Suppose f:A->B and g:B->C. Prove that if gof is one-to-one, then f is one-to-one.
I have no idea where to start. Any suggestions or comments that would help me are appreciated!
Assume the contrary and see what happens. If f is not one-to-one, there exist such that and . Then, while , which violates the fact that g(f( )) is one-to-one. Therefore it's impossible for f to not be one-to-one.
If you need to prove IF-THEN statements, always try assuming that it's not necessarily the case, and see if it leads to a contradiction. If it does, you have found a proof.
Thanks a lot! I tried to do it by a different approach. Here's what I came up with:
PF - - -
Suppose that f(x1)=f(x2) in A. Then
g(f(x1)) = g(f(x2))
g o f(x1) = g o f(x2)
x1=x2 since g o f is one-to-one.
Because f(x1) = f(x2), then x1 = x2. Therefore we can conclude that f is one-to-one.
Is this a logical proof?