# Thread: one to one function proof?

1. ## one to one function proof?

Let f: A-->B, g:B--->C be functions. Prove that if g o f is one to one then f is one to one.

Proof by contrapositive

Assume that f is not one to one then f(a1)≠f(a2) where a1,a2 belong to set A. We also assume a1≠a2. Let f(a1)=r and f(a2)=s where r and s belong to set B. Then that means r≠s. Hence (g o f)(a1)≠(g o f)(a2)=g(f(a1)≠g(f(a2)=
g(r)≠g(s). Thus g o f is not one to one.

Did I do this right? When a function is not one to one do i get to assume that f(x)
≠f(y) where x,y belong to domain and furthermore x≠y?

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Remark
I did this wrong. I forgot that when you negate the one to one definition it becomes A function f: A--->B is not one one when f(x)=f(y) where x,y are elements of A and x
≠y

Edit to proof
Assume that f is not one to one then f(a1)=f((a2) where a1,a2 belong to set A. We also assume a1≠a2. Let f(a1)=r and f(a2)=s where r and s belong to set B. Then that means f(a1)=f(a2). Hence (g o f)(a1)=(g o f)(a2)<=>g(f(a1)=g(f(a2)<=>
g(r)=g(s) but r≠s. Thus g o f is not one to one.

is it right now?

2. ## Re: one to one function proof?

Originally Posted by bonfire09
Let f: A-->B, g:B--->C be functions. Prove that if g o f is one to one then f is one to one.
Why not use the simple straightforward proof,
$\begin{gathered} f(a) = f(b) \hfill \\ g \circ f(a) = g \circ f(b) \hfill \\ a = b \hfill \\ \end{gathered}$

3. ## Re: one to one function proof?

Well the problem wants me to do it as many different ways I can. I already did a direct proof.

4. ## Re: one to one function proof?

Originally Posted by bonfire09
Well the problem wants me to do it as many different ways I can. I already did a direct proof.
Then that your problem not ours!
You should have made that clear in the OP.
Good luck!

5. ## Re: one to one function proof?

wait but i originally wrote to check if my proof was right in my remark. So please check it and give me any feedback.