Composition of functions that are onto or one-to-one I found part of my answer here: If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that...); however I wanted to flesh out the last two statements I had in a proposition in my notes.


  Proposition: 
  Let $f: A \rightarrow B$ and $g: B \rightarrow C$.  Then:
  
  (i) If $g \circ f$ is one-to-one, then $f$ is one-to-one.
  
  (ii) If $g \circ f$ is onto, then $g$ is onto.


Proof:
(i)  Suppose $f(x)=f(y)$ for some $x,y$.

Since $g \circ f$ is one-to-one: $$g\circ f(x) = g\circ f(y) \Rightarrow x=y,\forall x,y \in A.$$ 

Therefore $f$ must be one-to-one.

(ii)  Since $g \circ f (x)$ is onto, then for every $c \in C$ there exists an $a \in A$ such that $c=g(f(a))$.  Then there exists a $b \in B$ with $b=f(a)$ such that $g(b)=c$.  Thus g is onto.

I wanted to confirm that these proofs are both correct for my peace of mind (as they weren't proven in class).
Both of your proofs are correct.
On the first one i would say since g composed of f is one-to-one, if g(f(x))=g(f(y)) for all x,y in A then x=y. By definition of f then there exist some value f(x) and f(y) in B. Then because we already noted x=y. F(x) must equal F(y) by the definition of a function.