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Proving a surjective map iff the map of the inverse image is itself

  1. Sep 21, 2012 #1

    lpo

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    In the recommended format :)

    1. The problem statement, all variables and given/known data

    First we say that f:S→T is a map. If Y ⊆ T and we define f-1(Y) to be the largest subset of S which f maps to Y:
    f-1(Y) = {x:x ∈ S and f(x) ∈ Y}

    I must prove that f[f-1(Y)] = Y for every subset Y of T if, and only if, T = f(S).

    2. Relevant equations

    f-1(Y) = {x:x ∈ S and f(x) ∈ Y}

    f[f-1(Y)] = Y for every subset Y of T ⇔ T = f(S)

    3. The attempt at a solution

    So I understand the idea that f must be a surjective map from S to T because Y can be any subset of T so any inverse image of Y would need to map back to S. I also understand that if the map f of the inverse image of Y equals Y then f must be surjective from S to T. I drew the nice bubbles and arrows to help me out also but am having trouble proving this from a series of logical steps.

    (Symbols used: ∀ - for all, ∃ - there is, ∈ - is an element of, ⊆ - is a subset of)

    First I attempt to prove from right to left so:
    Assume f(S) = T meaning f is surjective
    ∀ t ∈ T ∃ at least one x ∈ S such that f(x) = t
    then by definition x ∈ f-1(Y)
    since the map is surjective then f-1(Y) = S and f(f-1(Y)) = f(S) = T = Y
    ∀ f(f-1(Y)) = Y

    Now I attempt from left to right:
    Assume f(f-1(Y)) = Y
    let y ∈ Y so y ∈ f(f-1(y))
    ∃ x ∈ f-1(y) such that f(x) = y ∈ Y ∀ Y ⊆ T
    ∀ y ∈ Y ∃ at least one x ∈ S such that f(x) = y meaning the map is surjective
    f(S) = T

    I don't feel like these are fully correct, there feels like there is some idea or connection I am missing.


    4. Note

    I live about an hour away from school and don't have any study partners to speak of and the professor is very very difficult to understand so any help and guidance would be extremely appreciated. Especially describing your thought process tackling this problem :)

    Thanks!
     
    Last edited: Sep 21, 2012
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