Mustn't a function map every element of its domain to range (but not codomain)? [Richard Hammack, P228] 
How to Prove It, D Velleman P226, P228:
Suppose $f$ is a relation from $A$ to $B.$ Then $f$ is a function from $A$ to $B$ means:
  $\forall \; \color{#009900}{a \in A}, \exists \; ! \; b \in B$ such that $(\color{#009900}{a},b) \color{#009900}{\in f}$. $\quad$ Ie/To wit : $ f = \{ \; (\color{#009900}a,b) \color{#009900}{\in A} \times B : f(a) = b \;\}$.
P228. Theorem 13.7: If $A$ is any set, then $|A|<|\mathscr P(A)|$.
  [Only an extracted proof here] … Next we need to show that there exists no surjection $f\colon A \to \mathscr P(A)$. Suppose for the sake of contradiction that there does exist a surjection $f\colon A \to \mathscr P(A)$. Notice that for any element $a\in A$, we have $f(x)\in\mathscr P(A)$, so $f(a)$ is a subset of $A$. Thus $f$ is a function that sends elements of $A$ to subsets of $A$.
  ${\Large{\color{red}{\blacklozenge}}} \;$ It follows that for any $a\in A, \begin{cases} a \in \quad f(a) \in \mathscr P(A) \\ a \notin \quad f(a) \in \mathscr P(A) \end{cases}.$
Using this idea, define the following subset $B$ of $A$: $B=\{a\in A:a\notin f(a)\}\subseteq A$.
  Now since $B\subseteq A$ we have $B\in\mathscr P(A)$, and since $f$ is surjective there is an $a\in A$ for which $f(a)=B$. Now, either $a\in B$ or $a\notin B$. We will consider these two cases separately, and show that each leads to a contradiction.

Source: (The inestimable, monumental, august) Book of Proof by Professor Richard Hammack. 
By Velleman's P228 definition above, a function maps every element in the function's domain to $\color{purple}{\text{exactly one element}}$ in the function's range. Per contra, a relation doesn't have to map every element in the relation's domain. For the elements that a relation does map, it can map to $\color{purple}{\ge 1 \text{ element}}$ in the relation's range.
I don't apprehend the sentence after the red lozenge. How can $a \notin f(a)$?
By Velleman's definition, isn't every $a \in A$ always mapped by $f$?
Supplementary dated Jan 3 2014:
I can't pinpoint why I still think $\color{#009900}{[ a \in A ] \in f(a)}$. Don't the green parts of Velleman's definition above mean and reveal $\color{#009900}{[ a \in A ] \in f(a)}$?  What am I misreading/misconceiving?
 A: The remarks on the distinction between function and relation are now correct, however the way the proof runs, relations are not involoved, only functions. I'll say a few things about your last paragraph:

"Thus, I don't apprehend the block quoted sentence in the proof of
  Theorem 13.7. How can x NOT be an element of f(x)? By Definition 12.1,
  isn't every x∈A always mapped by f?"

In one sense, it is usually the case for functions that $x$ is not an element of $f(x)$. For example given $f(x)=x^2$ we would say that $2$ is not an element of $f(2)=4$, since under a simple interpretation numbers are not regarded as sets. Even in the case $f(1)=1$ we wouldn't say that $1 \in 1$, again since in a simple interpretation $1$ is not a set. 
[Note: there is a "logic" interpretation of nonnegative integers in which each integer is the set of its predecessors, so that e.g. $4=\{0,1,2,3\}.$ With this view, one does have $x \in f(x)=x^2$ when $x=2$, since $2$ is a predecessor of $4$. But this logic interpretation would say for example that $1$ is not a member of $f(1)=1$ since $1=\{0\}$ in this "logic" interpretaion. A restriction of this logic interpretation is that it doesn't associate sets to real numbers other than nonnegative integers.]
So if you have, as in this proof, a context in which the question whether $x\in f(x)$ even makes sense to ask about specific $x$ in the domain, it at least should be the case that for $x$ in the domain the image of $x$ under $f$ should be a set. That is, if $A$ is the domain of $f$ then for each $a\in A$ the image $f(a)$ is a set, in the proof here the codomain of $f$ consists of the collection of subsets of $A$, and these are sets, so that asking if $x \in f(x)$ for specific $x \in A$ at least makes sense.
The final question of the quoted part above, "By Definition 12.1, isn't every $x \in A$ always mapped by $f$?" has the answer "yes", simply because to say $x$ is "mapped by $f$" only means that it is mapped to something. And since $A$ is the domain of $f$, naturally any $x$ in $A$ does get mapped to something. 
The question is, in the setup of the proof where one has assumed the existence of a map $f:A \to P(A)$ which is onto, whether a given $x$ in $A$ happens to map to a subset $f(x)$ of $A$ for which it happens that $x \in f(x).$ 
Here's a simple example in which $A=\{1,2,3\}.$ Suppose $f(1)=\{1,2\}$, $f(2)=\{3\}$, and $f(3)=\{1\}.$ In this case, we have $1 \in f(1)$ but $2 \in f(2)$ and $3 \in f(3)$ are each false.
The point of the proof is that, no matter how one sets the map $f$ up, it cannot wind up being an onto map from $A$ to the power set $P(A)$. Other answers (and the text you quote) have already covered this.  I'm only throwing these thoughts into an answer because you expressed in comments some remaining confusion about the situation, and hope this sheds some light on that.
Added material re. query in question supplement Jan 4.

I can't pinpoint why I still think $\color{#009900}{[ a \in A ] \in f(a)}$.
   Don't the green parts of Velleman's definition above mean and
  reveal $\color{#009900}{[ a \in A ] \in f(a)}$?  What am I
  misreading/misconceiving?

I think what you have is a confusion between the technical definition of $f$ as a collection of ordered pairs, as opposed to the notation $f(a)$, which refers to the (unique) second coordinate of the pair $(a,b) \in f$. From the definition, for each $a\in A$ there's a unique $b\in B$ for which $(a,b) \in f$, and the notation $f(a)$ is then used to denote that particular $b$. Note that $b$ is not in $f$, since $b$ is merely the second element of an ordered pair in $f$. 
If one looks at the particular pair $(a,b)$ as it occurs in the technical definition of $f$ as a collection of ordered pairs, and replaces $b$ by $f(a)$, what we have is that
$$(a,f(a)) \in f.$$
But one must keep in mind that in this statement $f$ is a collection of ordered pairs. If say I have the function $f=\{(1,7),(2,4)\}$ we can say that $(1,f(1))=(1,7)\in f.$ But we cannot say that $7 \in f$ because $7$ is not one of the pairs $(1,7),(2,4)$ which are the only two things in $f$. All we can say is that $7$ is the second coordinate of one of the pairs in $f$.
For usual functions, in which domain and range are collections of numbers, this confusion would be unlikely. My guess is that, in the present case where the range is a collection of sets, one might be tempted to think $f(a) \in f$ always. But the same thing happens, e.g. if $(2,\{1,2,5\}) \in f,$ we still cannot say $f(2) \in f$, only that $f(2)$ is the second coordinate of a pair in $f$.
I hope this clears up the supplementary question.
A: The confusion arises from mixing up the domain ($A$) of $f$ and the image of a specific point $x\in A$ under $f$.
What the theorem says, reworded, is the following:

Suppose $B$ is the set of all subsets of $A$, and suppose that $f:A\to B$ is a surjection. It doesn't matter what's $f$, the only important thing is that for any subset $Z$ of $A$ (as such, for any element of $B$) there is at least one $x\in A$ which is mapped to $Z$.
Now, every $x$ is mapped to some subset $f(x)$, and that subset may be the empty set, the whole $A$ or anything in the middle. Of course ($f$ is surjective) there is some $x_0$ which is mapped to the empty set, and some other $x_1$ which is mapped to $A$ itself (as an element of B).
So you can see that some $x$ are mapped to sets that contain $x$ ($x_1$ is an example) and some other are mapped to sets that do not contain $x$ ($x_0$ above is an example).
You can define from $f$ a subset $A_{\in f}$ of $A$ which contains all of the elements $x\in A$ for which $x \in f(x)$, and another subset, which is $A_{\notin f} = A \setminus A_{\in f}$, the set of $x\in A$ for which $x \notin f(x)$. It's not really important for the proof, but you can see that neither of those sets is empty, as $x_0 \in A_{\notin f}$ and $x_1 \in A_f$.
Suppose now that $f(x_?)$ is $A_{\notin f}$. Then we have (because of the equality)
    $$x_? \in A_{\notin f} \implies x_? \in f(x_?) \\
  x_? \notin A_{\notin f} \implies x_? \notin f(x_?)
  $$
    But because of the definition of $A_{\notin f}$, we have also
    $$x_? \in A_{\notin f} \implies x_? \notin f(x_?) \\
  x_? \notin A_{\notin f} \implies x_? \in f(x_?)
  $$
Given that exactly one of $x_? \in A_{\notin f}$ and $x_? \notin A_{\notin f}$ must be true, in any case we have a contradiction. So the initial assumption was wrong, and there is no $x_?$, so $f$ is not surjective.

A: For any $x \in A$ and any subset $X$ of $A$, either $x \in X$ or $x \not\in X$. Now two sentences before your block quote, you've already established that, for any $x \in A$, $f(x)$ is a subset of $A$.
Maybe the confusion lies in the fact that $f(x)$ contains, possibly, multiple elements of $A$. That may be true, but there is still exactly one image of $x$ under $f$ (as required by the definition of function), namely the set $f(x)$. (So it's not the elements of $f(x)$ that are the image; it's $f(x)$ itself).
