Thread: proofs with functions 1-1 and onto

1. proofs with functions 1-1 and onto

The problem is:
Suppose that f:A-->B, g:B-->C are functions. In the following cases, answer yes or no. If your answer is yes, prove it. If it is no, give a counterexample and say what additional hypotheses are needed to make the statement true, then prove.
1. if gof is 1-1, must f be 1-1?
2. if gof is 1-1, must g be 1-1?
3. if gof is onto, must f be onto?
4. if gof is onto, must g be onto?

I'm fine with 1 and 4, and for 2, I think the answer is yes, but I'm not sure how to go about proving it. I understand it, and I could explain it in English (if g isn't 1-1, then two elements in B go to the same element in C, and the two elements in A go to the two elements in B, which go to the same element in C) but I'm not sure how to say it in a proof. For 3, I'm just confused. The element that is not in the range of f could still be in the domain of g, so I think I have to say no, and add something to it, but I'm not sure what.

Thank you!

2. I'm not sure how to delete this post, but I figured it out.

3. you dont have to delete it..
anyways, you think that 2 is false? Ü

4. No, I thought that at first, but I drew lots of little pictures until I figured out that g doesn't need to be 1-1. My answers ended up being yes, no, no, yes.

5. consider $\displaystyle f(x) = \sqrt{x}$.. thus, domain is nonnegative reals.
suppose that $\displaystyle g(x) = x^2$..
therefore $\displaystyle (g o f)(x) = (\sqrt{x})^2 = |x| = x$ where $\displaystyle x$ is nonnegative reals (since the domain takes the domain of $\displaystyle f$ for which $\displaystyle g$ is defined), which makes $\displaystyle g o f$ to be 1-1..

however, $\displaystyle g(x) = x^2$ is not 1-1..

6. Yeah, so for number 2, g doesn't need to be 1-1 for gof to be 1-1, so the answer is no, right?

7. yup. the answer is no.