I am proving this, all improvements are good. If I am completely wrong tell me.
Let $h:M\rightarrow N$ and $i:N\rightarrow O$. Prove that if $i \circ h$ is injective, then $h$ is injective.
This is my proof for this. Lets assume that $i \circ h$ is one-to-one. This means that if $i(h(a_1))=i(h(a_2)) $ then $a_1=a_2$.
assume contradiction. $h$ is not injective, therefore if $h(b_1) = h(b_2)$ then $b_1 \neq b_2$. Then if you look at $i(h(a_1))=i(h(a_2)) $ now you will see that now $a_1 \neq a_2$. This contradicts $i \circ h$ being injective. Therefor $h$ must be injective