$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Proj{\mathbf{Proj}}$Let $S$ be a Noetherian scheme and $X$ finite $S$-scheme. The finite morphism $X \to S$ is projective in the sense of the definition in the Stacks Project (tag [01W8](https://stacks.math.columbia.edu/tag/01W8), part (1)).


This is proved in tag [0B3I](https://stacks.math.columbia.edu/tag/0B3I) of the Stacks Project, where a map $\sigma : X\to \textbf{Proj}\_S(f\_\*\mathcal{O}\_X)$ is constructed, associated to the surjective map
$$f^\*f\_\*\mathcal{O}\_X\to\mathcal{O}\_X.$$


However, when proving that $\sigma$ is a closed immersion, the Stacks Project invokes tag [01KT](https://stacks.math.columbia.edu/tag/01KT), saying that $\sigma$ is a section of a separated morphism.


Say $f : X\to S$ is also a locally free map for simplicity.



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> What morphism is $\sigma$ a section of?
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The proof seems to suggest there should be a morphism of $S$-schemes $\Proj\_S(f\_\*\mathcal{O}\_X)\to X$ and $\sigma$ should be a section to it, however, I don't see such a morphism.


Another question is this.



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> Suppose $g : X\to X$ is an $S$-endomorphism of $X$. Is there an $S$-endomorphism of the "proj" construction, $g' : \Proj\_S(f\_\*\mathcal{O}\_X)\to \Proj\_S(f\_\*\mathcal{O}\_X)$ that is compatible with $g$? That is, such that $g'\circ \sigma = \sigma\circ g$.
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To give some context, the questions came up while reviewing different definitions of projectivity for morphisms (in the Stacks Project, in R. Vakil's notes [The rising sea: Foundations of algebraic geometry](http://math.stanford.edu/%7Evakil/216blog/FOAGmay2523public.pdf), and in Hartshorne's book). The first question was about a slight inaccuracy in [0B3I](https://stacks.math.columbia.edu/tag/0B3I), now fixed. The second came up when R.Vakil's notes say, when $X=\text{Spec}(B)$ and $S=\text{Spec}(A)$, that there is a closed immersion of $X$ into $P:=$"$\text{Proj}(A\oplus B\oplus B\cdots)$" and I was confused as to how to extend endomorphisms of $X$ to $P$. Now it all makes sense.



Finite subschemes of projective bundles
Welcome new contributor. There is no such morphism from the relative Proj to $X$: it would contradict Stein factorization. However, there is a simple proof that $\sigma$ is a closed immersion that does not need this.


For every quasi-coherent $\mathcal{O}\_S$-module $\mathcal{F}$, denote by $$(p\_\mathcal{F}:\mathbb{P}\_S(\mathcal{F})\to S,\ \ \alpha\_{\mathcal{F}}:p\_{\mathcal{F}}^\*\mathcal{F} \twoheadrightarrow \mathcal{O}(1)),$$ a universal pair of an $S$-scheme and an invertible quotient of the pullback of $\mathcal{F}$ to that $S$-scheme. Associated to every invertible $\mathcal{O}\_S$-module $\mathcal{L}$ and every $\mathcal{O}\_S$-module homomorphism, $$\beta:\mathcal{L}\to \mathcal{F},$$ there is an induced morphism of invertible sheaves on $\mathbb{P}\_S(\mathcal{F})$; namely the composition, $$p\_{\mathcal{F}}^\*\mathcal{L} \xrightarrow{p\_{\mathcal{F}}^\*\beta} p\_{\mathcal{F}}^\*\mathcal{F} \xrightarrow{\alpha\_{\mathcal{F}}}\mathcal{O}(1). $$ The maximal open $U\_\beta$ of $\mathbb{P}\_S(\mathcal{F})$ on which this composition is an isomorphism is the complement of a "pseudo divisor", i.e., the ideal sheaf of the zero scheme of the composition is everywhere locally principal (perhaps locally equal to the zero ideal sheaf). Of course if $\beta$ is locally a direct summand of $\mathcal{F}$, then this closed subscheme is normally flat. If also $\mathcal{F}$ is a flat $\mathcal{O}\_S$\_module, then this closed subscheme is an $S$-flat relative hyperplane section.


Now let $\mathcal{F}$ be the locally free $\mathcal{O}\_S$-module $f\_\*\mathcal{O}\_X$, and let $\beta$ be the locally direct summand, $$f^\#:\mathcal{O}\_S\to f\_\*\mathcal{O}\_X.$$ The corresponding open subscheme $U\_{f^\#}$ is a relative affine space bundle whose sheaf of relative differentials is canonically isomorphic to the pullback of the cokernel of $f^\#$. By adjointness of pushforward and pullback, the following map of $\mathcal{O}\_X$-modules is an isomorphism, $$\mathcal{O}\_X \xrightarrow{f^\*f^\#} f^\*f\_\*\mathcal{O}\_X \xrightarrow{\text{nat}\_f} \mathcal{O}\_X.$$ Therefore the morphism $\sigma$ factors through $U\_{f^\#}$.


The induced morphism from $X$ to $U$ is a morphism of affine $S$-schemes. This morphism is a closed immersion if and only if the induced morphism of pushforward structure sheaves is a surjective morphism of quasi-coherent $\mathcal{O}\_S$-modules. Since both of these quasi-coherent sheaves is the target of compatible morphisms from $\mathcal{O}\_S$, the original morphism of quasi-coherent sheaves is surjective if and only if the induced map of quotients modulo $\mathcal{O}\_S$ is surjective. For an affine space bundle, up to the choice of a zero section (which exists locally), this quotient sheaf is the direct sum of all positive symmetric powers of the relative sheaf of differentials, i.e., all positive symmetric powers of $f\_\*\mathcal{O}\_X/\mathcal{O}\_S$. So already the restriction of the $\mathcal{O}\_S$-module homomorphism to the first symmetric power $f\_\*\mathcal{O}\_X/\mathcal{O}\_S$ is an isomorphism onto the quotient $f\_\*\mathcal{O}\_X/\mathcal{O}\_S$, and thus the full $\mathcal{O}\_S$-module homomorphism is surjective.


For your second question, since pushforward is functorial, the pushforward along $f$ of the pushforward along $g$ of $\mathcal{O}\_X$ is canonically isomorphic to the pushforward along $f$ of $\mathcal{O}\_X$. This canonical isomorphism defines an automorphism of the pushforward along $f$ of $\mathcal{O}\_X$ functorially associated to $g$. You can use this to prove that the induced automorphism of relative Proj is compatible with $g$ and $\sigma$.


