# prove that if g o f is surjective then g is surjective

• Jan 17th 2011, 03:52 PM
mremwo
prove that if g o f is surjective then g is surjective
Let f: A->B and g: B->C be functions
Prove that if g o f is surjective then g is surjective

Thanks!
• Jan 17th 2011, 04:04 PM
Chris11
Just crank it out using the definitions.
• Jan 17th 2011, 04:07 PM
emakarov
This is not a precise analogy, but still. Imagine that f is a post office and g is a letter carrier. The post office sorts letters and gives them to the letter carrier to deliver. Working together, they are able to deliver letters to every address in the neighborhood. Now, could this be the case if the letter carrier did not go to every address in the neighborhood?
• Jan 17th 2011, 05:31 PM
DrSteve
Let $c\in C$. You need $b\in B$ so that $g(b)=c$. Now see if you can use what's given to find $b$.
• Jan 17th 2011, 05:55 PM
Drexel28
Quote:

Originally Posted by mremwo
Let f: A->B and g: B->C be functions
Prove that if g o f is surjective then g is surjective

Thanks!

If $g$ is not surjective then $g(f(A))\subseteq g(B)\subsetneq C$...
• Jan 18th 2011, 09:58 AM
mremwo
I think I have it figured out, but im not sure:

Suppose that gof is surjective.
then for every c in C there exists an a in A such that g(f(a))= c, where f(a) is in B
so there must exist a b in B for every c in C such that g(b)= c. (b=f(a))
therefore g must also be surjective

?
• Jan 18th 2011, 10:21 AM
emakarov
You've got it!

A small detail. I would say it as follows.
Quote:

Suppose that g o f is surjective.
Then for every c in C there exists an a in A such that f(a) is in B and g(f(a)) = c. Therefore, g is surjective.
In your version, if you say "for every c in C" for the second time, you threw away some arbitrary c that you selected the first time, and, therefore, you threw away $a$ that depended on that first c. So, strictly speaking, one can't refer to that $a$ (by saying b=f(a)) when talking about the second arbitrarily chosen c. But I am probably being overly strict here.
• Jan 18th 2011, 10:41 AM
DrSteve
Emakorov, I don't think you're being overly strict. Mremwo, I'm just going to rewrite your argument and clean it up a bit (as emakorov said, you're argument is essentially correct, and would probably get close to full credit, if not full credit on an exam):

Let c be an arbitrary element of C. Since g o f is surjective, there exists an a in A such that (g o f)(a) = c. Let b=f(a). Then b is in B and g(b)=g(f(a))=(g o f)(a)=c. Since c was arbitrary, g is surjective.