# 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 ?

• For starters, they are two very different objects. The pre-image of a function is a subset of the domain and the inverse function is a function from the range back to the domain that satisfies certain properties. A function may not be invertible but we can always talk about pre-image of a function. – Anurag A Jul 7 '19 at 20:58
• @AnuragA I know now that they are not the same thing , I was just wondering if I was correct in identifying how they are different – excalibirr Jul 7 '19 at 21:00
• No, one peaks in almost all cases of the preimage of a subset of the target space. Take, for instance, the squaring function $f(x)=x^2$, and ask for the preimage of the set of odd numbers. Then this will be the set of all $\pm\sqrt{2k+1}$, with $k$ running through all integers. If one were to speak of the “preimage of a function”, presumably that would be the preimage of the whole target space, which is all of the domain of definition of the function. – Lubin Jul 7 '19 at 21:09

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.

• Thank you for such a brilliantly comprehensive answer :) there's literally nothing else I could think to ask . – excalibirr Jul 15 '19 at 7:34
• Actually , I did manage to think of question lol, (it's in the context of continuity in topology) say we have the function $f=x^2$ and a basis set $(-q,q), q \in \Bbb Q$ in our target. then I know that $f^{-1}((-q,q))=\{ x \in X| x\in (-\sqrt{q}, \sqrt{q})\}$ but why isn't it $(\sqrt{-q}, \sqrt{q})$ instead ?, also say we have the set (a,b) (a<b) with the same function then is $f^{-1}=(-\sqrt{b},-\sqrt{a}) \cup (\sqrt{a},\sqrt{b})$ ? – excalibirr Jul 15 '19 at 9:28

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.

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.