Proving a function to be one-to-one given 2 functions are one-to-one

Question

Suppose : → and : → . Prove or disprove your answer.

1. If f is one-to-one and ∘ is one-to-one, can you conclude that g is also one-to-one?

   Suppose ( ∘ ) is one-to-one. Show that g: A→ B is one-to-one
   
   Assume g(x) = g(y) for some x,y ∈ A and show that x = y
   
   f [g(x)] = f [g(y)]
   ( ∘ )(x) =  ( ∘ )(y)
   
   Since  ( ∘ ) is one-to-one, this will give a result of x = y. Therefore, this also shows that g is one-to-one.
   

2. If you know that g is one-to-one and ∘ is one-to-one, can you conclude that f is also one-to-one?

For the first question, how do I go about making use of f? I did not use the given statement "f is one-to-one" so I'd like to know how I can make use of f in my proof.

As for the second question, is the solution similar to the one to Q1?

Answer 1

Hint for question 1: let $g(x)=x^2$, where $A=\Bbb N$ and $B=\Bbb Z$. The rest should be easy to figure out.

Hint for question 2: If $f$ is NOT one-to-one, what does that mean? What does that imply for $f\circ g$?