The Stacks project

This proposition isn't hard to prove directly, but I wonder where you use : "and because ∏ is an exact functor on the category of families of abelian groups."?

This article is not well written. Namely, the Cech hypercohomology is NOT a limit, but a colimit or a direct limit. Why haven't you given the formula to show what you mean by a limit? You are confusing the reader. You should correct everywhere the deceptive reference to a limit.

Instead of saying that there is a module map $I/I^{n}\bigotimes_{R/I^{n}}M/I^{n}M\rightarrow(I\bigotimes_{R}M)/I^{n-1}(I\bigotimes M)$, it would be better to say that the map $I/I^{n}\bigotimes_{R/I^{n}}M/I^{n}M\rightarrow M/I^{n}M$ factors through the surjective map $I/I^{n}\bigotimes_{R/I^{n}}M/I^{n}M\rightarrow(I\bigotimes_{R}M)/I^{n-1}(I\bigotimes M)$

Is the map $\pi$ a splitting? If so it would be clearer to just deduce injectivity from that.

Suggestion: since this proposition uses multiple rings, maybe add a subscript for the tensor products indicating over what they are taken

Where is the induction here specifically done? Wouldn't it be enough just to replace R by R/I^k+1, M by M/I^k+1 and I by I/I^k+1 and apply the precceding lemma?

Maybe note in the almost last paragraph that you are using the snake lemma? Took me a moment to figure out what you were trying to do

I guess in the proof of (2) and (4) the notation was heavily relaxed, since why should $\varphi_{ii_j}(s) = \varphi_{i'i_j}(s')$ hold, actually it holds for some restriction of s and s'. Also it should be $\varphi_{ii_j}(U)$ or better of some $U_{i_j}$. I recommend to mention it or rework completly the proof.

In the proof, the minimality of $n$ is not needed, so "a finite set of generators" would be more to the point. Also, it would be worth pointing out that the $M_i$'s are finite modules, and perhaps that the last quotient $M/M_{n-1}$ is finitely presented if $M$ is.

I think Lemma 03GI is used to deduce $U_j$ is quasi-compact. It might worth to note it explicitly.

For the sake of having some reference that is actually instructive, here's a neat proof of the cartesianity of the square https://mathoverflow.net/a/80812/101848

Minor typo: instead of $\operatorname{Spec}((A \otimes_R B)/J)$ in "we see that $\Delta^{-1}(U \times_S V) \to U \times_S V$ can be identified with $\operatorname{Spec}((A \otimes_R B)/J)$," one should write $\operatorname{Spec}((A \otimes_R B)/J)\to\operatorname{Spec}(A \otimes_R B)$.

I think it could be nice to reference https://stacks.math.columbia.edu/tag/001U#comment-3413 to make explicit the formal nature of the argument (I believe such little hints are quite useful for those recently initiated to algebraic geometry—and often to category theory too—reading this webpage, like the version of myself one year ago).

An alternative reformulation of the proof would be by placing this lemma after remark https://stacks.math.columbia.edu/tag/01JW#comment-8464 and then just invoking the remark plus 26.17.6.

In the last sentence, I would give the hint "the third is a combination of the first two, plus the pasting law for pullbacks, https://stacks.math.columbia.edu/tag/001U#comment-3413 ."

I think the following could be an interesting remark to add after 26.18.3: Let $\mathcal{P}$ be a property of morphisms of schemes. Then $\mathcal{P}$ is preserved under arbitrary base change as in 26.18.3 (1) if and only if it is preserved under arbitrary base change as in 26.18.3 (2). To see ($\Rightarrow$), use the diagram of the proof of 26.18.2, where the top square is cartesian. To see ($\Leftarrow$), use again the same diagram but with $Y=S$, so $Y_{S'}=S'$.

In the proof of lemma 9.8.5, it should say that the elements are linearly dependent over K.

I think it would be informative if after «we usually just say “let $Z$ be a locally closed subscheme of $X$” since we may recover $U$ from the morphism $Z\to X$,» one added «namely, $U=X\setminus(\overline{Z}\setminus Z)=(X\setminus\overline{Z})\cup Z$.»

At the start of the proof of Lemma 02IJ, "Suppose that R⊂S′⊂S is a finitely generated R-subalgebra of S." is slightly odd in that it gives the initial impression that the string of inclusions is a finitely generated R-subalgebra of S (although it easy to decipher the meaning a moment later). I propose "Suppose that R⊂S′⊂S, where S' is a finitely generated R-subalgebra of S."

A typo: in the fourth line, "$\mathcal{F}=f_{\mathrm{small},\ast}\ast$" should be $\mathcal{F}=f_{\mathrm{small},!}\ast$.

Here $f_{small}$ is identified with the localization morphism $j$ in 66.18.11

I would like to ask how strict the equalities "$=$" in the section are?

To elaborate my question, first suppose that we have morphisms of topoi $f:\mathrm{Sh}(\mathcal{C})\rightarrow \mathrm{Sh}(\mathcal{D})$, $g:\mathrm{Sh}(\mathcal{D})\rightarrow \mathrm{Sh}(\mathcal{E})$ and $h:\mathrm{Sh}(\mathcal{C})\rightarrow \mathrm{Sh}(\mathcal{E})$. We say $g\circ f= h$ if $g_{\ast}\circ f_{\ast}=h_{\ast}$ and $f^{-1}\circ g^{-1}=h^{-1}$. However, if we only know $g_{\ast}\circ f_{\ast}=h_{\ast}$, then by adjunction we know that there exists a canonical isomorphism $f^{-1}\circ g^{-1}\cong h^{-1}$. Hence we get a canoical $2$-isomorphism $g\circ f\cong h$.

In this regard, for example in lemma 7.28.4, we have $u\circ j_U=j_V\circ u'$. Hence we have $_su\circ _sj_U=_sj_V\circ _su'$. However, we only have canonical isomorphisms $(j_U^p)^{\sharp}\circ (u^p)^{\sharp}\cong ((u\circ j_U)^p)^{\sharp}=((j_V\circ u')^p)^{\sharp}\cong (u'^p)^{\sharp}\circ (j_V^p)^{\sharp}$, for $(-)^p$ and $(-)^{\sharp}$ both involving taking inductive limits. Hence what seems to me is that we have a canonical isomorphism $f\circ j_U\cong j_V\circ f'$.

In the same manner, in lemma 7.28.1, we verified the equlity $(f')^{-1}j_V^{-1}=j_U^{-1}f^{-1}$ to conclude that $f\circ j_U=j_V\circ f'$. But again by adjunction of functors, we a priori only knows that $f\circ j_U\cong j_V\circ f'$ canonically.

Is there a remedy for this problem?