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Math Help - Surjective, Bijective

  1. #1
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    Surjective, Bijective

    Prove or disprove the following statement: if f is an injective function from A to B, and g is a surjective function from B to C, then gof is a bijective function from A to C.

    Really need help. I'm so lost.
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  2. #2
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    Consider A = \left\{ {1,2,3} \right\}\;,\;B = \left\{ {a,b,c,d} \right\}\;\& \;C = \left\{ {x,y} \right\}.
    Define functions: f:\left\{ {(1,a),(2,b),(3,c)} \right\}\;\& \;g:\left\{ {(a,x),(b,x),(c,y),(d,y)} \right\}.
    Use those to think about this problem.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Consider A = \left\{ {1,2,3} \right\}\;,\;B = \left\{ {a,b,c,d} \right\}\;\& \;C = \left\{ {x,y} \right\}.
    Define functions: f:\left\{ {(1,a),(2,b),(3,c)} \right\}\;\& \;g:\left\{ {(a,x),(b,x),(c,y),(d,y)} \right\}.
    Use those to think about this problem.
    but according to this example g0f(1) = x and g0f(2) = x so this violates the definition on a one-one function. so according to this example the function g0f is not 1-1..
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    Quote Originally Posted by doresa View Post
    but according to this example g0f(1) = x and g0f(2) = x so this violates the definition on a one-one function. so according to this example the function g0f is not 1-1..
    That's the point. It said prove or disprove.
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  5. #5
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    Injective and Surjective and Bijective (- oh my!)

    You might find Chapter 3 of Part I of Topology and the Language of Mathematics useful. It provides a number of example of problems like yours, with solutions.

    A free download of the first 50 pages (which includes Chapter 3 of Part I) is available here:
    Bobo Strategy - Topology

    Hope it's useful -
    Chris
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  6. #6
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    its somewhat usefull. but its more on the topology..
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  7. #7
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    Quote Originally Posted by extremely_lost View Post
    Prove or disprove the following statement: if f is an injective function from A to B, and g is a surjective function from B to C, then gof is a bijective function from A to C.

    Really need help. I'm so lost.
    Making the second of your maps the constant map while the first is a non-surjective injection will provide quite a neat solution.

    For instance, \phi: \mathbb{N} \rightarrow \mathbb{N}, x \mapsto x^2. Then, taking the constant map, \theta: \mathbb{N} \rightarrow \mathbb{N}, x \mapsto x.

    Here, the composition of \phi and \theta is just the first map, \phi. Clearly it is injective but not surjective.
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