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Working with the inverses of composite functions

  1. Sep 9, 2008 #1
    Let f:X->Y and g:Y->Z be functions. Suppose A is a subset of Z. I'm wondering whether (g o f)-1(A)=f-1(g-1(A)).

    I'm lost as to where to go with this problem. I know I need to do something using images, but manipulating the inverses of composite functions is proving to be very confusing; I don't know how to properly translate these functions into terms that match up with my definitions, which are as follows:

    1) Let f:X->Y be a function: for a subset A of X, the set f(A)={y in Y: y=f(x) for some x in A}.

    2) For a subset C of Y, the set f-1(C) = {x in X: f(x) is in C}.

    Can you please point me in the right direction? I'm sure that if I could just grasp how to prove this theorem, I'd understand inverses so much better.

    Thank you for your time!
     
  2. jcsd
  3. Sep 10, 2008 #2

    tiny-tim

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    Hi Esran! :smile:

    (have an exists: ∃ and an in/epsilon: ε :smile:)

    Hint: x ε f-1(g-1(A))

    means ∃ y ε g-1(A) with f(y) = x

    means ∃ z ε A with … ? :smile:
     
  4. Sep 10, 2008 #3
    x ε f-1(g-1(A))

    means ∃ y ε g-1(A) with f(y) = x (shouldn't this be f-1(y) = x?)

    means ∃ z ε A with g(y) = z.

    I need to show that somehow, the above implies x ε (g o f)-1(A), and vice versa.

    So, x ε (g o f)-1(A)

    means ∃ z ε A with (g o f)(x) = z or g(f(x)) = z?

    means ∃ y ε f(x) with g(y) = z?

    I'm lost again on how to translate this statement. What do I do when the inverse is on the outside of a composite function?
     
    Last edited: Sep 10, 2008
  5. Sep 10, 2008 #4

    tiny-tim

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    oops! :redface:

    yes … I meant to type: f(x) = y. :smile:
    That's right! :smile:, just stop there

    you have two expressions, (g o f)-1(A) and f-1(g-1(A)), and you want to prove that they're the same.

    So you rewrite each in the form "given x ε …"

    You have:
    i] ∃ z ε A with g(f(x)) = z

    ii] ∃ y ε g-1(A) with y = f(x)
    and ∃ z ε A with g(y) = z

    Don't you see, that once you write it out clearly like that, you have the answer? :wink:
     
  6. Sep 10, 2008 #5
    Alright! I see it now. Thank you. :smile:
     
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