Why $f: A \to A$ being injective not implies it is also surjective when $A$ is infinite? We know this is true for finite sets: If there is a function $f: A \to A$ and it injective then it is also surjective, hence bijective. It is really easy to understand with a graphical illustration but why the same not holds for infinite $A$?
 A: The successor-function $f: \mathbb N \to \mathbb N$ $$f(n)=n+1$$ is the classical counterexample. It is injective, but not surjective (since there is no $n$ providing us with $f(n) = 0$, yet $0 \in \mathbb N$). Another one would be $$f(n)=2n$$
A: This is true, not merely for transformations of a finite set $A,$ as you discussed above, but more generally for any function $f:A\to B,$ where $A,B$ are finite and $|A|=|B|.$ This generality helps you see what's doing the work (and not to be carried away by the red herring case when $A=B$). That is, we see that for finite sets, it is if they have equal cardinality that injection implies surjection.
What about when $A,B$ are infinite? First, assume that these sets again have equal cardinality. The notion of equinumerosity for infinite sets is quite weaker than for finite sets; hence, an infinite set can be equinumerous with some of its subsets -- this can never happen for finite sets. This is why it is not necessarily true for infinite sets, since even if $|A|=|B|,$ it doesn't follow that none of them is a proper subset of the other, as in the case of finite sets.
A: The way I thought about it when I was very young was:  If you remove a few things from an infinite set you will still have an infinite set.  So you can map from a set and skip a few things and still have a mapping from an infinite set to an infinite set that doesn't map them all.
Example:  $\{1,2,3,4,5,6,7,8,9.......\}$ is infinite.  
Remove the $3,4,$ and $6$ to get $\{1,2,5,7,8,9.....\}$ is still infinite.
Map:  $1\mapsto 1$ and $2\mapsto 2$ and $3\mapsto 5$ and $4\mapsto 7$ and $5\mapsto 8$ etc.  then that maps from the $\mathbb N \to \mathbb N$ but skips $3,4,$ and $6$ so is not surjective.
Or if you want to be silly:
$\mathbb N = \{1,2,3,4,5......\}$ is infinite.  And $\{\text{Babar; the elephant}, 1,2,3,4,5,.....\}= \{\text{Babar; the elephant}\} \cup \mathbb N$ is infinite.  ANd if $f:\mathbb N \to  \{\text{Babar; the elephant}\} \cup \mathbb N$ via $f(n) = n$ then there is no $f(n) = \text{Babar; the elephant}$ so it is not surjective.
[Not so silly:  $f: \mathbb N \to \mathbb Z$ via $f(n)$ is not surjective because for and $x \le 0$ there is no $f(n) = x$.]
More seriously.
Let $f: \mathbb N \to \mathbb N$ so that $f(n) = 2n$.  then there is no $f(n) = $an odd number.  So $f$ is not surjective.
Of $f(n) = n+1$ so that there is no $n$ so that $f(n) =1$, so $f$ is not surjective.
Which gets to the heart of it.
If $S$ and $R$ are both infinite it is possible that $S\subsetneq R$ so you can have injective but not surjective $f:S \to R$ by having a bijective $f: S \to Image(S) \subsetneq R$.
But if $A$ and $B$ are both finite and $|A| =|B|$ then it is impossible to have $A\subsetneq B$ and so you can't have an injective but not surjective $f: A\to B$.  [If $f:A \to Image(A)\subsetneq B$ then $|Image(A)| < |B| = |A|$ so then $f$ isn't injective.  And if $f:A \to IMage(A) = B$ is injective and... since $Image(A) =B$ that is, by definition, surjective.]
