3
$\begingroup$

I understand the concept, but I still can't figure out how to read the notation:

$$f^{-1}(E):=\{x\in A:f(x)\in E\}$$

I understood the concept due to the examples, not with the notation. Can someone translate/explain how to read it to me?

I'm thinking that it means: All numbers that when evaluated, will result in $f(x)$, I could find a inverse image in $f(x)=x^2+x$, for example: Consider $A=\{2,-3\}$ and $B=\{6\}$ where $A$ is the inverse image, for this I just took the procedure I found on wikipedia.

For example, for the function $f(x) = x^2$, the inverse image of $\{4\}$ would be $\{-2,2\}$.

But I got confused when I read this:

$x^2+x$ is not invertible as a function on R. Are you restricting the domain?

What's wrong?

$\endgroup$
2
  • $\begingroup$ Did you mean $f^{-1}(E)$??? $\endgroup$
    – copper.hat
    Apr 14, 2013 at 5:50
  • $\begingroup$ Oh, no. It's because I omited some details which are contained in the other question. Sorry. $\endgroup$
    – Red Banana
    Apr 14, 2013 at 5:55

7 Answers 7

4
$\begingroup$

If you have a function $f:A \to B$, you can always define a set-valued inverse $f^{-1}(E)$ as above, ie, all elements of $A$ that map into the set $E \subset B$.

It may be empty, as in $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$, then $f^{-1} [-2,-1] = \emptyset$.

It may have many values as in $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 1$, then $f^{-1} \{1\} = \mathbb{R}$.

It may be a singleton as in $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$, then $f^{-1} \{y\} = \{ \sqrt[3]{y} \}$.

Note that $f^{-1}(E) \subset A$ is a set.

However, if it turns out that $f^{-1} \{ y \}$ is a singleton for all $y \in B$, then you can define an inverse function, which (confusingly) is also denoted by the same symbol. In this case we have (the function) $f^{-1}(y) = x$, where $x \in f^{-1}\{y\}$ (this $f^{-1}$ is the set-valued inverse, remember that we are presuming that it is a singleton here).

$\endgroup$
2
  • $\begingroup$ Isn't it the set $E\subseteq B$? In the book it's stated in this form. Earlier in the book, the author mentions about defining a function "in pieces". I guess it's related somehow. $\endgroup$
    – Red Banana
    Apr 14, 2013 at 8:58
  • $\begingroup$ @Gustavo: Yes, it is, but $f^{-1}(B)=A$ almost by definition, so it doesn't make much difference if you consider $E=B$ or not. $\endgroup$
    – A.P.
    Apr 14, 2013 at 9:40
3
$\begingroup$

Recall that in modern mathematics functions are also relations, i.e. sets of ordered pairs. Given a relation $R$, we can talk about the inverse relation, $R^{-1}=\{\langle y,x\rangle\mid \langle x,y\rangle\in R\}$.

To understand the inverse relation one simply has to think about a relation as a bunch of arrows between points, and the inverse relation is merely the inversion of these arrows.

Now it should be simpler to understand $f^{-1}(E)$. This is the inverse relation of $f$. Note that if $f$ is injective, then $f^{-1}(x)$ is at most one point, and that defines a function (on a subset of the codomain); but this is irrelevant to the main point: "$f^{-1}(E)$ is the image of $E$ under the inverse relation of $f$"


On a general level, when we write $\{x\mid \varphi(x)\}$ (the $\mid$ is sometimes replaced by $\colon$) we define a collection, naively a set, which includes all the objects $x$ such that $\varphi(x)$ is true for them.

When we write $\{x\in A\mid\varphi(x)\}$ we mean $\{x\mid x\in A\land\varphi(x)\}$. This way we limit our collection to elements of $A$. This is a solution for several possible paradoxes of naive set theory when we assume that $A$ is a set, and the result is a set.

$\endgroup$
13
  • $\begingroup$ I loved your notation, it's way more comprehensible. But what confused me is the meaning of: $\{x\in A:f(x)\in E\}$ what did he meant with that? $\endgroup$
    – Red Banana
    Apr 14, 2013 at 9:03
  • $\begingroup$ Gustavo, this is simply the "direct image of the inverse relation". It is all the points in the domain of $f$ which are mapped into $E$. $\endgroup$
    – Asaf Karagila
    Apr 14, 2013 at 9:08
  • $\begingroup$ Yes. But I'm trying to figure out as if it were a programming languange, if I were to read $\{(\text{message 1}):\text{(message 2)}\}$ how would it be? Is it something like: "All the $x$'s that make the given $f(x)$'s? $\endgroup$
    – Red Banana
    Apr 14, 2013 at 9:11
  • 1
    $\begingroup$ In which programming language? In Common Lisp it would simply be defining message 2 as a predicate, and considering those in $A$ which satisfy it. Your question doesn't make much sense to me, because mathematics is not a programming language. $\endgroup$
    – Asaf Karagila
    Apr 14, 2013 at 9:13
  • 1
    $\begingroup$ Also, to push your idea further, try reading a block of code in Bovine, or in Brainfuck. Then try reading the same block of code in Scheme, Common Lisp, Clojure, C++, Visual Basic, Delphi, and lastly x86 assembler. Your ability to parse the same block of code depends on your ability to understand the language. Your question, if so, should not be "how to parse this block of code, which I already fully understand through examples" but "how do I understand the language of set theory?". $\endgroup$
    – Asaf Karagila
    Apr 14, 2013 at 10:52
2
$\begingroup$

Your definition is also called the preimage/pre-image of $E$: the subset of the domain $A$ that gets mapped to $E$. Draw a Venn-type diagram, a region for $A$, and an arrow from $A$ to a set $B$, the image of $A$ inside the range. Now imagine that $E$ is a set that has nonempty intersection with $B$. Which elements of $A$ get mapped into $E$ (or, more precisely, into its intersection with $B$)? You can refine your diagram by splitting $A$ into a part that gets mapped under $f$ into $E$, and another part that gets mapped into the rest of $B$. The former part is the pre-image of $E$ in $A$.

$\endgroup$
2
$\begingroup$

Take for example $f:R \rightarrow [-1,1]$ as $f(x)= sin(\pi x)$ let $E= \{1\}$ then $f^{-1}(E)=\{x\in R:f(x)\in E\}= \{\pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{9}{2},..\}$ Because $f(\pm \frac{1}{2})=1 \in E$, $f(\pm \frac{5}{2})=1 \in E$, and so forth.

but we cannot define a function $g:[-1,1] \rightarrow \mathbb R$ to be $f^{-1}$ on the entire domain of $f$ because then $g$ is not well-defined i.e $g(1)$ has infinitely many values, So you need to restrict $f$ on some interval like for example $[-\frac{1}{2},\frac{1}{2}]$ instead of $\mathbb R$ in this case $g(y)$ has only one image in $[-\frac{1}{2},\frac{1}{2}]$, $\forall y \in [-1,1] $.

$\endgroup$
1
$\begingroup$

The inverse of a point is a set. So you can not define one well defined point as the inverse image. Hence its not a function. such as $f^{-1}(6)=\{2,-3\}$ So which one you define as the inverse image ?

$\endgroup$
1
$\begingroup$

If you restrict the domain $A$ to be $[0,\infty)$, for example, then the function $f$ is invertible, because for each value $f(x)$, there exists a unique $x$ in the (now restricted) domain $[0,\infty)$ that generates that value.

Therefore, the function $f(x)=x^2+x$ is not invertible as a function on $\mathbb R$, because $\mathbb R$ also includes negative numbers.

$\endgroup$
1
$\begingroup$

Emended picture from http://yorkporc.wordpress.com/2011/05/28/transforms-pre-images-and-kernels-and-null-spaces/

enter image description here

The original picture from Khan Academy has still more colours which I've lessened.

The pink should avail to answer your question:
It's the preimage/inverse image of $S \subseteq Y$ (though this picture shows $S \subsetneq Y$) under $T$ $= T^{-1}[S] = \{x \in X : T(x) \in S\}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .