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Math Help - Surjection

  1. #1
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    Surjection

    Ok so it's been a long time since I've done this

    Let f:A-->B and g:B-->C be functions

    B) Show that if g(f(x)) is surjective, then g is surjective


    This is what I've attempted:


    => g(f(x)):A->C and g(f(x))=C

    g(F(A))=C


    I get stuck here, I can't assume F(A)=B can I?
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  2. #2
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    So from

    G(F(A))= C

    F(A)= G ^{-1}(C)

    F(A)=B

    plug that in and get G(B)=C which states that g is surjective, does that work?
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  3. #3
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    Hello,

    No it doesn't work,
    because you cannot assume F(A)=B
    (implying that F too is surjective, which is not necessarily true.)

    Since F(A)\subset B,
    G(F(A))\subset G(B).
    Thus,
    C=G(F(A))\subset G(B)\subset C
    implies G(B)=C.

    If G were not surjective,
    there would be an element c of C
    which cannot be expressed in the form c=G(b)
    (b: element of B).
    This c cannot be expressed in the form c=G(F(a))
    (a: element of A). (Otherwise, let b=F(a).)
    It contradicts the assumption that GF is surjective.

    Bye.
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  4. #4
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    Blah, I was doing that for extra help and for some reason got stuck on it.
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