The biggest difference between a preimage and the inverse function is that the preimage is a subset of the domain. The inverse (if it exists) is a function between two sets.
In that sense they are two very different animals. A set and a function are completely different objects.
So for example: The inverse of a function $f$ might be: The function $g:\mathbb R \to \mathbb R: g(x) = \sqrt[3]{x-9}$. Whereas the preimage of a set $B$ of the function might be $[1,3.5)\cup \{e, \pi^2\}$.
Now $g(x) = \sqrt[3]{x-9}$ and $[1,3.5)\cup \{7, \pi^2\}$ are completely different types of things.
This will be the case if $f$ is $f:\mathbb R \to \mathbb R: f(x) = x^3 + 9$ and $B= [10, 51.875) \cup \{352, \pi^6 + 9\}$.
The inverse $f^{-1}(x)$ (if it exist) is the function $g$ so that if $f(x) = y$ if and only if $g(y) = x$. So if $f(x) = x^3 + 9 = y$ then if such a function exists it must be that $g(y)^3 + 9 = y$ so $g(y)^3 = y-9$ and $g(y) = \sqrt[3]{y-9}$ so $g(x) = \sqrt[3]{x-9}$.
That's that.
The pre-image of $A= [10, 51.875) \cup \{352, \pi^6 + 9\}$ is the set $\{x\in \mathbb R| f(x) \in [10, 51.875) \cup \{352, \pi^6 + 9\}\}=$
$\{x\in \mathbb R| x^3 + 9 \in [10, 51.875) \cup \{352, \pi^6 + 9\}\}=$
$\{x\in \mathbb R| x^3 \in [1, 42.875) \cup \{343, \pi^6 \}\}=$
$\{x\in \mathbb R| x \in [1, 3.5) \cup \{7, \pi^2 \}\}=$
$[1, 3.5) \cup \{7, \pi^2 \}\}$.
And that's the other.
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Now that's not to say the inverse of a function and the pre-image of a set under the function aren't related. They are. But they refer to different concepts. This is similar to how a rectangle and its area are related. But one is a geometric shape... the other is a positive real number. THey are two different types of animals.
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I'll add more in an hour or so but I have to take the dog for a walk. I'll be back.
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It occurred to me as I was walking the dog that maybe what is confusing you is that the inverse function (if it exists) and the preimage of a set have very similar notation and the only way to tell them apart is in context.
If $f$ is invertible then the inverse function is written as $f^{-1}$ so if $f(x) = x^3 + 9$ then $f^{-1}(x) = \sqrt[3]{x-9}$.
But the preimage of $B$ under $f$ whether $f$ is invertible or or not is writen as $f^{-1}(B)$.
So if $f(x) = x^3 + 9$ then $f^{-1}(17) = 2$ means that if you enter $17$ into the function $\sqrt{x -9}$ you get $2$. But $f^{-1}(\{17\})=\{3\}$ and $f^{-1}(\{36,17\}) = \{2,3\}$ means that set of values that will output $\{17\}$ is the set $\{2\}$ and the set of values that will output $\{36,17\}$ is the set $\{2,3\}$.
A few things to note:
If $f$ is invertible then the preimage of a set is the same thing as the image of the set under the inverse function and that means the notation is compatible.
If $f(x) = x^3 + 9$ then $f^{-1}([1,36)) = [1,3)$ can be interpretated as both the the image of the set under the inverse function: $f^{-1}([1,36))= \{f^{-1}(x) = g(x) = \sqrt[3]{x-9}| x\in [1,36)\}$
OR it can be interpreted as the preimage for $f$: $f^{-1}([1,36)) = \{x\in \mathbb R| f(x) \in [1,36)\}$.
but this is not the case if $f$ is not invertible.
Say $f:\mathbb R \to [-1,1]; f(x)\to \sin x$. This is not invertible.
The pre-image of$B= \{\frac {\sqrt 2}2\}$ is $\{...-\frac {11\pi}4, -\frac {9\pi}4,-\frac{3\pi}4,-\frac \pi 4, \frac \pi 4, \frac {3\pi}4, \frac {9\pi}4, \frac {11\pi}4,....\}$ this is still written as $f^{-1}( \{\frac {\sqrt 2}2\})$ even though there is no function $f^{-1}:[-1,1]\to \mathbb R$.
Another thing to note is that not all the elements in $B$ have to have pre-image values.
If $f= x^2+9$ then $f^{-1}(\{8\}) = \emptyset$. This is because $\{x\in \mathbb R| f(x) = x^2 + 9 \in \{8\}\} = \emptyset$.
And some elements may have many preimages.
And $\sin^{-1}(\{\frac {\sqrt2} 2}$ showed.
That's a perfectly well defined function from $B \to \mathscr P(A)$ but it is not a function from $B\to A$ at all. So that defeats the purpose.
If we try to argue that a dolphin is a fish and then say, well, a dolphin is a mammal that lives the entirety of its life in water, so we can use that instead, is to .... be totally irrelevant.
You seem to think $g{c}$ vs. $g(c)$ vs $g(\{c\})$ is significant in whether we are mapping a set with a single element or single value. And I suppose conceptually and in strict definition it is. But in praticality, there isn't much difference (and in ZFC where everything is a set, there is no difference) between expressing somthing in terms of a single value or in terms of a set containing just the single value. The actual notation $f(x)=c$ is shorthand for $(x, c)\in f \subset A\times B$.
Although $f^{-1}: B \to \mathscr P(A)$ is a well defined function it does not map subsets of $B$ to subsets of $A$. It maps single elements (or equivalently singleton sets) of $B$ to subsets of $A$. To map subsets of $B$ to subset of $A$ we must define $f^{-1}(C)$ where $c \subset B$ as $f^{-1}(C)$ as $\cup_{c\in C} \{a\in A| f(a) = c\}$.
With that definition $f^{-1}: \mathscr P(B) \to \mathscr P(A)$
Best Answer
Inverse image preserves unions, intersections and complements. In topology and analysis we often work with families of subsets of the space, "open sets" or "Borel sets", and so on. These families are often closed under unions, or intersections - or some variation thereof.
Using the fact that inverse images preserves set operations means that we can use it to characterize "good" functions. Continuous functions are those that the preimage of an open set is open; Borel functions are those that the preimage of a Borel set is Borel; and so on.
For this reason we are usually more interested in the inverse image than in the direct image; however that too has its uses (but often coupled with some condition associated with inverse images). For example a bijection which is both open and continuous is in fact a homeomorphism.