This question is based on the following proposition (EGA IV, 2.7.1)

Let $f: X \rightarrow Y$ be a $S$-morphism of $S$-schemes, $g: S'\rightarrow S$ a faithfully flat and quasi-compact morphism. Denote $X \times_S S'$ by $X'$, and denote $Y \times_S S'$ by $Y.$ We have a natural morphism $f': X\rightarrow Y.$ Consider the following properties of morphisms:

(i) separated

(ii) quasiseparated

(iii) locally of finite type

(iv) locally of finite presentation

(v) finite type

(vi) finite presentation

(vii) proper

(viii) isomorphism

(ix) monomorphism

(x) open immersion

(xi) quasi-compact immersion

(xii) closed immersion

(xiii) affine

(xiv) quasi-affine

(xv) finite

(xvi) quasi-finite

(xvii) entire (I'm not sure exactly what a "morphisme entier" is, but some reading of the french wikipedia gave me the impression that it's an integral morphism)

If $P$ is one of the preceding properties, then $f$ has property $P$ if and only if $f'$ has property $P.$

My impression of the proof is that there are quite a few ingredients needed to prove this proposition, where different ingredients are needed for different properties, and it surprises me that there seems to be no "unifying principle" that covers all the proofs.

What I would like is something like this theorem: Let $P$ be a property closed under composition and base change. Then, if $f:X \rightarrow Y$ is a map of $Z$-schemes, such that the structure morphism $X \rightarrow Z$ is in $P$ and the diagonal morphism of the structure morphism $Y \rightarrow Z$ is in $P$, then $f$ is in $P$. I like this theorem because it explicitly states the conditions the property needs to satisfy, and the conditions are fairly loose. Then, you have essentially the same proof of this theorem for every such property $P$.

Is there any such unification of the proofs for this result? To me, it seems like this result is some incredibly mysterious miracle. I would appreciate any intuition behind this result.