Proof of properties of injective and surjective functions. I'd like to see if these proofs are correct/have them critiqued.

Let $g: A \to B$ and $f: B \to C$ be functions.  Then:
  (a) If $g$ and $f$ are one-to-one, then $f \circ g$ is one-to-one.
  (b) If $g$ and $f$ are onto, then $f \circ g$ is onto.
  (c) If $f \circ g$ is one-to-one, then $f$ is one-to-one?
  (d) If $f \circ g$ is one-to-one, then $g$ is one-to-one?
  (e) If $f \circ g$ is onto, then $f$ is onto?
  (f) If $f \circ g$ is onto, then $g$ is onto?  

(a)  Let $a,b \in A$.  If $(f \circ g)(a)=(f \circ g)(b)$, then $f(g(a))=f(g(b))$.  Since $f$ is one-to-one, we know that $g(a)=g(b)$.  And, since $g$ is one-to-one is must be that $a=b$.  Hence $f \circ g$ is one-to-one.  
(b) Since $f$ is surjective we know that for all $c \in C$ there is a $b\in B$ such that $f(b)=c$.  Since $g$ is surjective, there is a $a \in A$ such that $g(a)=b$.  Hence, for all $c \in C$ there is an $a \in A$ such that $(f \circ g)(a)=f(g(a))=f(b)=c$.  
(c)  This is false.  Let $A=\{1\}$, $B=\{1,2\}$, and $C=\{1\}$. Define $g(1)=1$ and $f(1)=f(2)=1$. Then $f \circ g$ is injective, but $f$ is not injective.  
(d)  This is true.  Let $a,b \in A$ where $a \neq b$.  If $(f\circ g)(a)\neq(f\circ g)(b)$, then $f(g(a))\neq f(g(b))$.  Hence $g(a) \neq g(b)$.
(e)  Since $f \circ g$ is surjective, then for every $ c \in C$ there is an $a \in A$ such that $f(g(a))=c$.   Let $b = g(a)$.  Then $b \in B$ and $f(b)=g(f(a))=c$.  Thus $f(b)=c$.  Hence $f$ is surjective.
(f)  Using the same set up as part (c), we have that $g$ is not surjective since there is nothing in $A$ such that $g$ will go to 1.

For a), By the definition of an injection take $a\neq b$, then $f(g(a))\neq f(g(b))$.
But you started assuming they are equal which is confusing.
Also, when you start by saying since f is one-to-one, that means $g(a) = f(a)$, because $g(a)$ comes in as input to the function $f$. I.e. $f(g(a))$ in other words. 
Not what you said: "Since f is one-to-one, we know that $g(a)=g(b)$"
The way you wrote it is backwards (for me) and not very clear. But your final answer is right, so hey! You passed right? You asked for critique, so this is mine. No hard feelings :p.
Also, it sounds like you are assuming contradiction when using '$=$' sign, but you did not state it if that is what you did. Again, check the definition of an injection(one-to-one).
