The Stacks project

I think it's also interesing to think about the presheaf on $\text{Sch}/S$ that the object $\pi:\underline{\text{Spec}}_S\mathcal{A}\to S$ represents. Let $\mathcal{C}$ be a category, and let $c$ be an object of $\mathcal{C}$. Let $G:\mathcal{C}^\text{opp}/c\to\text{Sets}$ be a presheaf on $\mathcal{C}/c$. Consider the induced functor $G':\mathcal{C}^\text{opp}\to\text{Sets}$ that sends $d\in\text{Ob}(\mathcal{C})$ to $\coprod_{f \in \mathcal{C}(d,c)} G(f)$. Then $G$ is representable if and only if $G'$ is representable (not difficult to show). In our situation, $\mathcal{C}=\text{Sch}$, $c=S$, the functor $G$ sends $f:T\to S$ to $\text{Mor}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T)\cong\text{Mor}_{\mathcal{O}_T\text{-Alg}}(f^*\mathcal{A},\mathcal{O}_T)$, and $G'=F$ is 27.4.0.1.

That is, in Görtz & Wedhorn words (see (11.2) of the 2nd ed.), this construction can be regarded as a globalized version of the natural isomorphim $$\text{Hom}_{\operatorname{Spec}R}(T,\operatorname{Spec} A) \cong \text{Hom}_{R\text{-Alg}}(A,\Gamma(T,\mathcal{O}_T)),$$ where $A$ is an $R$-algebra.

I'm a little confused about the statement of the lemma and its proof. In the statement, we see $U$ is an algebraic space étale over $X$. But in the proof, after citing lemma 17.5.2, we see that $U$, having an underlying topological space, is a scheme.

I guess that previous lemma 66.20.1 is used to prove the space case. Since the $\mathcal{F}\vert_U \in U_{\mathrm{space,étale}}$, we may assume $U=X$. Then consider the sheaf $\mathcal{G}$ assigning $V\rightarrow X$ to a singleton if $\sigma\vert_V=0$ and assigning $V$ to $\emptyset$ otherwise. Then using 66.20.1 we get the open subspace $W$.

There is also a typo for the conclusion $(3)$: where $\bar{s}=(U\rightarrow X)\circ \bar{u}$.

I think the expression $(\underline{\text{Spec}}_S(\mathcal{A}), \varphi) \in F(\underline{\text{Spec}}_S(\mathcal{A}))$ should be $(\pi, \varphi)\in F(\underline{\text{Spec}}_S(\mathcal{A}))$ instead. Here's an alternative way to phrase the proof that doesn't require invoking $X$: "Let $U\subset S$ be open affine. Since the functor $F_U$ is represented by $\pi^{-1}(U)$, let $(\pi^U,\varphi^U)\in F_U(\pi^{-1}(U))$ be the universal element. By the proof of 26.15.4 (we verified the hypotheses in the proof of 27.4.3), it suffices to see that $(\pi,\varphi)|_U=(\pi^U,\varphi^U)$. This follows from the construction of $\pi$ and $\varphi$."

In line 4, it should read "such that (X_{K'})_{red} is geometrically reduced"

Okay, sorry, I have just realized that my last comment is just the first paragraph of 27.2.3.

Here is an elegant argument to argue the uniqueness of $X$ up to unique isomorphism over $S$: Define a relative gluing datum over $S$ to be a pair $(\{f_U\}_{U\in\mathcal{B}},\{\rho_V^U\mid U,V\in\mathcal{B}, V\subset U\})$ as in (1), (2) and satisfying (a), (b). Define a morphism of relative gluing data over $S$ $(\{f_U:X_U\to U\},\{\rho_V^U\})\to(\{g_U:Y_U\to U\},\{\eta_V^U\})$ to be a collection of maps $h_U:X_U\to Y_U$ over $U\in\mathcal{B}$ such that for $U,V\in\mathcal{B}$ with $V\subset U$ the following diagram commutes: $$\require{AMScd} \begin{CD} X_V@>{h_V}>> Y_V\\ @V{\rho_U^V}VV@VV{\eta_U^V}V\\ X_U@>>{h_U}> Y_U \end{CD}$$ Denote $\text{GD}/S$ to the category of relative gluing data over $S$. We have an obvious functor $\text{Sch}/S\to\text{GD}/S$ that is easily verifiable to be fully faithful. The lemma can be reformulated to "this functor is essentially surjective."

In the last paragraph, to give more details for why $i_W$ is an isomorphism, maybe one could say "with the notations of [https://stacks.math.columbia.edu/tag/001L#comment-8427], verify that $F_\iota=F_W$, where $\iota:f^{-1}(W)\to X$ is the inclusion."

Minor typo: in the last paragraph, $(U_i,\xi'_{U_i})$ should be $(U_i,\xi'|_{U_i})$. On the other hand, to give more details in "on the overlaps $V_i\cap V_j$ the morphisms $g_i$ and $g_j$ agree," one could write: first note that a morphism $h:T\to U_i$ factors through $U_i\cap U_j$ if and only if $h^*\xi'|_{U_i}\in F_j(T)$. Thus, the identity $g_i^*(\xi'|_{U_i})|_{V_i\cap V_j}=\xi|_{V_i\cap V_j}=g_j^*(\xi'|_{U_j})|_{V_i\cap V_j}$ implies $g_k(V_i\cap V_j)\subset U_i\cap U_j$, for $k=i,j$. Hence, $g_i^*(\xi'|_{U_i\cap U_j})|_{V_i\cap V_j}=g_j^*(\xi'|_{U_i\cap U_j})|_{V_i\cap V_j}$ and using [https://stacks.math.columbia.edu/tag/001L#comment-8427] we get $g_i|_{V_i\cap V_j}=g_j|_{V_i\cap V_j}$.

Maybe one could add to this section the definition of subfunctor (given in 26.15.3) stated in a general setting and also add the following remark: Let $\mathcal{C}$ be a category and let $f:A\to X$ be a morphism in $\mathcal{C}$. Suppose $F:\mathcal{C}^\text{opp}\to\text{Sets}$ is a functor with a representation $\text{Mor}(-,X)\cong F$. For an object $Y\in\text{Ob}(\mathcal{C})$, define $F_f(Y)$ to be the image of the composite $\text{Mor}(Y,A)\xrightarrow{f_*}\text{Mor}(Y,X)\xrightarrow{\cong}F(Y)$. Then $F_f$ is a subfunctor of $F$ and the natural transformation $\text{Mor}(-,A)\to F_f$ is an isomorphism if and only if $f$ is a monomorphism. Moreover, if these conditions hold and $\xi\in F(X)$ is the universal object for $\text{Mor}(-,X)\cong F$, then $F(f)(\xi)\in F_f(A)$ is the universal object for $\text{Mor}(-,A)\cong F_f$.

See 6.30. You can use the search function to find stuff like this.

In the penultimate paragraph, "Let $d={\rm deg}(g_1)$": should be $d_1$ instead of $d$.

Would you please give me the reference for the remark that we can define the sheaf on basis in which we have the restriction maps and gluing properties? Thank you!

Typo: instead of “$F(X)\to F(X')$ has a section” it should be “has a retraction.” Also, I am having trouble understanding the proof: the first problem I have is with “hence $F(X') = F(X) \oplus E$ for some object $E$ of $\mathcal{D}'$ such that $E \to F(X' \oplus Y) \to RF(X'\oplus Y) = RF(X \oplus Y)$ is zero (Lemma 13.4.12).” I do understand why $F(X)\to F(X')$ having a retraction implies that it must be isomorphic to an inclusion $F(X)\to F(X)\oplus E$ (this is a property of monomorphisms in pre-triangulated categories), but where does the latter condition on the vanishment of $E$ comes from? On the other hand, taking for granted this vanishment condition, I think I understand everything what comes after until “$F(t')\to E$ annihilates $E$” (this last thing I understand too). But why is it that after “it follows that $F$ is essentially constant on $X/S$ with value $F(X)$”?

The proof can be simplified.

By virtue of Lemma 01TH, it is sufficient to prove that: Let $Y$ be a locally Noetherian scheme, and $x \in Y$. Then $x$ is isolated in $Y$ if and only if ${\rm dim}_x Y=0$.

Proof. If $x$ is an isolated point, $\{x\}$ is open (by Definition 06RM). Hence {\rm dim}_x Y = 0 (by Definition 0055). Conversely, if ${\rm dim}_x(Y) =0$, then there exists an open subset $x\in U\subset Y$ such that ${\rm dim} U = 0$. We can take $U$ Noetherian because $Y$ is locally Noetherian. Since $U$ is 0-dimensional Noetherian scheme, it is discrete. Hence $x$ is isolated point of $Y$. q.e.d.

I think in the diagram of the statement one should change $D^+(\mathcal{A})$ by $K^+(\mathcal{A})$.

To prove (2), is there some reason by which one needs to invoke 13.20.1 and 13.16.8 instead of just saying "(2) is a particular case of (1)"?

To justify the short exactness of $0\to I^\bullet\to E^\bullet\to C^\bullet\to 0$, maybe one could link to Homology, Section 12.6 (specifically, to what I comment in https://stacks.math.columbia.edu/tag/010I#comment-8412 )

To justify "there exists a quasi-isomorphism $\alpha : K^\bullet \to L^\bullet$ with $\beta \circ \alpha$ homotopic to zero," one could link the result I propose between parentheses in https://stacks.math.columbia.edu/tag/05RW#comment-8372

It took me a while to find out how the map $d^{n-1}(L^{n-1})\to I^n$ is defined. Here is a proof by chasing elements: it suffices to show that the composite $\operatorname{Ker}d_L^{n-1}\xrightarrow{\beta} I^{n-1}\to I^n$ vanishes. Let $y\in\operatorname{Ker}d_L^{n-1}$. Since $H^{n-1}(K^\bullet)\to H^{n-1}(L^\bullet)$ is an isomorphism, there are $x\in\operatorname{Ker}d_K^{n-1}$ and $z\in L^{n-2}$ such that $\alpha(x)=y+d_L^{n-2}z$. Hence, the image of $x$ along $K^{n-1}\to I^n$ equals $(d_L^{n-1}\circ\beta)(y)$ and vanishes by commutativity of the solid diagram, for $x\in\operatorname{Ker}d_K^{n-1}$.

The correct proof is not much different from the actual one: suppose there is $n$ and that, for each $i\leq n$, there are maps $h^i:K^i\to I^{i-1}$ such that $\alpha^i=d^{i-1}h^i+h^{i+1}d^i$ for all $i<n$. We are going to construct a map $h^{n+1}:K^{n+1}\to I^n$ such that $\alpha^n-d^{n-1}h^n=h^{n+1}d^n$. For this, note that $$\begin{gather} (\alpha^n-d^{n-1}h^n)d^{n-1}=\alpha^nd^{n-1}-d^{n-1}h^nd^{n-1}=\\ \alpha^nd^{n-1}-d^{n-1}(\alpha^{n-1}-d^{n-2}h^{n-1})=\alpha^nd^{n-1}-d^{n-1}\alpha^{n-1}=0. \end{gather}$$ Hence, $0=(\alpha^n-d^{n-1}h^n)|_{d^{n-1}(K^{n-1})}$, and we can reuse the already existing argument.