# Set Theory Proof

anonymity
Prove that if g$\circ$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$\in$Z. Since g $\circ$ f is surjective, there exists x$\in$X such that g$\circ$ f(x) = z. Equivalently, for f(x) = y, we have that g(y) = z. Now for any z$\in$Z there exists y = f(x) such that z = g(y), and that g is surjective.

mtayab1994
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.

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Homework Helper
Prove that if g$\circ$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$\in$Z. Since g $\circ$ f is surjective, there exists x$\in$X such that g$\circ$ f(x) = z. Equivalently, for f(x) = y, we have that g(y) = z. Now for any z$\in$Z there exists y = f(x) such that z = g(y), and that g is surjective.
Yes, this is a perfectly valid proof. Although, I would say "let y= f(x)" rather than "for f(x)= y".

mtayab1994
Yes, this is a perfectly valid proof. Although, I would say "let y= f(x)" rather than "for f(x)= y".

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?