So from
G(F(A))= C
F(A)= G
F(A)=B
plug that in and get G(B)=C which states that g is surjective, does that work?
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?
Hello,
No it doesn't work,
because you cannot assume F(A)=B
(implying that F too is surjective, which is not necessarily true.)
Since ,
.
Thus,
implies .
If G were not surjective,
there would be an element c of C
which cannot be expressed in the form c=G(b)
(b: element of B).
This c cannot be expressed in the form c=G(F(a))
(a: element of A). (Otherwise, let b=F(a).)
It contradicts the assumption that GF is surjective.
Bye.