Thread: 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!

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

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

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.

consider .. thus, domain is nonnegative reals.
suppose that ..
therefore where is nonnegative reals (since the domain takes the domain of for which is defined), which makes to be 1-1..

however, is not 1-1..

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?

yup. the answer is no.