1. ## Surjection

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?

2. So from

G(F(A))= C

F(A)= G$\displaystyle ^{-1}(C)$

F(A)=B

plug that in and get G(B)=C which states that g is surjective, does that work?

3. 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 $\displaystyle F(A)\subset B$,
$\displaystyle G(F(A))\subset G(B)$.
Thus,
$\displaystyle C=G(F(A))\subset G(B)\subset C$
implies $\displaystyle G(B)=C$.

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.

4. Blah, I was doing that for extra help and for some reason got stuck on it.