# how is my proof on composition functions

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

Your proof is correct. If you want to be nitpicky observe that if $h(b_1)=h(b_2)$ then we better have that $i(h(b_1))=i(h(b_2))$ or else $i$ is not a well-defined function. So if you start with $b_1\neq b_2$ and $h(b_1)=h(b_2)$ you cannot have $i(h(b_1))\neq i(h(b_2))$