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Show that if f: A → B is injective and E is a subset of A, then f −1(f

  1. Aug 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that if f: A → B is injective and E is a subset of A, then f −1(f(E) = E

    2. Relevant equations



    3. The attempt at a solution

    Let x be in E.
    This implies that f(x) is in f(E).
    Since f is injective, it has an inverse.
    Applying the inverse function we see that x is in f-1(f(E)).
    Since x in a implies x is in f-1(f(E)),
    E is a subset of f-1(f(E)).

    Conversely, let x be in f-1(f(E)).
    Then f(x) is in f(E).
    Since f is again injective it has an inverse; applying the inverse shows that
    x is in E.
    Therefore, since x in f-1(f(E)) implies that x is in E,
    f-1(f(E)) is a subset of E.

    Therefore, since f-1(f(E)) is a subset of E and vice versa, the two sets are equal.

    QED :3

    I'm pretty sure this is right, but there might be a mistake in the 2nd part. Thanks guys :3
     
  2. jcsd
  3. Aug 11, 2013 #2
    You seem to have misunderstood inverses and their existence a bit. Let us fix a function [itex]f : A \to B[/itex]. Let a function [itex]g : B\to A [/itex] be given, then we say that g is the inverse of f if
    [tex]f(g(b)) = b \qquad g(f(a)) = a[/tex]
    for all [itex]a \in A[/itex] and [itex]b \in B[/itex]. Sometimes only one of these conditions is satisfied in which case we call g a right inverse or a left inverse. In particular if
    [tex]g(f(a)) = a[/tex]
    for all [itex] a\in A[/itex], then we say that g is a left inverse of f.

    Now in your case f is injective so you conclude that f must have an inverse, but this is not true in general. Consider for instance the case
    [tex]A = \{0\} \qquad \textrm{and}\qquad B = \{0,1\}[/tex]
    [tex]f : A \to B[/tex]
    [tex]f(0) = 0[/tex]
    Then f is injective, but you cannot construct an inverse. If you don't see why you cannot construct an inverse, then try. There are only a few options and you should quickly see what condition fails if you just experiment a bit.

    What however is true is that if f is injective, then f has a left inverse g. This statement is not trivial so you can't use it unless you have a reference for it in your book. I would advice you to try something else as this is not necessary and would overcomplicate the problem even if your book has such a result.

    Instead recall that for [itex]x \in A[/itex] and F a subset of B we have that [itex]x \in f^{-1}(F)[/itex] by definition is equivalent to [itex]f(x) \in F[/itex]. Try to apply this equivalence to your statement
    [tex]x \in f^{-1}(f(E))[/tex]
    and see if the resulting equivalence could replace the use of inverses in the first part of your argument (this part does not need injectivity).

    For the second part you need a little more to replace your use of inverse functions, but not a lot. If you have trouble you can post here again and I will be happy to provide a bit more assistance.
     
  4. Aug 12, 2013 #3
    O dear I see my mistake; I confused the definitions of injective and bijective :/

    Then for the part of the proof you mentioned,

    x is in f-1(f(E)).
    Thus f(x) is in f(E).

    Then it seems obvious that x is in E at this point, but I'm not sure if it's trivial or not.
    So if this step works we have f-1(f(E)) a subset of E.

    I went over the definitions in the chapter several times put I can't come up with anything clever :( Thanks for bearing with me, I'm a complete noob at proofs :3
     
  5. Aug 12, 2013 #4
    Intuitively I can tell that it's not obvious and that step isn't allowed, but in that case I've run into a break wall.
     
  6. Aug 13, 2013 #5
    Well it should only seem obvious if you remember that f is injective. This should suggest that you need injectivity in some way. Can you state what [itex]f(x) \in f(E)[/itex] actually means in terms of elements? Do you see any way to apply injectivity to this expression?
     
  7. Aug 13, 2013 #6
    O of course, since f(x) is in of f(E), then x must be an element of E because injectivity implies that each value of f(x1) corresponds to a unique element of A.

    I think this works, but at any rate I'm still stuck on the other half of the proof..

    if I try to suppose x is in E I get back to the same argument that f(x) is in f(E). Gosh, the earlier proofs were so much easier but this one's making me feel badly about myself
     
  8. Aug 13, 2013 #7
    Just do it one step at a time.
    [tex]f(x) \in f(E)[/tex]
    means that
    [tex]f(x) =f(e)[/tex]
    for some [itex]e \in E[/itex]. You can now apply injectivity to get x = e, from which you can conclude [itex]x \in E[/itex].


    So the problematic part was:
    "This implies that f(x) is in f(E).
    Since f is injective, it has an inverse.
    Applying the inverse function we see that x is in f-1(f(E)). "
    Without applying the inverse you want to conclude that f(x) is in f(E) implies x is in [itex]f^{-1}(f(E))[/itex], but these are equivalent by the definition of [itex]f^{-1}[/itex]. If you just look up what the second statement means in the definitions then you get the first statement. Therefore in particular you get the implication you were looking for.
     
  9. Aug 14, 2013 #8
    Simpler than I expected! Thanks , I understand now :3
     
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