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Preimage proof?

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Define f:S→T, where B[itex]\subseteq[/itex]T. Let f-1(B)={x[itex]\in[/itex]S:f(x)[itex]\in[/itex]B} be the preimage of B.

    Demonstrate that for any such map f, f(f-1(B))=B.

    My main question is, would I prove this using set inclusion both ways?

    I was going to begin by letting an element be in the preimage of B, and explain what that means, then mapping that element to B. Would this be correct?

    I just need a push in the right direction.

    Thank you.
     
  2. jcsd
  3. Sep 16, 2012 #2

    SammyS

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    Generally, yes. To show set equality you need to show inclusion in both directions.
     
  4. Sep 16, 2012 #3

    micromass

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    Just to let you know, but that equality is false in general. You need extra assumptions on f.
     
  5. Sep 16, 2012 #4
    SammyS- Thank you

    Micromass- Yes, the function needs to be surjective, right? I'm not sure why there were no assumptions...
     
  6. Sep 16, 2012 #5

    SammyS

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    I'm very reluctant to question you, micromass.

    But isn't the extra assumption of f needed for [itex]f^{-1}\left(f(A)\right)=A[/itex] rather than for [itex]f\left(f^{-1}(B)\right)=B\ ?[/itex]
     
  7. Sep 16, 2012 #6

    micromass

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    An extra assumption is needed for both. Consider [itex]f(x)=x^2[/itex] and B=[-1,0]. Then [itex]f^{-1}(B)=\{0\}[/itex] and thus [itex]f(f^{-1}(B))=\{0\}[/itex]. The problem is that f is not surjective.
     
  8. Sep 16, 2012 #7

    SammyS

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    Well, my reluctance was well founded!

    Thanks as always.
     
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