2. Suppose that $f:X \mapsto P(X)$ is any mapping from a set to its power set.
Now define $A = \left\{ {y \in X:y \notin f(y)} \right\}$. Clearly $A \in P(X)$.
If $f$ were a surjection then $\left( {\exists a \in X} \right)\left[ {f(a) = A} \right]$.