Show that the composition function and its inverse are one to one and onto 
Let both $f: A \mapsto B$ and $g: B \mapsto C$ be one-to-one and onto. Show that the composition $g \circ f : A \mapsto B$ and the inverse $f^{-1} : B \mapsto A $ are one-to-one and onto. 

Here is what I have so far: 
Show onto: By definition of composition 
$$g \circ f (x) = g(f(x))$$
So for any $b \in B$, there exist an $a \in A$ such that $f(a)=b$, so $$f(a)=b$$ thus $g(b)= c$ for all $c\in C$ which to me is not correct. 
Showing one to one: For all $a \in A$ there exist a $b \in B$ such that $f(a) = f(b)$ where $a=b$. 
Can anyone point me in the right direction?
 A: What you have written seems pretty confused. Let's look at the "onto" portion given bijections (one-to-one and onto functions) $A \overset{f}{\to} B \overset{g}{\to} C$.
You say that 

By definition of composition $g \circ f (x) = g(f(x))$ so for any $b \in B$ there exists an $a \in A$ such that $f(a) = b$ [...]

But this gives the impression that somehow composition implies the existence of such an $a \in A$. Such an $a \in A$ is guaranteed to exist, but it's because the function $f$ maps $A$ onto $B$.
Similarly, your statement that 

For all $a \in A$ there exist a $b \in B$ such that $f(a)=f(b)$ where $a=b$

shows that you haven't really understood the definition of one-to-one functions. It's really an "If... Then..." statement; if we know that $f(x_1) = f(x_2)$ and that $f$ is one-to-one, then we know that $x_1 = x_2$ (otherwise, two distinct elements of the domain, $x_1$ and $x_2$, would get mapped to the same element).
You will absolutely not be able to prove things like this if you don't have a firm understanding of the definitions. It's also good practice, at first, to write down exactly what you need to prove: The assumptions, and the conclusion you're looking for. Justify each claim you make, and pay close attention to what you've written. For example...

I would argue that $f \circ g$ maps $A$ onto $C$ as follows.
We need to show that, given any $c \in C$, there exists some $a \in A$ so that $(g \circ f)(a) = g(f(a)) = c$. 
Since $g$ maps $B$ onto $C$, we know that there exists some $b \in B$ so that $g(b) = c$. But then, since $f$ maps $A$ onto $B$, we can find some $a \in A$ so that $f(a) = b$ hence 
$$g(f(a)) = g(b) = c,$$
as desired.
To show that $g \circ f$ is one-to-one, you'll pick $a_1, a_2 \in A$ and assume that $g(f(a_1)) = g(f(a_2))$. Your job will be to show that $a_1 = a_2$, and I'll leave that to you (Hint: Since $g$ is one-to-one, what can we say about $f(a_1)$ and $f(a_2)$?) .
A: Remember: 
For  onto: ‘$g\circ f$ onto’ means every element in $C$ has a pre-image  in $A$.
For one-to-one: it means different elements in $A$ have  different images in $C$.
