Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Composition of Inverse Functions
  1. Composite function (Replies: 4)