The proof below is due to George Boolos, and appears in his article *Constructing Cantorian Counterexamples*, **Journal of Philosophical Logic**, Vol. 26, No. 3 (Jun., 1997), pp. 237-23.  

The proof gives an explicit proof that there cannot exist a one-to-one function from the powerset of a set into the same set.

A similar proof appears as Corollary 1.3 of the following paper (a copy of which is available here).

Akihiro Kanamori, David Pincus, D*oes GCH imply AC locally?*, in **Paul Erdős and his mathematics, II** (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.

Suppose $f: \mathcal{P}(A)\rightarrow A$. We want to explicitly define subsets $B$ and $C$ of $A$ such that $f(B) \neq f(C)$.

For any binary relation $r$, let $r_{x} = \{y: (y,x)\in r \land y\notin x \}$, and let $\mathrm{field}(r)$ be the *field* of $r$, i.e., the set of objects that appear as the first or second coordinate of an ordered pair in $r$.

Call a relation $r$ to be *good* iff $r$ is a reflexive well-ordering of a subset of $A$ and for every $x$ in $\mathrm{field}(r)$, $f(r_{x}) = x$. Let $R$ be the union of all good $r$. 

If $r$ and $r'$ are good, then one of $r$ and $r'$ is an initial segment of the other; therefore $R$ is itself good. Let $C = \mathrm{field}(R)$. For $C \subseteq A$, Let $x = f(C)$, and let $B = R_{x}$. 

Note that $C$, $x$, and $B$ are all explicitly defined from $f$. 

If $x \notin C$, then $R \cup \{(y, x): y E C \lor y = x\}$ is good, and therefore $x \in C$. So $x\in C$. 

Since $x \notin \{y: yRx \land y\neq x} = B$, $B\neq C$. 

Since $R$ is good, $x = f(R_{x}) = f (B)$. But $x = f (C)$. Thus $f$ is not one-one. 

As noted by Boolos, since $Rx \subset \mathrm{field}(R)$, the above proof shows a nontrivial strengthening of the nonexistene of an injection of the powerset of a $A$ into $A$, namely, if $f: \mathcal{P}(A) \rightarrow A$, then there subsets $B$ and $C$ of $A$ such that $B\subsetneq C$ and $f (B) = f (C)$.