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Functions real analysis

  1. Sep 19, 2011 #1
    Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g). Prove that a g = f^-1

    Pf/

    How would you go about starting this besides saying

    Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g).

    isn't obvious that the functions would have to be the inverse of each other, How else could you get the identity? So how do you prove it, can I just say clearly it is.
     
    Last edited: Sep 19, 2011
  2. jcsd
  3. Sep 19, 2011 #2

    benorin

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    Homework Helper

    What does g = f^-1 mean? My calc text gives this definition:

    g = f^-1 if, and only if, f(x) is a one-to-one function with domain D(f) and range D(g) such that

    g(y)=x <=> f(x)=y for all y in D(g).

    Start with the given and prove this definition must be true.
     
  4. Sep 19, 2011 #3
    g= f^-1 means that g is the inverse of f
     
  5. Sep 19, 2011 #4
    How do you prove a definition true. If its a definition shouldn't it always be true?
     
  6. Sep 20, 2011 #5
    This is what I have been able to get to

    (g[itex]\circ[/itex]f)(x)=x for all x ϵD(f) and (f[itex]\circ[/itex]g)(y)=y for all y ϵD(g).


    So choose
    g(y)=x and f(x)=y

    so

    x=g(y)=f^-1(y)

    ans since f and g are bijections we can conclude g=f^-1


    does that work
     
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