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Inverse composite proof (wording of the proof)

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Let f:A->B and g:B->C be invertible mappings. Show (g o f)^-1 = f^-1 o g^-1.


    2. Relevant equations
    A mapping is invertible iff it is bijective


    3. The attempt at a solution
    I understand why these are equivalent statements; however, I can't figure out the wording of the proof.

    The best I can think of is:

    Suppose f and g are invertible mappings defined f:A->B and g:B->C. Let a, b, and c be elements of sets A, B, and C respectively such that f(a)=b and g(b)=c. Since f and g are bijective, f^-1(b)=a and g^-1(c)=b. So (g o f)^-1(c) = a = f^-1(g^-1(c)).
     
  2. jcsd
  3. Nov 9, 2011 #2

    Office_Shredder

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    You haven't actually proven this to be the case.


    We want to show for all [itex]c\in C[/itex] that [itex](g\circ f)^{-1}(c) = f^{-1}(g^{-1}(c))[/itex]. Let [itex]b\in B[/itex] be the unique element of B such that [itex] g^{-1}(c)=b[/itex] and [itex]a\in A[/itex] the unique element of A such that [itex]f^{-1}(b) = a[/itex]. These both exist because f and g are invertible.

    Clearly [itex]f^{-1}(g^{-1}(c)) = a[/itex] just by how a and b were defined. Since c is arbitrary, it suffices to prove that [itex](g\circ f)^{-1}(c) = a[/itex] as well. How can you do that?
     
  4. Nov 10, 2011 #3
    "Since c is arbitrary, it suffices to prove that (g∘f)−1(c)=a as well. How can you do that?"

    I don't quiet understand your question, or the statement before it.
     
  5. Nov 10, 2011 #4

    Deveno

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    what other statement involving a and c can you use, here?

    where did a and c come from?

    can we get c from a, somehow using f and g? how?

    our assumptions have been as follows:

    g-1:c-->b

    f-1:b-->a

    what is f(a)?
    what is g(b)?
     
  6. Nov 10, 2011 #5

    Office_Shredder

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    To prove that two functions a(x) and b(x) are equal, you can prove that given any possible input x0, a(x0)=b(x0) (this shouldn't be particularly surprising) So to prove that [itex] (g\circ f)^{-1} = f^{-1}\circ g^{-1}[/itex], you can prove that given any [itex]c\in C[/itex], [itex] (g\circ f)^{-1}(c) = f^{-1}(g^{-1}(c))[/itex]. We know that the right hand side of this last equation is a by how we defined b and a in my post... how can you prove that the left hand side is equal to a as well?
     
  7. Nov 11, 2011 #6
    This is the part that confuses me. Should I be saying something to the effect:
    For any a[itex]\in[/itex]A there exists a unique b[itex]\in[/itex]B and a unique c[itex]\in[/itex]C such that f(a)=b and g(b)=c. It follows that g(f(a))=c. Thus (g o f)^(-1)(c)=a
     
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