Cantor's Theorem (surjection vs bijection) Let me state Cantor's Theorem first: Given any set $A$, these does not exist a function $f:A \rightarrow P(A)$ that is surjective.
I understand the proof of this theorem, but I'm wondering why it's enough to show that the function $f$ is not surjective, but rather not bijective?  Can't you map one element of $A$ to multiple elements of $P(A)$ such that you cover the whole set?  Shouldn't it be that you can't produce a bijective function?
 A: A map, by definition, maps an object in the domain to (at most) one object in the range. If you were to be able to do that, then we'd get results such as $\mathbb{R}$ being countable.
Also, if there is no surjection, that implies that there is no bijection.
A: A surjective map from $A \rightarrow B$ maps each element of $A$ to one element of $B$, such that every element of $B$ is mapped to by at least one element of $A$. 
So the Cantor proof does show that there is no surjective mapping, because surjective does not allow one element of $A$ to be mapped to multiple elements of $P(A)$.
A: A bijective function is defined as a function that is both injective and surjective. If a function is not surjective, then it certainly cannot be bijective.

Can't you map one element of A to multiple elements of P(A) such that you cover the whole set?

No, if you assigned multiple elements of $P(A)$ to a single element of $A$, then it wouldn't be a function. A function by definition assigns exactly one element from the codomain set to each element in the domain set.
