Understanding the difference between pre-image and inverse I am a little confused as to what the difference between the pre-image and the inverse of a function are and how to find each given a particular function ( I had though they were essentially the same thing but I realise now I was mistaken in that thought ).
From Wikipedia the definitions given are :
Inverse
Let $f$ be a function whose domain is the set $X$, and whose image (range) is the set $Y$. Then $f$ is invertible if there exists a function $g$ with domain $Y$ and image $X$, with the property: 
$f ( x ) = y \iff g ( y ) = x . $
If $f$ is invertible, the function $g$ is unique, which means that there is exactly one function $g$ satisfying this property (no more, no less). That function $g$ is then called the inverse of $f$, and is usually denoted as $f ^{−1}$
Pre-image 
Let $f$ be a function from $X$ to $Y$. The preimage or inverse image of a set $B\subseteq Y$ under $f$ is the subset of $X$ defined by 
$f ^{− 1 }[ B ] = \{ x \in X \mid f ( x ) \in B \}$.
So using an example I want to see if I have this about right ( please correct any mistakes I make)
Example 1:
Lets say $f:\Bbb R \rightarrow \Bbb R$
$f(x)=x^2$
For this example, clearly $f$ cannot be invertible (hence no inverse) as there exists no function $g$ which will satisfy $ f ( x ) = y \iff g ( y ) = x$. (as it would only map to positive values of $\Bbb R$ (i.e. not the whole set))
The pre-image of this function I believe is related to the inverse except it does not require that we map to the whole set $\Bbb R$, but rather just a subset of it. Therefore we can find the function $f^{-1}$ in an analogous way to to finding the inverse we just have to be more considerate about what the co-domain of this function is. 
So if $f(x)=x^2 \Rightarrow y=x^2  $swap variables to get $x=y^2 \Rightarrow \sqrt{x}=y=f^{-1} $
So the pre-image is the set $f ^{− 1 }[ \Bbb R_+ ] = \{ x \in X \mid f ( x ) \in \Bbb R_+, f^{-1}=\sqrt{x} \}
$
Example 2:
An example of an invertible function would be $f:\Bbb R \rightarrow \Bbb R$
$f(x)=5x$, as a function $g(x)=x/5$ has domain $\Bbb R$ and range $\Bbb R$ and satisfies $f ( x ) = y ⇔ g ( y ) = x .$
The pre-image in this case will be equal to the inverse.
Could anyone please explain to me any mistakes I'm making here ?
 A: 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.
........
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.
....
I'll add more in an hour or so but I have to take the dog for a walk.  I'll be back.
.....
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.
A: When we talk about pre-image then it has two components: a set and a function. So when we say $f^{-1}[B]$, then we want the pre-image of the set $B$ (a subset of the co-domain) under the function $f$. So asking about pre-image of a function is a bit ambiguous. 
Let us consider $f:\{1,2,3\} \rightarrow \{a,b,c\}$ such that $f(1)=a, f(2)=a$ and $f(3)=c$. Then 
\begin{align*}
f^{-1}\left[\{a\}\right] & =\{1,2\}\\
f^{-1}\left[\{c\}\right] & =\{3\}\\
f^{-1}\left[\{a,c\}\right] & =\{1,2,3\}\\
f^{-1}\left[\{b\}\right] & =\emptyset\\
f^{-1}\left[\{a,b,c\}\right] & =\{1,2,3\}
\end{align*} 
The function $f$ (as described above) is not invertible. So relating the inverse function (which doesn't exist) to the inverse image is meaningless in this case.
One can check invertibility of a function $f: A \rightarrow B$ by checking the inverse images of singleton subsets of the co-domain. 
What it means is that: if we can ensure that for every $b \in B$, the inverse image set $f^{-1}\left[\{b\}\right]$ has exactly one element (this is to ensure both one-one and ontoness), then $f$ is invertible.
A: So, there's no such thing as the preimage of a function. Functions can have inverses; functions do not have preimages.
An inverse is something that certain functions have, and the inverse of a function is another function.
Given a function, a preimage is something that sets have, and the preimage of a set is another set.
Specifically:
Given a function $f : A \to B$, that function may or may not have an inverse. If it does, then that inverse is a function $B \to A$.
Given a function $f : A \to B$, and a set $s$ which is a subset of $B$, that set always has a preimage under $f$. That preimage is a subset of $A$.
That might clear up some of your confusion.
