The Relationship Between Inverse and Image Functions Explained

[quantum123]

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?

 

[HallsofIvy]

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!)

 

[CompuChip]

Example: Consider f(x) = x^2.

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

 

[quantum123]

>>>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.

 

[CompuChip]

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 f^{-1} as the function that satisfies
f^{-1}f = f f^{-1} = \mathrm{id} (\ast)
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 (\ast) is satisfied.

 

[tronter]

Example: Consider f(x) = x^2.

If it's defined on the whole real axis, then f^{-1}(\{4\}) = \{ +2, -2 \}.
If we restrict it to the positive axis, then If it's defined on the whole real axis, then f^{-1}(\{4\}) = \{ 2 \}.
But now the function is injective, so invertible (you can in fact write down an explicit inverse, namely g(x) = \sqrt{x}. The inverse satisfies g(4) = 2. 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 f: \mathbb{R}^{+} \to \mathbb{R}^{+} defined by f(x) = x^{2}, then f^{-1}(x) = \sqrt{x} (positive square root) is bijective.

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

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

 

[ZioX]

A function doesn't have to be bijective to be invertible, only injective. Essentially, one-to-one means that f^{-1} 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 f is one-to-one and then start invoking f^{-1}, since it exists.

 

[tronter]

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 \sin: [-\pi/2, \pi/2] \to [-1,1] is bijective and thus invertible. But \sin: \mathbb{R} \to \mathbb{R} is not invertible. It essentially hits the image, and is one-to-one (first one). Usually we say that the inverse of \sin(x) is \arcsin(x). But we make assumptions about the domain and codomain (i.e restrictions).

 

[quantum123]

Tronter: I seldom see such rigor in textbooks. Wonder where you get that?

 

[fopc]

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 f: \mathbb{R}^{+} \to \mathbb{R}^{+} defined by f(x) = x^{2}, then f^{-1}(x) = \sqrt{x} (positive square root) is bijective.

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

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

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.

 

[quantum123]

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}}.