Suppose $f \circ g$ is one-to-one and $f$ is not one-to-one. Show that $g$ is not onto. Given $A,B,C$ are sets and $g$ and $f$ are functions such that $g: A\rightarrow B, f: B\rightarrow C$.
Intuitively, I can ambiguously argue why this is true with a couple of sketches, but I'm having trouble proving this formally.
My proof goes as follows: (by contradiction)

Suppose $f \circ g$ is injective, $f$ is not injective and $g$ is surjective. I want to show this is impossible.
Let's assume we found $a,a'\in A$ such that $(f\circ g)(a) = (f\circ g)(a')
\iff f\big(g(a)\big) = f\big(g(a')\big)$
Well, we know $g$ is surjective, that means for all $b\in B$, there exists $a \in A$ such that $g(a) = b$, thus we can conclude that both $g(a),g(a')$ are in B.
So we know there exists $b, b'\in B$ such that $b = g(a)$ and $b' = g(b') \iff f(b) = f(b')$

Now this is where I feel like I can create the contradiction by saying there exists $b$ and $b'$ such that $f(b) = \big(f(b')\big)\implies b \not= b'.$ (Given $f$ is not $f$ is not injective).
But how do I do this? I can show that $b \not= b'$, but how can I use that to show that there exists $a, a' \in A$ such that $(f \circ g)(a) = (f\circ g)(a') \implies a \not= a'$, thus raising that contradiction?
Thanks,

If $f$ is nor one-to-one, then there exists $a\ne b, f(a)=f(b).$  If we assume by way of contradiction that $g$ is surjective, then there are $c,d$ such that $g(c)=a, g(d)=b.$
Take it from here. 
