The Relationship Between Inverse and Image Functions Explained

In summary: Basically, a function is a relation that has a unique output for every input. The inverse of a function is when the input and output are switched, so the output becomes the input and vice versa. This can be written as f^{-1}. However, not all functions have an inverse. A function must be injective (one-to-one) for it to have an inverse. In other words, each input must have a unique output. If a function is not injective, it can still have a left or right inverse, but not a two-sided (or full) inverse.
  • #1
quantum123
306
1
The inverse image of a function, and the image of the inverse of a function is written in the same way right?
It is so confusing.
Are there any better ways to write such stuff?
 
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  • #2
No, it's not confusing. If f has an inverse, then the image of the inverse and the inverse image are exactly the same.

(I once made a fool of myself when, giving a proof about "inverse images", I assumed (naturally!) that the function had an inverse!)
 
  • #3
Example: Consider [tex]f(x) = x^2[/tex].

If it's defined on the whole real axis, then [tex]f^{-1}(\{4\}) = \{ +2, -2 \}[/tex].
If we restrict it to the positive axis, then If it's defined on the whole real axis, then [tex]f^{-1}(\{4\}) = \{ 2 \}[/tex].
But now the function is injective, so invertible (you can in fact write down an explicit inverse, namely [itex]g(x) = \sqrt{x}[/itex]. The inverse satisfies [tex]g(4) = 2[/tex]. Actually, the inverse in a point is the pre-image of that point (which consists of just one element).
 
  • #4
>>>If f has an inverse, then the image of the inverse and the inverse image are exactly the same.
I have proven this. But it should be a theorem as it is not trivial or obvious to me.
 
  • #5
It's not? Look at my example. The inverse image of y contains all the points that get mapped by f to y. A function has an inverse if it's injective, that is: the inverse image contains just one point x_y. If we define the inverse [itex]f^{-1}[/itex] as the function that satisfies
[tex] f^{-1}f = f f^{-1} = \mathrm{id}[/tex] ([itex]\ast[/itex])
we must assign x_y to y. We could also define the inverse function as the map that does this, and then it's almost trivial to check that ([itex]\ast[/itex]) is satisfied.
 
  • #6
CompuChip said:
Example: Consider [tex]f(x) = x^2[/tex].

If it's defined on the whole real axis, then [tex]f^{-1}(\{4\}) = \{ +2, -2 \}[/tex].
If we restrict it to the positive axis, then If it's defined on the whole real axis, then [tex]f^{-1}(\{4\}) = \{ 2 \}[/tex].
But now the function is injective, so invertible (you can in fact write down an explicit inverse, namely [itex]g(x) = \sqrt{x}[/itex]. The inverse satisfies [tex]g(4) = 2[/tex]. Actually, the inverse in a point is the pre-image of that point (which consists of just one element).

Just wanted to point out that a function has to be bijective to be invertible. So it has to be both injective and surjective. So if we have the function [tex] f: \mathbb{R}^{+} \to \mathbb{R}^{+} [/tex] defined by [tex] f(x) = x^{2} [/tex], then [tex] f^{-1}(x) = \sqrt{x} [/tex] (positive square root) is bijective.

Also maybe you are talking about the following: [tex] \overrightarrow{f}: \mathcal{P}(X) \rightarrow \mathcal{P}(Y)[/tex] is defined by [tex] \overrightarrow{f}(A) = \{f(x) | x \in A \}[/tex] for [tex] A \in \mathcal{P}(X) \}[/tex] and [tex] \overleftarrow{f}: \mathcal{P}(Y) \rightarrow \mathcal{P}(X)[/tex] is defined by [tex] \overleftarrow{f}(B) = \{x \in X | f(x) \in B \}[/tex] for [tex] B \in \mathcal{P}(Y) }[/tex]. These two functions are basically extensions of [tex] f [/tex] and [tex] f^{-1} [/tex].

So if [tex] f [/tex] is a bijection with inverse [tex] f^{-1} [/tex], then [tex] f(x) = y_0 [/tex] iff [tex] x = f^{-1}(y_0) [/tex] so that [tex] \overleftarrow{f}(\{y_0\}) = \{f^{-1}(y_0)\} [/tex]. So the RHS could contain more than 1 element, or none at all for certain values of [tex] y [/tex].
 
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  • #7
A function doesn't have to be bijective to be invertible, only injective. Essentially, one-to-one means that [itex]f^{-1}[/itex] is a function with domain Im(f). Surjectiveness will tell you that the domain of the inverse (exists if f is one-to-one) is precisely the codomain.

Just look at any analysis text that needs to use inverses: they will show that [itex]f[/itex] is one-to-one and then start invoking [itex]f^{-1}[/itex], since it exists.
 
  • #8
Look at this: http://en.wikipedia.org/wiki/Inverse_function. For the inverse function to exist and be valid, it must be bijective.

So the function [tex] \sin: [-\pi/2, \pi/2] \to [-1,1] [/tex] is bijective and thus invertible. But [tex] \sin: \mathbb{R} \to \mathbb{R} [/tex] is not invertible. It essentially hits the image, and is one-to-one (first one). Usually we say that the inverse of [tex] \sin(x) [/tex] is [tex] \arcsin(x) [/tex]. But we make assumptions about the domain and codomain (i.e restrictions).
 
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  • #9
Tronter: I seldom see such rigor in textbooks. Wonder where you get that?
 
  • #10
tronter said:
Just wanted to point out that a function has to be bijective to be invertible. So it has to be both injective and surjective. So if we have the function [tex] f: \mathbb{R}^{+} \to \mathbb{R}^{+} [/tex] defined by [tex] f(x) = x^{2} [/tex], then [tex] f^{-1}(x) = \sqrt{x} [/tex] (positive square root) is bijective.

Also maybe you are talking about the following: [tex] \overrightarrow{f}: \mathcal{P}(X) \rightarrow \mathcal{P}(Y)[/tex] is defined by [tex] \overrightarrow{f}(A) = \{f(x) | x \in A \}[/tex] for [tex] A \in \mathcal{P}(X) \}[/tex] and [tex] \overleftarrow{f}: \mathcal{P}(Y) \rightarrow \mathcal{P}(X)[/tex] is defined by [tex] \overleftarrow{f}(B) = \{x \in X | f(x) \in B \}[/tex] for [tex] B \in \mathcal{P}(Y) }[/tex]. These two functions are basically extensions of [tex] f [/tex] and [tex] f^{-1} [/tex].

So if [tex] f [/tex] is a bijection with inverse [tex] f^{-1} [/tex], then [tex] f(x) = y_0 [/tex] iff [tex] x = f^{-1}(y_0) [/tex] so that [tex] \overleftarrow{f}(\{y_0\}) = \{f^{-1}(y_0)\} [/tex]. So the RHS could contain more than 1 element, or none at all for certain values of [tex] y [/tex].



Nice post. Extending your remarks a bit, the following is a useful theorem about inverses:

Let f:A->B, with A not empty. Then

f is injective iff f has a left-inverse,
f is surjective iff f has a right-inverse,
f is bijective iff f has a two-sided inverse (a left and right inverse that are equal).

This two-sided inverse is called the inverse of f.
 
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  • #11
Nice theorem.
A function is a special type of relation R in which every element of the domain appears in exactly one of each x in the xRy. A relation is a subset of a Cartesian product. A Cartesian product AXB is a set of (a,b) tuple where a belongs to A, and b belongs to B. A (a,b) tuple is actually the set {a,{a,b}}.
 
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What is the relationship between inverse and image functions?

The relationship between inverse and image functions is that they are two sides of the same coin. Inverse functions undo the original function, while image functions map the output of the original function onto a new set of values.

What is an inverse function?

An inverse function is a function that reverses the output of the original function. It essentially undoes the original function, returning the input value.

What is an image function?

An image function is a function that maps the output of the original function onto a new set of values. It essentially takes the output of the original function and uses it as the input for a new function.

How are inverse and image functions related?

Inverse and image functions are related in that they both involve manipulating the output of the original function. Inverse functions undo the original function, while image functions map the output onto a new set of values.

Why are inverse and image functions important?

Inverse and image functions are important because they allow us to solve for unknown values and manipulate data in a variety of ways. They are also essential in many mathematical and scientific applications, such as in calculus and linear algebra.

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