How should I understand $f^{-1}(E):=\{x\in A:f(x)\in E\}$?

Question

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?

Answer 1

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).

Answer 2

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 $f$"

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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.

Answer 3

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 $A$.

Answer 4

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] $.

Answer 5

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 ?

Answer 6

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.

Answer 7

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

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 $T$ $= T^{-1}[S] = \{x \in X : T(x) \in S\}$.