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Math Help - urgent homework-proving surjectivity from conditions

  1. #1
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    urgent homework-proving surjectivity from conditions

    I am arguing from the given supposition:
    if g(f(x))=h(f(x)) then g(y)=h(y)
    (f maps from A to B, g and h map from B to C)

    to the conclusion that f is surjective

    I'm trying to argue to a contradiction by supposing that f isn't surjective...do you think that will work? Any suggestions?
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  2. #2
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    Quote Originally Posted by blackmustachio View Post
    I am arguing from the given supposition:
    if g(f(x))=h(f(x)) then g(y)=h(y)
    (f maps from A to B, g and h map from B to C)

    to the conclusion that f is surjective

    I'm trying to argue to a contradiction by supposing that f isn't surjective...do you think that will work? Any suggestions?
    f: A\mapsto B
    g: B\mapsto C
    h: B\mapsto C

    Say f(x) = b for all x\in A and specific b\in B. I constructed a function from A to B which collapses all its elements into a single element. Now f is surly not surjective (unless B has a single element, so say B has more than just b. Why?) And let g,h be the exact same function from B to C. So we have g(f(x)) = h(f(x)) for all x\in A, and we also have g(y) = h(y) for all y\in B by construction. But f is not surjective. Meaning what you are trying to prove is completely wrong.
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  3. #3
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    ddddddddd

    Let C = {0,1}.

    Assume f is not surjective (toward contradiction). Thus there is some b in B such that b is not in the range of f.

    Define g:B to C by g(x) = 0.

    Define h:B to C by h(b)= 1, and h(x) = 0 whenever x is not b.

    g(f(x)) = h(f(x)) = 0 for all x in A.

    but g(b) = 0 and h(b) =1, thus h is not g.
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