I was trying to prove the following scheme morphism exercise. If $f: X \rightarrow Y, g: Y \rightarrow Z $ are such that $g \circ f $ is locally of finite type then $f$ is locally of finite type. I believe such a property on morphisms is called elimination otherwise I will edit this post.
I understand how to prove the case where all the schemes are affine. It comes down to a commutative algebra exercise. I wanted to reduced to this case by using the fact that being locally of finite type is local on the target. However, I believe this only lets me assume that the intermediate scheme $Y$ is affine.
When trying to apply the definition, I start with an affine open $U \subseteq Y $ and I am struggling to find an affine open on $Z$ such that $U$ is contained in its preimage.I tried writing $U= \cup[ U \cap g^{-1}(Z_i)]$ where the $Z_i$ are the affine open cover on $Z$ but then I am not sure if this helps.
Thanks in advance !