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Preimage and function composition?

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f:S→T and let A[itex]\subseteq[/itex]T. Define the preimage of A as f-1(A)={x in S: f(x) is in A}.
    Demonstrate that for any such map f and B[itex]\subseteq[/itex]ran(f), f(f-1(B)) = B.

    I am going to use set inclusion to prove this, but can I use function composition in the portion in red? I was going to say an element y is in f(f-1(B)) and then was thinking to apply function composition so as to map an element x back to S then apply f to it and map the new element to B.
    Does this seem right?

    Thanks.
     
  2. jcsd
  3. Sep 18, 2012 #2

    jbunniii

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    It's not exactly function composition, because [itex]f^{-1}[/itex] isn't necessarily a function. This is because, for a given [itex]y[/itex] in the image of [itex]f[/itex], there may be more than one [itex]x[/itex] satisfying [itex]y = f(x)[/itex].

    The way to interpret the expression in red is that [itex]f^{-1}(B)[/itex] is the set of all [itex]x[/itex] such that [itex]f(x) \in B[/itex]. And f(f^{-1}(B)) is the image of that set under [itex]f[/itex].
     
  4. Sep 19, 2012 #3
    For what it's worth, it's good to remember that the notation f-1 is actually being abused here. This is the normal use, but it's good to remember what it means.

    f is a function from S -> T; and at some other time in class we may have occasion to talk about its inverse function f-1 : T -> S.

    However here we're not doing that. f-1 is being defined as a function from the power set of T to the power set of S; that is,

    f-1 : P(T) -> P(S)

    For each subset of T, we define P(T) as a particular subset of S.

    You have to be careful to remember that f goes left to right from points to points, and f-1 goes from right to left from subsets to subsets.
     
  5. Sep 19, 2012 #4
    I see, thank you! I was a little hesitant with using composition, but now I know to just use the definition of preimage and I think that should do the trick.
     
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