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

Prove that if g o f is surjective then g is surjective

Thanks!

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- January 17th 2011, 03:52 PMmremwoprove 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! - January 17th 2011, 04:04 PMChris11
Just crank it out using the definitions.

- January 17th 2011, 04:07 PMemakarov
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?

- January 17th 2011, 05:31 PMDrSteve
Let . You need so that . Now see if you can use what's given to find .

- January 17th 2011, 05:55 PMDrexel28
- January 18th 2011, 09:58 AMmremwo
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

? - January 18th 2011, 10:21 AMemakarov
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.

- January 18th 2011, 10:41 AMDrSteve
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.