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Inverse images

  1. Jan 27, 2012 #1
    Hey all,

    How is it that [itex] f(f^{-1}(E))\subset E[/itex] for some induced metric [itex]E\subset Y[/itex], can be a proper subset of E, rather than E itself?

    Thanks!
     
  2. jcsd
  3. Jan 27, 2012 #2
    The function f might not be surjective. If it were not, there would be an x in E such that f-1({x})=∅. Then f(f-1({x}))=∅, which is a proper subset of {x}.
     
  4. Jan 27, 2012 #3
    Thank you.
     
  5. Jan 27, 2012 #4
    One more question, say [itex]E\subset X[/itex] s.t. [itex]f^{-1}(f(E))\supset E[/itex]. How can this occur? Is this the case when a function is non-bijective, such as where f(a)=f(b) for [itex]a \neq b,a \in E, b\notin E [/itex]
     
    Last edited: Jan 27, 2012
  6. Jan 30, 2012 #5

    HallsofIvy

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    Let's take a specific example: [itex]f(x)= x^2[/itex] which is neither surjective ("onto" the real numbers) nor injective ("one to one").
    Let E= [-1, 4]. Then [itex]f^{-1}(E)= [-2, 2][/itex] and [itex]f(f^{-1}(E))= [0, 4][/itex] which is a proper subset of E.

    Let E= [0, 2]. Then [itex]f(E)= [0, 4][/itex] and [itex]f^{-1}(f(E))= [-2, 2][/itex] which contains E as a proper subsert.
     
  7. Feb 11, 2012 #6
    Thank you.
     
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