1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor

    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:



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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook