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Finding Inverse the Function of multiple variable functions

  1. Oct 14, 2012 #1
    I understand that given two functions

    f:X→Y and g:Y→X, to say that f is the inverse function of g means

    g o f:X→X is defined by g(f(x))=idx

    and to say g is the inverse function of f means

    f o g: Y→Y is defined by f(g(x))=idy

    I understand how to find inverses of one variable functions and I'm able to apply this definition. However I'm having difficulty of finding examples of finding inverses of maps such as f:ℝ^2→ℝ^2 or something like f:ℝ^3→ℝ^3. Similarly I'm not sure if we can find inverses of functions with such mappings as f:ℝ^2→ℝ. Specific examples would highly be appreciated. Thanks.
     
    Last edited: Oct 14, 2012
  2. jcsd
  3. Oct 15, 2012 #2

    haruspex

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    More accurately, you should distinguish between left inverses and right inverses. An onto function will have one, a 1-1 function the other (I forget which is which of left/right). Only a 1-1 onto function will have an inverse that works both ways.
    For an invertible function on R2, you can have any rotation, translation, magnification, and lots besides.
    A map f from R2 to R cannot be 1-1, so can only have a one-sided inverse, namely, a function from R to R2 that f will then undo.
     
  4. Oct 15, 2012 #3
    right only bijections will work both ways and right now that's what i'm asking for specific examples of mappings from the plane or in space. but thanks for the clarification, that totally makes sense that a map from R2 to R cannot be 1-1.

    what do you mean magnification and lots besides? these do not sound like these functions will preserve distance. i understand rotation and translations because those are direct isometries. either way would you have a specific example of finding the inverse of a function in either the plane or space?
     
  5. Oct 15, 2012 #4

    micromass

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    Well, actually, there do exist bijections between [itex]\mathbb{R}^2[/itex] and [itex]\mathbb{R}[/itex]!! This is very unintuitive and the map will be very ugly, but it will exist.
     
  6. Oct 15, 2012 #5

    haruspex

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    E.g. map (x, y) to (2x, 2y). Clearly a bijection. No requirement to preserve distance or area, they're only meaningful concepts in a metric space; here we're just treating R2 as a set.
     
  7. Oct 15, 2012 #6

    haruspex

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    Yes, sorry, you're right. They're both aleph 1.
     
  8. Oct 15, 2012 #7

    haruspex

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    Maybe not that ugly. Can map [0,1)x[0,1) to [0,1) bijectively by interlacing digits of the decimal expansions.
     
  9. Oct 15, 2012 #8
    i guess my question is really dumb in a way, the example you gave me, would it be fair to say that given a function in euclidean space, that function has an inverse if each of its components are invertible?
     
  10. Oct 15, 2012 #9

    haruspex

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    Yes, if f and g are invertible R1 to R1 then h(x, y) = (f(x), g(y)) is invertible.
    Wrt invertible between R1 and R2, my simple example based on [0, 1) doesn't help that much. It's much harder to map invertibly between (0, 1)x(0, 1) and (0, 1).
     
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