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.

Nicely done! In part (f), I think you mean that $g$ is not surjective since there is nothing in $A$ that $g$ will take to $2,$ but the idea is spot on.
A minor critique for your proof of (b): I would instead suggest that you take an arbitrary $c\in C$, use surjectivity of $f$ to conclude that there is some $b\in B$ such that $f(b)=c$, then use surjectivity of $g$ to conclude that there is some $a\in A$ such that $g(a)=b,$ whence $(f\circ g)(a)=c$ as you showed. Since $c$ was an arbitrary element of $C,$ then for all $c\in C$ there exists $a\in A$ such that $(f\circ g)(a)=c$. This is basically the same as the approach you took, but the connection is a bit clearer and more justified, to my mind.
You mentioned that you weren't sure about part (e). You can clean it up a bit by again taking an arbitrary $c\in C,$ concluding from surjectivity of $f\circ g$ that there is some $a\in A$ such that $(f\circ g)(a)=c,$ and noting that $g(a)\in B$ and $f\bigl(g(a)\bigr)=c.$ Since $c$ was an arbitrary element of $C,$ then for all $c\in C$ there exists $b\in B$ such that $f(b)=c$. As an alternate approach, you could let $h$ be the restriction of $f$ to the range of $g$--that is, $h:g(A)\to C$ is defined by $h(b)=f(b)$ for all $b\in g(A)$. Note/prove that $h$ is surjective if and only if $f\circ g$ is. Since $h$ is a restriction of $f$ and is surjective, then $f$ is surjective. I think your approach is better, personally, but you mentioned you weren't sure about it.
