There exists no surjective function from a set to its power set. To see why, let $X$ be a set and assume there exists such a $f:X\to\mathcal{P}(X)$. Since $f$ is surjective, we have: $$ \forall y\in\mathcal{P}(X),\, \exists x\in X\,( f(x)=y ) $$ Let $D=\{x\in X\mid x\notin f(x)\}$. Since $D\in\mathcal{P}(X)$, we have: $$ \exists x\in X\,( f(x)=D ) $$ Let $d\in X$. Then, we have: $$ f(d)=D $$ Clearly, either $d\in D$ or $d\notin D$. If $d\in D$, then $d\notin f(d)$. Otherwise, $d\in f(d)$. Overall, since $f(d)=D$, we have: $$ d\in D \iff d\notin D $$ Since this is a contradiction, the assumption that $f$ exists is false.