Ok so it's been a long time since I've done this

Let f:A-->B and g:B-->C be functions

B) Show that if g(f(x)) is surjective, then g is surjective

This is what I've attempted:

=> g(f(x)):A->C and g(f(x))=C

g(F(A))=C

I get stuck here, I can't assume F(A)=B can I?