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Math Help - proof with functions

  1. #1
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    proof with functions

    So I am pretty sure this statement is false so my proof should be incorrect by I cannot see where the error is.

    The statement is: If g o f is 1 -1 then g is 1 -1

    Let (f(x),g[f(x)]), (f(z),g[f(z)]) be elements of g
    to prove 1 -1 assume g[f(x)] = g[f(z)]
    so we must prove f(x) = f(z) in order to be 1 - 1

    (x, g[f(x)]), (z, g[f(z)]) are elements of g o f
    since g[f(x)] = g[f(z)] and g o f is 1 - 1, x = z

    (x, f(x) ), (z, f(z) ) are elements of f
    since x = z and f is a function, f(x) = f(z)
    therefore g is 1 - 1.

    Can someone help?
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  2. #2
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    Re: Proof with functions

    Quote Originally Posted by stuffthings View Post
    The statement is: If g o f is 1 -1 then g is 1 -1
    If f:A\to B and g:B\to C, you need f to be surjective (onto), otherwise the statement is not true. However I see from your working that you are implicitly assuming f to be surjective. Still it's always better to state everything clearly at the beginning.
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    Re: proof with functions

    I was not assuming it was onto. Thank you for the clarification.
    Last edited by stuffthings; May 18th 2012 at 08:46 AM.
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    Re: proof with functions

    Quote Originally Posted by stuffthings View Post
    [FONT=Verdana,Arial,Helvetica][SIZE=2][FONT=Verdana,Arial,Helvetica][SIZE=2]So I am pretty sure this statement is false so my proof should be incorrect by I cannot see where the error is.
    The statement is: If g o f is 1 -1 then g is 1 -1
    f=\{(a,1),(b,3)\}~\&~g=\{(1,x),(2,x),(3,y)\}. Then g\circ f=\{(a,x),(b,y)\}
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    Re: proof with functions

    would a similar proof work to prove f is 1 -1 when g o f is 1 -1.

    (x, f(x) ), (z, f(z) ) are elements of f
    f(x) = f(z)

    (f(x), g[f(x)] ) (f(z), g[f(z)]) are elements of g
    since f(x) = f(z) and g is a function g[f(x)] = g[f(z)]

    (x, g[f(x)] ), (z, g[f(z)] ) are elements of g o f
    since g[f(x)] = g[f(z)] and g o f is 1 -1, x = z

    therefore f is 1 -1 .
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  6. #6
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    Re: proof with functions

    this is how it works:

    if gof is 1-1, then f is 1-1 (g might not be 1-1).
    if gof is onto, then g is onto (f may not be onto).

    here is how a proof of the two works:

    suppose that f(x) = f(y), and that gof is 1-1.

    then gof(x) = g(f(x)) = g(f(y)) = gof(y) (since f(x) and f(y) are the same point in the image of f).

    since gof is 1-1, x = y, so f is 1-1.

    ***

    now, suppose that we have that gof is onto. let z be any point in the image of g.

    we need to find a y such that g(y) = z.

    because gof is onto, there IS an x in the domain of gof, with gof(x) = z.

    let y = f(x). then g(y) = g(f(x)) = gof(x) = z, as desired.

    ***

    a counter-example, to show that even if gof is 1-1, g might not be 1-1, and even if gof is onto, f might not be onto:

    let g = {(r,x),(s,y),(t,y)} that is:

    g(r) = x
    g(s) = y
    g(t) = y

    then g is not 1-1, because g(s) = g(t), but s ≠ t.

    let f = {(a,r), (b,s)} that is:

    f(a) = r
    f(b) = s.

    note that gof is well-defined, we have:

    gof(a) = g(f(a)) = g(r) = x
    gof(b) = g(f(b)) = g(s) = y

    so gof = {(a,x), (b,y)}, which is 1-1, even though g ISN'T.

    ******

    if X = {x,y}, then the same example shows that gof is onto X:

    we have a such that gof(a) = g(f(a)) = g(r) = x,

    we have b such that gof(b) = g(f(b)) = g(s) = y.

    however, f is NOT onto S = {r,s,t}, because there is no element of dom(f) = {a,b} that f maps to t.
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