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Is this proof ok?

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    If f and g are both onto show g°f is onto.

    f:X→Y and g:Y→Z



    2. Relevant equations



    3. The attempt at a solution

    Since f is onto then there exists an x in X such that f(x) = y, for all y in Y.

    Since g is onto then there exists a y in Y such that g(y) =z for all z in Z.

    Hence, if g°f is not onto then there exists a z in Z such that there is no corresponding x in X. But since g and f are onto this is not possible. Therefore, g°f is onto.
     
  2. jcsd
  3. Mar 18, 2012 #2
    Your basic idea is correct. However, I would be careful with how exactly you write this out. Your first statement, "Since f is onto then there exists an x in X such that f(x) = y, for all y in Y" is not quite correct. The way you have it arranged, it reads as if there is one specific x in X that maps to all values of y in Y.

    Of course this is not what you mean, but if you're starting out with proofs (which I'm assuming you are) then it is good to be precise at first. "For a given y in Y, there exists some x in X such that f(x) = y."

    Hope that helps!
     
  4. Mar 18, 2012 #3
    Ok thanks that makes alot more sense, noted!
     
  5. Mar 18, 2012 #4

    SammyS

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    Just pick an arbitrary z ∊ Z . It's pretty straight forward to show that there is some x ∊ X , such that g(f(x)) = z .
     
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