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Set Theory Proof

  1. Nov 2, 2011 #1
    Prove that if g[itex]\circ[/itex]f is surjective, then g must be surjective.

    I know that one valid proof of this statement is acquired via the contrapositive, what I am not sure of is if the following proof is flawed (if it is, please say why):

    Suppose z[itex]\in[/itex]Z. Since g [itex]\circ[/itex] f is surjective, there exists x[itex]\in[/itex]X such that g[itex]\circ[/itex] f(x) = z. Equivalently, for f(x) = y, we have that g(y) = z. Now for any z[itex]\in[/itex]Z there exists y = f(x) such that z = g(y), and that g is surjective.
     
  2. jcsd
  3. Nov 2, 2011 #2
    I prefer using sets to prove it.

    If gof is surjective : f:A to B and g: B to C so therefore;

    g(f(A)) = C. And we also know that f(A)⊆B, now try to complete the proof using what u know in sets.
     
    Last edited: Nov 2, 2011
  4. Nov 2, 2011 #3

    HallsofIvy

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    Yes, this is a perfectly valid proof. Although, I would say "let y= f(x)" rather than "for f(x)= y".
     
  5. Nov 2, 2011 #4
    I believe this is easier proof if he's familiar to sets.
    g(f(A)) = C.
    We also know: f(A)⊆B, which means: g(f(A))⊆g(B). g(B)⊆C, by definition.

    All together:
    C = g(f(A)) ⊆ g(B) ⊆ C which implies: C = g(B)

    That's correct i believe right?
     
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