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 ?
Best Answer
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.