Thread: 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?

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,

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

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