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Math Help - Transitive Surjection proof

  1. #1
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    Nov 2010
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    Transitive Surjection proof

    Hey all, can't figure out how to lay out the proof for the following problem:

    If f: A -> B is surjective and g: B -> C is surjective then g o f : A -> C is surjective.

    well by definition, for all b ∈ B there exists a ∈ A such that f(a) = b
    and, for all c ∈ C there exists b ∈ B such that g(b) = c

    Any help would be appreciated!
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  2. #2
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    203

    Re: Transitive Surjection proof

    well you already have started the proof. first identify the goal precisely.
    The goal is to prove that g\circ f:A\to C is surjective. i.e. For all
    c\in C there exists a\in A such that (a,c)\in g\circ f.
    So first let c\in C to be arbitrary. But as you pointed out , there exists
    b\in B such that g(b)=c , which means that
    (b,c)\in g. Again as you said, for all b'\in B there exists
    a\in A such that f(a)=b'. Since b\in B, we have
    f(a)=b, which means that (a,b)\in f.

    So what can you conclude from this ?
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