Results 1 to 4 of 4

Math Help - Inverse and composition functions

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    26

    Inverse and composition functions

    I have this problem from my discrete math comp sci class about inverse functions

    show that for the functions f, g and h

    (fog^-1oh)^-1 = h^-1 o g o f^-1

    All i have so far is that

    h^-1 o g o f^-1 = h^-1 o g o f^-1

    because I have in my notes that (gof)^-1 = f^-1 o g^-1

    If that right? And if so is there any other way I can show more work?
    Last edited by guyonfire89; September 21st 2010 at 07:43 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,561
    Thanks
    785
    Hint: to produce f^{-1} and f\circ g, write f^{-1} and f\circ g.

    All i have so far is that

    h^-1 o g o f^-1 = h^-1 o g o f^-1
    This is hardly surprising since equality is reflexive

    If you know that (f\circ g)^{-1}=g^{-1}\circ f^{-1} and (f^{-1})^{-1}=f (e.g., when all functions are bijections), then you can write (f\circ g^{-1}\circ h)^{-1}, parenthesize it fully (one of the \circ has to be done first; does it matter which?) and then apply the equalities above to gradually rewrite this into h^{-1}\circ g\circ f^{-1} (I believe there are three steps).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by guyonfire89 View Post
    I have this problem from my discrete math comp sci class about inverse functions

    show that for the functions f, g and h

    (fog^-1oh)^-1 = h^-1 o g o f^-1

    All i have so far is that

    h^-1 o g o f^-1 = h^-1 o g o f^-1

    because I have in my notes that (gof)^-1 = f^-1 o g^-1

    If that right? And if so is there any other way I can show more work?
    You are not asked to derive it or anything hard, you are just asked to `show' it. Therefore, just multiply them together. If you get the constant map then clearly it is your inverse. You do, so it is.

    (f \circ g^{-1} \circ h) \circ( h^{-1} \circ g \circ f^{-1})=\ldots

    (Also, in LaTeX (`math mode') write f^{-1} to make sure that the whole -1 is put up there. Use \circ instead of `o' for composition too.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,441
    Thanks
    1862
    "inverses" are [b]defined[\b] by "f(x)= y, if and only if f^{-1}(y)= x".

    Suppose that fg^{-1}h(x)= f(g^{-1}h)(x)= y. Then f^{-1}(y)= g^{-1}(h(x)) where I have taken "x" in the definition to by g^{-1}h(x). from g^{-1}(h(x))= f^{-1}(y), now with "x" on the right side of the definition taken as f^{-1}(y) and "y" as h(x), we have g(f^{-1}(y))= h(x). Finally, taking g(f^{-1}(y)) as "y", x= h^{-1}(g(f^{-1}(y))= (h^{-1}gf^{-1})(y)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Composition of Two Functions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 14th 2011, 12:26 PM
  2. Composition of an Inverse Hyperbolic Function
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: August 22nd 2010, 04:26 AM
  3. Inverse & Composition Function Problem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: February 3rd 2010, 12:55 PM
  4. Composition & Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 13th 2008, 01:44 PM
  5. Inverse Composition of Functions Proof
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: September 17th 2007, 06:26 AM

Search Tags


/mathhelpforum @mathhelpforum