Proofs of mathematical functions

Assuming that f: A --> B and g: B --> C Prove If g o f is one to one then g may or may not be one to one.

Let g(x) = x^2 and f(x) = sqrt(x). Then let A = all real numbers, B = C = all nonnegative real numbers.

g(f(x)) = (sqrt(x))^2 = x which is one-to-one

Normally this would equal |x| but since C = nonnegative real numbers |x| = x.

g(x) = x^2 is not one-to-one because g(-1) = 1 = g(1) but -1 =/= 1.

Counterexample:

A = {-2,-1,0,1,2}

f(-2) = 0
f(-1) = -1 or 1
f(0) = -2 or 2
f(1) = -3 or 3
f(2) = -4 or 4

B = {-4... 4}

g(x) = |x| - 2

C = A

So g(f(x)) is one-to-one, but g(x) is not.