1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook