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Composition of Inverse Functions

  1. Sep 11, 2008 #1
    In Micheal C. Gemignani, "Elementary Topology" in section 1.1 there is the following exercise

    If [tex]f:S \rightarrow T[/tex] and [tex]G: T \rightarrow W [/tex], then [tex](g \circ f)^{-1}(A) = f^{-1}(g^{-1}(A))[/tex] for any [tex]A \subset W [/tex].

    I think the above is only true if A is in the image of g yet the book says to prove the above. I have what I believe is a counter example. Any comments? I will give people two days to prove the above or post a counter example. After this time I'll post my counter example for further comment.
  2. jcsd
  3. Sep 12, 2008 #2


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    How is [tex] \mathbf{W} [/tex] used here - does [tex] g [/tex] map to all of [tex] \mathbf{W} [/tex] or only into [tex] \mathbf{W} [/tex]? That could explain the possible confusion.
  4. Sep 12, 2008 #3
    The way the problem has been written you can only prove that:

    .........f^-1[g^-1(A)] IS a subset of (g*f)^-1(A) i.e the right hand side of the above is a subset of the left hand side

    FOR the above to be equal we must have that: f(S)={ f(x): xεS } MUST be a subset of

  5. Sep 14, 2008 #4
    xε[tex]f^{-1}(g^{-1}(A))[/tex]<====> xεS & f(x)ε[tex]g^{-1}(A)[/tex]====> xεS & g(f(x))εA <====> xε[tex](g\circ f)^{-1}(A)[/tex]

    since [tex]g^{-1}(A)[/tex] = { y: yεT & g(y)εΑ}

    ΙΝ the above proof all arrows are double excep one which is single and for that arrow to become double we must have :

    .........................................f(S)[tex]\subseteq g^{-1}(A)[/tex].....................................

    and then we will have ;

    [tex](g\circ f)^{-1}(A) = f^{-1}(g^{-1}(A))[/tex]
  6. Sep 14, 2008 #5
    Let [itex]x \in (g \circ f)^{-1}(A)[/itex]. Then [itex]x \in S[/itex] with [itex]g(f(x)) \in A[/itex]. This means [itex]f(x) \in g^{-1}(A)[/itex] and thus [itex]x \in f^{-1}(g^{-1}(A))[/itex].

    The other direction has been shown.

    How are those not all double arrows, evagelos? If [itex]g(f(x)) \in A[/itex], then certainly [itex]f(x) \in g^{-1}(A)[/itex] by definition. We already know that [itex]f(x) \in T[/itex].

    I'm curious as to what this supposed counter-example is.
  7. Sep 14, 2008 #6
    Your proof looks correct. There appears to be a mistake in my counter example. I'll spend a few futile minutes anyway trying to think up a counterexample anyway.
  8. Sep 15, 2008 #7
    What definition,write down please
  9. Sep 15, 2008 #8


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    g-1(A) is defined as the set of all x such that g(x) is in A. If g(f(x)) is in A, then, by that definition, f(x) is in g-1(A).
  10. Sep 15, 2008 #9
    write a proof where you justify each of your steps ,if you wish.

    The above proof is not very clear
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