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Suppose that f is an injection. Show that f-1(f(x)) = x for all x in D

  1. Aug 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose that f is an injection. Show that f-1(f(x)) = x for all x in D(f).


    2. Relevant equations



    3. The attempt at a solution

    Let z be in f-1(f(x)).
    Then f(z) is in f(x) by definition of inverse functions.
    Since f is injective, z = x for some x in D(f).
    Thus z is a subset of x, and therefore
    f-1(f(x)) is a subset of x.

    Since f(x) = R(f) by definition, we have that f-1(f(x)) = f-1(R(f)).
    But the range of a function is equivalent to the domain of its inverse, thus f-1(R(f)) = f-1(D(f-1)).
    The range of a function's inverse is the domain of the function.
    Thus f-1(D(f-1)) = D(f).
    Since x is in D(f), we have that x is in f-1(x(x)).
    Therefore x is a subset of f-1(f(x)).

    I fairly confident that the logic of my proof works out but there are a few stylistic errors. If you guys could check over this proof I'd appreciate it :3
     
  2. jcsd
  3. Aug 15, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    f-1 should be f-1, otherwise it looks like the difference between f and 1.

    How did your course define inverse functions?
    f(x) is a single value, not a set.

    That does not need injective functions. As z is in the image of f-1, it has to be in D(f), independent of f.
    x is not a set, and if you let z be an element of something, it is not a (relevant) set either.

    That does not make sense. f(x) is an element of R(f).

    I could continue like that... you are confusing sets and elements of sets in such a way that it is hard to understand what you mean.
     
  4. Aug 15, 2013 #3
    Yes, great question. This will dictate how to proceed as what you are trying to prove may basically be directly derivable from definitions.
     
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