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(a_{1})≠f(a_{2}) where a_{1},a_{2} belong to set A. We also assume a_{1}≠a_{2}. Let f(a_{1})=r and f(a_{2})=s where r and s belong to set B. Then that means r≠s. Hence (g o f)(a_{1})≠(g o f)(a_{2})=g(f(a_{1})≠g(f(a_{2})=

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(a_{1})=f((a_{2}) where a_{1},a_{2} belong to set A. We also assume a_{1}≠a_{2}. Let f(a_{1})=r and f(a_{2})=s where r and s belong to set B. Then that means f(a_{1})=f(a_{2}). Hence (g o f)(a_{1})=(g o f)(a_{2})<=>g(f(a_{1})=g(f(a_{2})<=>

g(r)=g(s) but r≠s. Thus g o f is not one to one.

is it right now?

Re: one to one function proof?

Quote:

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,

$\displaystyle \begin{gathered} f(a) = f(b) \hfill \\ g \circ f(a) = g \circ f(b) \hfill \\ a = b \hfill \\ \end{gathered} $

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.

Re: one to one function proof?

Quote:

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!

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.