The Stacks project

I think what the proof does might not be enough to conclude that $K^\bullet\to I^\bullet$ is homotopic to zero, https://mathoverflow.net/questions/185366/is-such-a-map-null-homotopic

I think one can refer the whole proof to 13.15.5. For part (1), one uses the fact that an injective resolution of $A$ is the same as a quasi-isomorphism into a complex with injective terms $A[0]\to I^\bullet$ such that $I^n=0$ for $n<0$. Part (2) is 13.15.5, (2).

I agree with #372 that maybe one could mention the results from which it follows the short exactness of the pullback of $E$ via $B'\to B$ (dually, the short exactness of the pushout of $E$ via $A\to A'$). These are 12.5.12 and 12.5.13.

Small typo: in the Proof of $1\implies 8$ , the second to last line should read, "By construction $\tau^{-1}(\mathfrak{m}) = \mathfrak{q}$" instead of "By construction $\tau^{-1}(\mathfrak{q}) = \mathfrak{m}$".

A comment on the cohomology $H^1(G, M)$ used on this page. By Definition 59.57.2 it is given as the derived functors of taking invariants on the abelian category $\text{Mod}_G$ of discrete $G$-modules defined in Definition 59.57.1. If $M$ is an object of this category, then $M$ has the discrete topology and the action of $G$ on $M$ is continuous. This means that every element of $M$ has an open stabilizer. If $G$ is $S^1$ this means that every element is fixed by all of $S^1$ and hence $H^0(S^1, M) = M$. Thus for $S^1$ we are taking the derived functors of an exact functor and it follows that $H^1(S^1, M) = 0$ for all $M$ in $\text{Mod}_{S^1}$.

I would say finite ring extension usually means a finite injective ring map. But injectivity is not used in Lemma 00GK.

(Minor typo: in the slogan, "on an Abelian categories" should be "on abelian categories".)

When the proof says "$H_k$ holds trivially for $k\leq n_0$", I think it should be $k<n_0$. For the sake of greater clarity, I propose: (i) in the definition $H_k$, replacing "$RF$ is defined at $J^n$" by "$J^n\in\mathcal{I}$", (ii) changing "using Proposition 13.14.8, we see that $RF$ is defined at $J^{n + 1}$" to "using Lemma 13.14.12, we see that $J^{n + 1}\in\mathcal{I}$", (iv) substituting the two instances of $R^0F(J^n)$ and $R^0F(J^{n+1})$ by $F(J^n)$ and $F(J^{n+1})$, respectively.

The "$f_{\ast}$" that appears in the commutative diagram is more like "$f^{\ast}$", since it is induced by taking base change of covers, and $f_{\ast}$ is already the map of étale fundamental groups.

In the first paragraph, when we say "by Lemma 13.16.4", wouldn't it suffice to say "by Definition 13.15.3"? (or simply "by definition"). Also, "distinguished" is misspelled just before expression $(RF(\sigma_{\geq i + 2} A^\bullet), RF(A^\bullet), RF(\sigma_{\leq i + 1} A^\bullet))$.

In the first sentence, I believe it should be $D^+(\mathcal{A})$, not $K^+(\mathcal{A})$.

I don't know if it is worth to add any of this to the already existing proof, but in case it helps anyone (and maybe even my future self), here are additonal details to conclude that $F\to R^0F$ is an isomorphism provided that $F$ is left-exact: by what is already written (and maybe 17.5), there is a quasi-isomorphism $A[0]\to K^\bullet$, with $K^n=0$ for $n<0$ and a map $RF(A[0])\to F(K^\bullet)$ satisfying Categories, Definition 4.22.1, (1). In particular, the composite $RF(A[0])\to F(K^\bullet)\to RF(A[0])$ is the identity and there are $K^\bullet\to L^\bullet$ and $A[0]\to L^\bullet$ in $A[0]/\text{Qis}^+(\mathcal{A})$, with $L^n=0$ for $n<0$, such that, inside $\mathcal{D}^+(\mathcal{B})$, the map $F(A[0])\to F(L^\bullet)$ equals the composite $F(A[0])\to RF(A[0])\to F(K^\bullet)\to F(L^\bullet)$. We claim that the image under $H^0$ of each of the maps in the last composite is an isomorphism (the first one gives $F(A)\to R^0F(A)$). Note that by what is already written in the proof, the images under $H^0$ of $F(A[0])\to F(L^\bullet)$ and $F(K^\bullet)\to F(L^\bullet)$ are isos. This implies that the image under $H^0$ of $RF(A[0])\to F(K^\bullet)$ must be an epimorphism. But it is also a split monomorphism; hence, it is an iso, and we win.

Typos: instead of «Thus $RF$ is» I think it should be «Thus $RF(K^\bullet)$ is». After, instead of «for these $s$ it follows» it should be «for these $s$. It follows» On the other hand, in «it follows that $H^i(RF(K^\bullet))=0$ for $i<a$», maybe one could give more detail? I guess saying «there is a quasi-isomorphism $s:K^\bullet\to L^\bullet$ with $L^n=0$ for $n<a$ and a map $RF(K^\bullet)\to L^\bullet$ as in Categories, Definition 4.22.1, (1). In particular $RF(K^\bullet)\to L^\bullet$ is monic, so $L^\bullet\cong RF(K^\bullet)\oplus E^\bullet$ for some $E^\bullet$ by Lemma 13.4.12, and the claim follows.»

In the third paragraph, the second line, the assumption used to refine the covering should be $(2)(c)$?

A little (maybe pedantic) remark: At the beginning of the proof I think we should say "we may add $0$ to $\mathcal{I}$ if necessary and suppose that $\mathcal{I}$ is closed under isomorphisms" (i.e., if $P\to R$ is an isomorphism in $\mathcal{A}$ with $P$ in $\mathcal{I}$, then also $R\in\mathcal{I}$), equivalently, we could just say "we may replace $\mathcal{I}$ by the essential image of $\{0\}\cup\mathcal{I}\to\mathcal{A}$ if necessary". I say this because after only adding $0$ to $\mathcal{I}$, then we may consider a short exact sequence $0\to 0\to P\to R\to 0$, hence the condition "closure under isomorphisms" for (2) to still hold.

Typo: in "such that $\alpha$ induces isomorphisms $H^j(K^\bullet) \to \text{Ker}(d^j)/\text{Im}(d^{j - 1})$", the domain and codomain should be reversed.

I think that maybe we can give a little bit more detail in "thus it is clear that (1) holds": Consider the commutative diagram: $$\require{AMScd} \begin{CD} X/\text{Qis}^+(\mathcal{A})@>>> X/\text{Qis}(\mathcal{A})\\ @V{F}VV@VV{F}V\\ \mathcal{D}^+(\mathcal{B})@>>>\mathcal{D}(\mathcal{B}) \end{CD}$$ Suppose the left map is essentially constant. Then we still have essentially constancy after postcomposition with the bottom map, Lemma 4.22.8. Hence, since the top map is cofinal, the right map is essentially constant. Conversely, suppose that the right map of the diagram is essentially constant. This means that there is a cocone under $X/\text{Qis}(\mathcal{A})\to\mathcal{D}(\mathcal{B})$ with vertex $Y\in\mathcal{D}(\mathcal{B})$, an object $X\to X'$ of $X/\text{Qis}(\mathcal{A})$ and a map $Y\to F(X')$ in $\mathcal{D}(\mathcal{B})$ satisfying Categories, Definition 4.22.1, (1). Since there is a quasi-isomorphism $X'\to X''$ with $X''\in K^+(\mathcal{A})$ (Lemma 13.11.5), by https://stacks.math.columbia.edu/tag/05PT#comment-8394 we can substitute $X'$ by $X''$ and assume $X'\in K^+(\mathcal{A})$. We claim that $Y\in\mathcal{D}^+(\mathcal{B})$. Since $Y\to F(X')$ is a (split) monomorphism, we have $Y\oplus Z\cong F(X')$ in $\mathcal{D}(\mathcal{B})$, by the proof of Lemma 13.4.12. Hence, since $\mathcal{D}^+(\mathcal{B})$ is strictly full and saturated in $\mathcal{D}(\mathcal{B})$ (13.6.4), we get $Y\in\mathcal{D}^+(\mathcal{B})$. Lastly, condition (2) of Categories Definition 4.22.1 for the left functor of last diagram follows from the fact that the top functor is cofinal.

I think it would be interesting to add the following remark to this section (maybe placing it after Definition 4.22.1): "Let $M:\mathcal{I}\to\mathcal{C}$ be a diagram, with $\mathcal{I}$ filtered (resp., cofiltered). Suppose that $M$ is essentially constant with value $X$ with respect to a morphism $X\to M_i$ (resp., to a morphism $M_i\to X$). Then, for any morphism $i\to j$ (resp., $j\to i$) in $\mathcal{I}$, the diagram $M$ is also essentially constant with value $X$ with respect to the composite morphism $X\to M_i\to M_j$ (resp., $M_j\to M_i\to X$). Verifying this fact is an easy exercise that uses the (co)filteredness of $\mathcal{I}$ and we leave it to the reader."

The lemma statement should say, ''and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.''

It's better to use $B'/B = (A'/A) \otimes _ A B$ instead of $B'/B = A'/A \otimes _ A B$ as the latter may be confused with $B'/B = A'/(A \otimes _ A B)$.