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Math Help - proof about composition

  1. #1
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    proof about composition

    Let f : A to B and g : B to C.
    if g o f is surjective then g must be surjective

    how do you prove this
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  2. #2
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    Quote Originally Posted by leinadwerdna View Post
    Let f : A to B and g : B to C.
    if g o f is surjective then g must be surjective

    how do you prove this

    Let c\in C : since g\circ f is surjective there exists a a\in A\,\,s.t.\,\,\,g\circ f(a)=c\Longrightarrow denote b=f(a)\in B\Longrightarrow g(b)=g(f(a))=c\Longrightarrow g is surjective.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Let c\in C : since g\circ f is surjective there exists a a\in A\,\,s.t.\,\,\,g\circ f(a)=c\Longrightarrow denote b=f(a)\in B\Longrightarrow g(b)=g(f(a))=c\Longrightarrow g is surjective.

    Tonio
    can you elaborate a little more on this
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  4. #4
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    Quote Originally Posted by leinadwerdna View Post
    can you elaborate a little more on this


    Since g o f is surjective and is a transformation from A->C, any c in C must have at least one element a in A such that g o f(a) = c. It could have more, but it has at least one.

    Now let b be in B and be the result of f(a), thus f(a)=b. Since g o f(a) = c is surjective, g(b) is by substitution. So for every c in C, there exists b in B such that g(b)=c, thus g is a surjection.

    That was just writing tonio's post into words. Hopefully it has no errors, but the idea is there.
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