one-to-one functions?

Question

Prove the following assertion or give a counterexample. Given any set x and any functions f, g, and h mapping x to x, if h is one-to-one and f º h = g º h then f = g.

Please help I am having a lot of trouble...thanks!

iluxa · Accepted Answer

Disproving.

Idea is this: since h may no be onto, there's no telling how f and g will behave for values which h does not produce.

Counter-example: take 
x = real numbers, 
h(x) = e^x
f(x) = x
g(x) = |x|

f º h = f(h(x)) = f(e^x) = e^x
g º h = g(h(x)) = g(e^x) = |e^x| = (since e^x > 0 for any x) e^x

yet f is not equal to g.

Anonymous · Answer

Well since the question says "prove" that is true it can't be false.  If the statement were false it would have said prove or disprove.

one-to-one functions?

2 Answers