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Math Help - functions

  1. #1
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    functions

    Prove or disprove
    f:A-->B
    g:C-->A
    h:C-->A
    if f o g = f o h, then g=h,

    note: o denotes the composite function.
    No idea on how to do this question.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by hugo84 View Post
    Prove or disprove
    f:A-->B
    g:C-->A
    h:C-->A
    if f o g = f o h, then g=h,

    note: o denotes the composite function.
    No idea on how to do this question.
    A function f is defined such that an element in the domain of f gets mapped to exactly one element in the codomain of f.

    In other words, f(a) = f(b) if and only if a = b.

    See how that applies?
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  3. #3
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    Quote Originally Posted by undefined View Post
    A function f is defined such that an element in the domain of f gets mapped to exactly one element in the codomain of f.
    In other words, f(a) = f(b) if and only if a = b.
    The statement in red is not true.
    Consider this example: A=\{a,b\},~B=\{1\},~\&~C\{x,y\}
    and f=\{(a,1), (b,1)\},~g=\{(x,a),(y,b)\},~\&~h=\{(x,b),(y,a)\}
    Is it clear that f \circ g = f \circ h?
    But is it true that g=h?
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Plato View Post
    The statement in red is not true.
    Consider this example: A=\{a,b\},~B=\{1\},~\&~C\{x,y\}
    and f=\{(a,1), (b,1)\},~g=\{(x,a),(y,b)\},~\&~h=\{(x,b),(y,a)\}
    Is it clear that f \circ g = f \circ h?
    But is it true that g=h?
    Sorry, the "if" is true but not the "only if." Dumb mistake on my part. Thanks for correcting me.

    Edit: Another very simple counterexample is:

    f(x) = x^2
    g(x) = x
    h(x) = -x
    Last edited by undefined; April 21st 2010 at 08:09 AM.
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