The Stacks project

Typo: in the equality $C_n=B^n\oplus Q_3^n$, in the LHS the subscript should be a superscript.

Also, regarding the reason of why we have morphisms of triangles $$\begin{gather} (A^\bullet \to B^\bullet \to Q_1^\bullet \to A^\bullet [1]) \longrightarrow (A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet [1]), \\ \\ (A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet [1]) \longrightarrow (B^\bullet \to C^\bullet \to Q_3^\bullet \to B^\bullet [1]). \end{gather}$$

and why the morphism $\delta: Q_3^\bullet\to Q_1^\bullet[1]$ equals the composite $Q_3^\bullet\to B^\bullet[1]\to Q_1^\bullet[1]$. I think one could make the argument more explicit by simply mentioning https://stacks.math.columbia.edu/tag/011J#comment-8369. The case for the morphism of triangles is clear, whereas the claim on $\delta$ follows from application of the comment to the morphism $$\require{AMScd} \begin{CD} 0@>>> B^\bullet@>>> C^\bullet@>>> Q_3^\bullet@>>> 0\\ @.@VVV@VVV@|@.\\ 0@>>>Q_1^\bullet@>>> Q_2^\bullet@>>> Q_3^\bullet@>>> 0 \end{CD}$$ of termwise splitting s.e.s.

Suggestion: Maybe one could add to the statement "Moreover, the morphism $\delta$ is natural in the following way: If we have a commutative diagram of complexes $$\require{AMScd} \begin{CD} 0@>>> A^\bullet@>>> B^\bullet@>>> C^\bullet@>>> 0\\ @.@VV{\alpha}V@VV{\beta}V@VV{\gamma}V@.\\ 0@>>>\tilde{A}^\bullet@>>> \tilde{B}^\bullet@>>> \tilde{C}@>>> 0 \end{CD}$$ where the rows are termwise split short exact sequences and we are given termwise splittings $\widetilde{s}^n:\widetilde{C}^n\to\widetilde{B}^n$ and $\widetilde{\pi}^n:\widetilde{B}^n\to\widetilde{A}^n$ such that the diagram $$\require{AMScd} \begin{CD} A^n@<{\pi^n}<< B^n@<{s^n}<< C^n\\ @V{\alpha^n}VV@V{\beta^n}VV@VV{\gamma^n}V\\ \widetilde{A}^n@<<{\widetilde{\pi}^n}< \widetilde{B}^n@<<{\widetilde{s}^n}< \widetilde{C}^n \end{CD}$$ commutes for all $n$, then the following diagram commutes: $$\require{AMScd} \begin{CD} C^\bullet@>{\delta}>> A[1]^\bullet\\ @V{\gamma}VV@VV{\alpha[1]}V\\ \widetilde{C}^\bullet@>>{\widetilde{\delta}}>\widetilde{A}[1]^\bullet \end{CD}$$ where $\widetilde{\delta}^n=\widetilde{\pi}^n\circ d^n_{\widetilde{B}}\circ\widetilde{s}^n$."

The proof is immediate from commutativity of the second-to-last diagram plus the fact that $\beta$ is a cochain map.

Small point, but in the definition of $\sigma_{a}$ I think you want to exclude the case $j=a$

Unless I missed a convention somewhere, all modules are denoted as left modules, so strictly speaking the notation $xb$ is not defined. I understand all rings are commutative, and the right module notation is convenient here, but maybe a remark would be in order.

In definition 00D1, the M should be N.

In lemma 0583, the second equality in (2) should be the result of the lemma 07JY.

Is the second condition actually the sheaf $h_S\times F'$ is an algebraic space over $S$? Since the via the identification $Sh((Sch/S) _ {fppf}) \cong Sh((Sch/S') _ {fppf})/h_S$ the functor $j^{-1}:Sh((Sch/S') _ {fppf})\rightarrow Sh((Sch/S) _ {fppf})$ is identified with $Sh((Sch/S') _ {fppf}) \ni F\mapsto (h_S\times F\rightarrow h_S) \in Sh((Sch/S') _ {fppf})/h_S$ according to Lemma 7.25.9. Maybe the citation of Lemma 7.25.4 should also be Lemma 7.25.9?

a typo:

"The following immediate consquence of the smooth base change theorem is what is often used in practice."

consquence > consequence

As $e$ is a terminal set presheaf on $\mathscr{C}$ and not an abelian presheaf on $\mathscr{C}$, for every abelian sheaf $\mathcal{F}$ on $\mathscr{C}$, it does not seem that $\Gamma(\mathscr{C},\mathcal{F}) \stackrel{\text{def}}{=} {\operatorname{Hom}_{\operatorname{PSh}(\mathscr{C})}}(e,F(\mathcal{F}))$ has the structure of an abelian group, where $F$ here denotes the forgetful functor from $\operatorname{AbSh}(\mathscr{C})$ to $\operatorname{PSh}(\mathscr{C})$. 1. What exactly is the abelian category $\mathsf{A}$ we are considering that contains all $\Gamma(\mathscr{C},\mathcal{F})$ where $\mathcal{F}$ is an abelian sheaf on $\mathscr{C}$? 2. For abelian sheaves $\mathcal{F}$ and $\mathcal{G}$ on $\mathscr{C}$, how is ${\operatorname{Hom}_{\mathsf{A}}}(\Gamma(\mathscr{C},\mathcal{F}),\Gamma(\mathscr{C},\mathcal{G}))$ equipped with the structure of an abelian group?

Thank you very much!

There's no need to quote lemma 10.15.2

Tag 00MC doesn't need $(R,\mathfrak m)$ to be local. The result holds as long as $I$ is contained in the Jacobson radical of $R$. Perhaps it could be stated this way since the proof does not require any serious modifications?

Is U not supposed to represent the affine specR from the statement of the lemma? Whereas the U in the proof I think refers to the intersection with X_s

A small remark, but the last "equivalence" in the statement should be "anti-equivalence".

This lemma follows trivially from Lemma 52.15.6 whose proof is a lot cleaner as well.

A minor problem. In the proof for $(3)$, let $\mathcal{F}=\mathcal{G}$ and $a=b=\mathrm{id}$, we have $\mathcal{F}\times _{a,\mathcal{G},b}\mathcal{F}=\mathcal{F}$ which is not isomorphic to $\mathcal{F}\times \mathcal{F}$. Maybe we can use the fact that $a=b$ iff $\mathrm{ker}(a,b)\rightarrow \mathcal{F}$ is an isomorphism?

Edit: I see now... it is an exact sequence of sheaves and the cech functor is only a delta functor on the pre-sheaf category... Perhap it's worth starting the proof with "note that lemma 20.10.2 does not apply here" to avoid confusion for future readers.

Can you not deduce this immediately from the exact sequence of Cech cohomology and the fact that the 0'th cohomology is just the global sections for sheaves?

Suggestion: substituting "we omit the verification that we obtain a long exact sequence" by "the exactness of the long sequence is the exactnesss in part (2) of 12.5.17", and substituting "we omit the verification of the properties mentioned at the end of the statement of the lemma" by "the functoriality is Lemma 12.5.18, whereas compatibility with shifts comes from Definition 12.14.8".

It seems that (1) and (2) hold as long as $\varphi$ is finite type and of finite Tor dimension.

I think we should consider the case that $Y_{n}=0$ for some $n$. Whenever $Y_{n}=0$ for some $n$, we can prove $X_{n} \rightarrow X$ is isomorphism, which means we can stop the step for $n$.