- #1

anonymity

- 163

- 0

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.