How are image and pre-image different from range and domain respectively? How are image and pre-image different from range and domain respectively, in Layman's terms (as simple as possible)?
Are they basically just keywords that often indicate more nuanced subsets of the domain and range respectively when studying esoteric properties?
Is that basically what image and pre-image are? Words that just often help mathematicians drill down definitions to more specific subsets of the domain and range?
 A: The domain of a function is simply the set on which it's defined, so one naturally expects the domain of $x^2$ to be the reals $\mathbb{R}$ and of $1/x$ to be everything but zero $\mathbb{R}\setminus\{0\}$. But it's also possible to restrict the domain, and consider for instance $x^2|_{\mathbb{Q}}$ the restriction of $x^2$ to the rational numbers $\mathbb{Q}$; this is just the function that squares a rational number, where you forget that it's also possible to square real numbers (et cetera.) 
On the other side, the range of a function is now most often understood, although this is not consistent, to be a specified set in which the function lands. For instance, $x^2$ naturally lands in $[0,\infty)$, but many people would also be happy to name $\mathbb{R}$ as its range; certainly all values of $x^2$ are in the range $\mathbb{R}$, even though not every elements of $\mathbb{R}$ is $x^2$ for some $x$.
So domain and range have both been divorced from the earlier notions of the largest set on which a function can be defined, respectively, the set of all values of the function. Thus we frequently name a function along with a choice of domain and range. Thus $x^2:\mathbb{Q}\to \mathbb{R},x^2:\mathbb{R}\to[0,\infty),x^2:\mathbb{R}\to\mathbb{R}$ may be considered as three different functions, even though we calculate them all in the same way. 
However, most people probably consider the last two to be the same function, since they're defined on the same set and take the same values. Such a person is less likely to bother specifying the range of a function, leaving it implicit from the definition (as $[0,\infty)$ is naturally associated to $x^2$.) This "natural" range of a function is what we call the image, namely, the set of all values the function actually takes on. As I said when range first came up, the confusing fact is that "range" and "image" are still sometimes synonyms, depending on context. So, a newer word that definitely only means range as I used it in the second and third paragraphs is codomain. 
Preimage is really something different. Here you have a function $f$ and a subset $S$ of the image, range, or codomain of $f$. We say the preimage $f^{-1}(S)$ is the set of all points $x$ in the domain of $f$ such that $f(x)\in S$. So the preimage of $[1,\infty)$ under $f(x)=x^2$ is $(-\infty,-1]\cup[1,\infty)$. 
We can also define the image of any subset $T$ of the domain of $f$, similarly: it's the set $f(T)$ of all points in the image/range/codomain which are $f(x)$ for some $x\in T$.
OK, so, to summarize: we have the domain $D$, which is a nearly arbitrary* choice of set on which we're going to consider our function to be defined, and which determines images $f(T)$ for each $T\subset D$. In particular we get an image for the whole function $f$ as $f(T)$. Some people call this image $f(T)$ the range, while others give a range for $f$, a nearly arbitrary* choice of set in which $f$ will be considered to take its values. This last nearly arbitrary choice can also be called the codomain. Finally, each subset of the codomain/range/image has a preimage, and in particular the preimage of the image should be the whole domain.
*The domain is arbitrary except that $f$ must make sense on all of $D$, so we can't give $1/x$ the domain $\mathbb{R}$ without changing the defintion; the codomain is arbitrary except that it must contain the whole image $f(D)$.
A: Before answering the question we recall that a function from $A$ to $B$ is a relation $f\subset A\times B$ which has $A$ as its domain and $(x,y),(x,z)\in f$ implies $y=z$. It allows us then to write $f:A\to B$ and $y=f(x)$. Thus, a function has three intrinsic elements: a domain, a counter-domain and an association rule.
Now, given $S\subset B$,
$$ f^{-1}(S) = \{x\in A: f(x)\in S\} .$$
It does not coincide with the domain $A$ unless $S$ is the range
$$Im(f):=\{y\in B: y=f(x),\ x\in A\}. $$
On the other hand, given $C\subset A$, its image by $f$ is the range of the restriction $f_{|C}:C\to B$, defined by $f_{|C}(x) = f(x)$, that is,
$$f(C):= \{y\in B: y=f(x), x\in C\} .$$
And, in general, it does not coincide with the $Im(f)$. Therefore, unlike the pre-image of a set, the domain is intrinsic to the function, and the range is the domain's image by $f$.
