Suppose that $g: A \rightarrow B$, $h: B \rightarrow C$. Prove that if $g$ is onto and $h$ is not one-to-one, then $h \circ g$ is not one to one. How can I start to approach this question? I know that if $g$ and $h$ are one-to-one, then $h \circ g$ is one-to-one, and also if $h$ and $g$ are both onto, and $h \circ g$ is onto. But this question didn't fit into both cases...
Since $h$ is not one to one, there is $b_1, b_2 \in B$ distinct such that $h(b_1) = h(b_2)$. Since $g$ is onto, there is $a_1, a_2 \in A$ such that $g(a_1) = b_1$ and $g(a_2) = b_2$.
Then notice that $h(g(a_1)) = h(b_1) = h(b_2) = h(g(a_2))$. So then $h\circ g$ is not one to one.
I'd start with considering that is $h$ is not one to one considering the $x\ne y$ so that $h(x) = h(y)$ and see what happens.
$g$ is onto so there are $a,\alpha \in A$ so that $g(a) = x$ and $g(\alpha) = y$.  And.... well, seems the obvious thing to try is to  $h(g(a)) = h(x)$ while $h(g(\alpha) = h(y) = h(x)$.
So if $a \ne \alpha$ then $h\circ g$ is not one-to-one.  And clearly $a \ne \alpha$ because if $a = \alpha$ then $g(a) = g(\alpha)$ but $g(a) = x\ne y = g(\alpha)$.
So that's that.