Let there are two functions f and g such that

f: A → B and g: B → C

Now gof: A → C 

Given that gof is one-one.  

To prove that f: A → B is one-one, we have to prove that

f(x) = f(y) ⇒ x = y for all x, y belongs to A.

Now let x, y ∈ A such that f(x) = f(y)

Then gof(x) = g(f(x)) = g(f(y))

=> gof(x) = gof(y)

=> x = y                    (since gof(x) is one-one)

Sine gof is one-one, hense it shows that f is one-one.

Again g may or may not be one-one.

So gof is one-one does not imply that both f and g has to be one-one.