
In the basic Fraenkel Model (section 4.3 in Jech's Axiom of Choice) there 
is an infinite set where there is no surjection as in the question.
Let $A$ be the infinite set of atoms in the model.
Step 1. Every subset of $A$ in the model is finite or cofinite.
Let $X\subseteq A$ and let $E\subseteq A$ be finite such that
$\operatorname{fix} E\subseteq\operatorname{sym} X$.
If $X\subseteq E$ then $X$~is finite.
If $X\not\subseteq E$ then $A\setminus E\subseteq X$: 
fix $x\in X\setminus E$ and let $y\in A\setminus E$ be arbitrary;
the permutation~$\pi$ that interchanges $x$ and $y$ is in $\operatorname{fix} E$,
so $\pi X=X$, but this means that $y=\pi(x)\in\pi X=X$.
Step 2. Let $P$ be partition of $A$ in the model and
let $E\subseteq A$ be finite with $\operatorname{fix} E\subseteq \operatorname{sym} P$.
Assume there are $a$ and $b$ in $A\setminus E$ that are distinct
and $P$-equivalent.
Let $c\in A\setminus E$ be arbitrary and not equal to~$a$. 
Let $\pi$ be the permutation that interchanges $b$ and $c$;
then $\pi\in\operatorname{fix} E$, hence $\pi P=P$.
Then $a,b\in X$ for some $X\in P$, and hence $a,c\in\pi X$.
As $\pi X\in\pi P=P$ this shows that $a$ and $c$ are $P$-equivalent as well.
It follows that either there is an element of $P$ that 
contains $A\setminus E$, or all elements of $A\setminus E$ determine
one-element members of $P$.
Step 3. Assume $f:\mathcal{P}(A)\to\operatorname{Part}(A)$ is an order-preserving
surjection.
And let $E\subseteq A$ be finite such that $\operatorname{fix} E\subseteq\operatorname{sym} F$.
Assume $X$ and $Y$ are finite such that $X\cap E=Y\cap E$ 
and $|X\setminus E|=|Y\setminus E|$, and such that
$f(X)$ and $f(Y)$ are partitions that contain $\{\{a\}:a\in A\setminus E\}$.
Then $f(X)=f(Y)$.
To see this let $\pi$ be a permutation that 
maps $X\setminus E$ to $Y\setminus E$ and vice versa and leaves all other
points in place.
Then $\pi\in\operatorname{fix} E$, so that $\pi(f(X))=f(X)$ and $\pi(f(Y)=f(Y)$.
The ordered pair $\langle X,f(X)\rangle$ is in $f$, hence 
$\langle \pi(X),\pi(f(X))\rangle$ is in~$\pi f$, but $\pi f=f$ and $\pi X=Y$
so that $\langle Y,f(X)\rangle\in f$, that is, $f(Y)=f(X)$.
A similar statement holds when $X$ and $Y$ are infinite, 
hence cofinite, and $X\cap E=Y\cap E$, and 
$|A\setminus(E\cup X)|=|A\setminus(E\cup Y)|$.
Step 4. Now consider the set $\Pi(E)$ of partitions of~$A$ that contain
$\{\{a\}:a\in A\setminus E\}$; this is essentially the set of partitions
of~$E$ where each is augmented with $\{\{a\}:a\in A\setminus E\}$.
The argument above also shows the following: if $X$ and $Y$ are finite
such that $X\cap E=Y\cap E$, $|X\setminus E|\le|Y\setminus E|$
and $f(X),f(Y)\in\Pi(E)$ then $f(X)\le f(Y)$.
This is so because we can find a permutation $\pi$ in $\operatorname{fix} E$ such that
$\pi(X)\subseteq Y$.
Likewise: if $X$ and $Y$ are infinite such that 
$X\cap E=Y\cap E$, 
$|A\setminus(X\cup E)|\le|A\setminus(Y\cup E)|$
and $f(X),f(Y)\in\Pi(E)$ then $f(X)\ge f(Y)$.
Finally: if $X$ is finite and $Y$ is infinite such that 
$X\cap E=Y\cap E$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\ge f(Y)$.
In conclusion: each subset of $E$ determines a chain in the set~$\Pi(E)$.
Step 5. If we enlarge $E$ then $\operatorname{fix} E$ becomes smaller, so we can choose
$E$ as large as we please.
Let $n$ be a natural number.
The number of partitions of the set $\{1,2,\ldots,4n\}$ into four sets
of size $n$ is equal to
$$
\binom{4n}{n}\binom{3n}{n}\binom{2n}{n}
$$
This number is larger than
$$
3^n\cdot 2^n\cdot\frac{4^n}{2n+1} = \frac{24^n}{2n+1}
$$ 
For $n\ge9$ we have $24^n/(2n+1)>16^n$.
This gives us our contradiction: if necessary enlarge $E$ so that $|E|=4n$
for some $n\ge9$.
Then, by 4 above, the map $f$ divides $\Pi(E)$ into, 
at most, $2^{4n}=16^n$ chains.
On the other hand the partitions of $E$ into four pieces of size $n$ form an 
antichain of cardinality more than $24^n/(2n+1)$, which in turn is
larger than $16^n$.
A: It may be clarifying to work with equivalence relations $E$ on $X$ rather than partitions on $X$. The two are in natural bijection, with $E$ inducing a partitioning quotient map $q: X \to X/E$, and $X/E$ refines $X/E'$ iff $E \subseteq E'$ as subsets of $X \times X$. 
Next, there is a surjective order-preserving map $P(X \times X) \to \text{Equiv}(X)$ where a general relation $R \in P(X \times X)$ is mapped to the equivalence relation $E_R$ that it generates. This is clearly surjective since an equivalence relation $E$ is mapped to itself. 
Finally, if $X$ is infinite, there is a bijection $X \cong X \times X$, which induces an isomorphism of orders $P(X) \to P(X \times X)$. The composite 
$$P(X) \to P(X \times X) \to \text{Equiv}(X)$$ 
provides what you want. 
