Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse and Image

  1. Jul 3, 2007 #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?
     
  2. jcsd
  3. Jul 3, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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!)
     
  4. Jul 3, 2007 #3

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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).
     
  5. Jul 3, 2007 #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.
     
  6. Jul 3, 2007 #5

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Jul 9, 2007 #6
    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].
     
    Last edited: Jul 9, 2007
  8. Jul 10, 2007 #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.
     
  9. Jul 10, 2007 #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).
     
    Last edited: Jul 10, 2007
  10. Jul 10, 2007 #9
    Tronter: I seldom see such rigor in text books. Wonder where you get that?
     
  11. Jul 10, 2007 #10


    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.
     
    Last edited: Jul 10, 2007
  12. Jul 10, 2007 #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}}.
     
    Last edited: Jul 10, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Inverse and Image
  1. Inverse gamma (Replies: 1)

  2. Image of a set (Replies: 1)

Loading...