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Let $f: X\rightarrow Y$and $g: Y\rightarrow Z$ be two morphism between schemes, we assume $g\circ f$ is projective and $g$ is separated, then I want to prove $f$ is projective.

What I have done:

Since $g$ is separated, base change $Y\rightarrow Y\times_Z Y$, we get $X\times_YY\rightarrow X\times_ZY\times_YY$ is projective, then by two isomorphisms, $X\rightarrow X\times_ZY$ is projective.

Since $g\circ f: X\rightarrow Z$ is projective, then base change, we get $X\times_Z Y\rightarrow Y$ is projective.

To prove $f$ is projective, I want to prove that the composition of $X\rightarrow X\times_ZY$ and $X\times_Z Y\rightarrow Y$ is exactly $f$. Actually, the definition of the first morphism contain a base change and two isomorphisms, the second morphism is just a projection of fiber product. I have the intuition that the composition is $f$, since these constructions are natural, but when I want to give a strict proof by using the uniqueness of the morphism induced by the universal property of fiber product, I get a complicated commutative diagram and I don't know how to continue.

So could anyone help me explain why the composition of the two morphisms is $f$?

In addition, if the following diagram is a cartesian diagram, can we prove that $h$ is exactly the identity map? $$X\overset{f}{\rightarrow} Y\\{h} \downarrow \,\,\,\,\,\,\, \downarrow{id} \\ X\underset{f}{\rightarrow} Y$$

Thanks very much!

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  • $\begingroup$ For your last question, the answer is no. Take $X$ to be 2 points, $Y$ to be a point, $f$ to be the fold map, and $h$ the isomorphism that switches the two points. Then you can check that the diagram is cartesian. All you can deduce is that $h$ is an isomorphism such that $fh=f$. $\endgroup$ – jgon Oct 22 at 0:32
  • $\begingroup$ @jgon Thanks, I think your example is right. But in general, how to deduce $h$ is an isomorphism? $\endgroup$ – Sate Oct 22 at 0:53
  • $\begingroup$ well, the square with the identity is cartesian, and any other pullback must be isomorphic to the identity one. $\endgroup$ – jgon Oct 22 at 5:37
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First, we should be careful when we base change $\Delta : Y\to Y\times_Z Y$: you're taking the base change over $Y,$ but there are two natural maps $Y\times_Z Y\to Y$ (it doesn't change anything which you use, but it's something to be aware of).

To answer the question, observe that the following diagram is commutative, and that the two bottom squares are Cartesian:

$$ \require{AMScd} \begin{CD} X @>f>> Y @>\operatorname{id}_Y>> Y\\ @V\operatorname{id}_X\times f VV @VV\Delta V @VV\operatorname{id}_YV\\ X\times_Z Y @>>> Y\times_Z Y @>p_2>> Y\\ @V{p_1}VV @Vp_1VV @VVgV\\ X @>f>> Y @>g>> Z. \end{CD} $$ The map you want to consider is then one of the maps from $X$ in the upper left hand corner to either $Y$ in the lower-right square, and the diagram shows that either composition is in fact $f.$

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