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My sketchy proof on functions and mappings. I need help. ( Latex Fixed!)

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] f : X \rightarrow Y[/tex] , [tex]g : Y \rightarrow Z[/tex] and [tex]B \subset Z[/tex]
    Prove that
    [tex]\left(g \circ f\right)^{-1}\left(B \right) = f^{-1} \left({g^{-1} \left(B\right)\right)[/tex].

    What is wrong with this proof ?

    3. The attempt at a solution

    [tex] x_{0} \in \left(g \circ f\right)^{-1}\left(B\right) \Rightarrow g \left(f\left(x_{0} \right)\right)\in B \Rightarrow f \left(x_{0}\right) \in g^{-1} \left(B\right)[/tex]
    [tex]f \left(x_{0} \right) \in g^{-1} \left(B\right) \Rightarrow x_{0} \in f^{-1} \left(g^{-1} \left(B \right) \right)[/tex]

    [tex]\left(g \circ f\right)^{-1}\left(B \right) \subset f^{-1} \left({g^{-1} \left(B \right)\right)[/tex].

    I feel uneasy about the proof. I believe my inferences are correct but there is something unsettling about what I did.

    Part 2.

    Suppose [tex]x_{0} \in f^{-1}\left(g^{-1}\left(B\right)\right)[/tex]

    [tex]x_{0}\in f^{-1}\left(g^{-1} \left(B\right)\right) \Rightarrow f \left(x_{0}\right) \in g^{-1}\left(B\right) \Rightarrow g\left(f\left(x_{0}\right)\right) \in B[/tex]

    [tex]g \left(f \left(x_{0} \right) \right) \in B \Rightarrow x_{0} \in \left(g \circ f\right)^{-1} \left(B \right)[/tex]

    [tex]f^{-1} \left({g^{-1} \left(B\right)\right) \subset \left(g \circ f\right)^{-1}\left(B\right)[/tex].


    [tex]\left(g\circ f\right)^{-1}\left(B\right) = f^{-1}\left({g^{-1}\left(B\right)\right)[/tex].

    Some feedback would be appreciated. I would like to know if this proof is valid and good enough.
    Last edited: Sep 15, 2010
  2. jcsd
  3. Sep 15, 2010 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    You need to put a space between latex commands and the next character.

    For example \circg will not show up. It needs to be \circ g

    Based on your title maybe you made this change? At any rate it's not coming through on my computer
  4. Sep 15, 2010 #3


    Staff: Mentor

    Fixed some more of your LaTeX. The codomains of the functions weren't showing, plus a few other problems.
  5. Sep 15, 2010 #4
    The funny thing is that I fixed it all and checked to see if it was okay. When I logged back to read the replies it was unchanged.

    The forum always acts weird when I use my phone to post threads. I think the mobile version of the forum has some bugs.

    Thanks Mark.

    Any feedback ?
  6. Sep 15, 2010 #5
    Does anyone have any suggestions are comments ?

    I would like to know if this proof supplied here is valid . I would like to know if I have to make adjustments or rewrite the entire thing.
  7. Sep 16, 2010 #6
    Yes, your proof is correct. It's not dodgy.

    I think that the reason you think it may possibly be dodgy (correct me if I'm wrong) is because you keep using the fact that [tex]x\in h^{-1}(A)[/tex] is equivalent to [tex]h(x)\in A[/tex].

    The notation [tex]h^{-1}[/tex] when applied to a set (instead of an element) is not the inverse function, it's the preimage, and so that statement is actually true by definition
  8. Sep 16, 2010 #7
    Great. You are correct about my uneasiness.

    I thought I was making some claims that may put me into trouble.
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