The Stacks project

Typo in the first paragraph: In “since $\text{Tot}(L^{\bullet, \bullet}) = \text{Tot}(F(I^{\bullet, \bullet})) = F(\text{Tot}(I^{\bullet, \bullet}))$ computes $H^n(RF(K^\bullet))$,” I think it should be “computes $RF(K^\bullet)$” instead.

The statement of Lemma 13.18.9 doesn't guarantee that $I^{\bullet,\bullet}$ is concentrated in the upper-half plane (i.e., that $I^{p,q}=0$ if $q<0$), although its proof does, see #9944.

One could add rewrite the last claim of the statement to be:

Moreover: 1. If $A^{n}$ (resp., $B^{n}$, $C^n$) vanishes for all values of $n$ less than $n_1$ (resp., $n_2$, $n_3$), then $I_1^\bullet$, $I_2^\bullet$, $I_3^\bullet$ might be chosen so that $I_k^n=0$ for all $n<n_k$. 2. Given any injective resolution $A^\bullet\to I^\bullet$ we may assume $I_1^\bullet=I^\bullet$.

And we can assume 1+2 provided that $I^n=0$ for all $n<n_1$.

In (3) of the statement I think it should say "...for all $U \in \operatorname{Ob}(\mathcal{B})$"

For (2), in order to conclude that $\alpha_0$ and $\beta_0$ are homotopic we also need to define a map $h:M\to N_0$ such that $h\circ d_{N,1} + d_{F,1}\circ h_0 = \gamma_0$. Luckily this is a simple matter of rearragning this equation into $h(d_{N,1}(x)) := \gamma_0(x) - d_{F,1}(h_0(x))$, which is well-defined because $d_{N,1}$ is surjective.

In Definition 12.24.1, second paragraph, after “we assume, for the moment, that $\mathcal{A}$ is an abelian category which has countable direct sums and countable direct sums are exact” one could add “i.e., $\mathcal{A}$ is AB4” (mentioned as well in #9940).

After Definition 12.23.1, in “we assume, for the moment, that $\mathcal{A}$ is an abelian category which has countable direct sums and countable direct sums are exact” one could add “i.e., $\mathcal{A}$ is AB4, see Injectives, Definition 19.10.1.”

Okay, yes. It's 12.19.1. Maybe it is worth linking this?

How are the notations $\bigcap Z_r$ and $\bigcup B_r$ defined? Do they mean minimum and supremum in the poset of subobjects of $E_1$?

In case it's helpful for anyone, here's a sketch of proof.

We can use the Yoneda lemma to prove that monomorphisms coincide with injections, as in the proof of Tag 7.3.2.

An isomorphism of sheaves is clearly both injective and surjective. Conversely, a subsheaf $\mathcal{F} \subseteq \mathcal{G}$ which satisfies Condition (2) of Tag 7.11 must equal all of $\mathcal{G}$, by the sheaf conditions applied to $\mathcal{F}$ and $\mathcal{G}$.

Given two morphisms of sheaves $g_1, g_2 \colon \mathcal{G} \rightarrow \mathcal{H}$, their equalizer is given by an injection of sheaves $\mathcal{K} \rightarrow \mathcal{G}$. We thus find that a surjection of sheaves is an epimorphism.

Conversely, suppose we are given an epimorphism of sheaves $f \colon \mathcal{F} \rightarrow \mathcal{G}$. Without loss of generality, we may assume $f$ is injective by replacing $\mathcal{F}$ with its image under $f$. Now we may apply the argument in the proof of Tag 7.3.2, upon noting that the presheaf pushout $\mathcal{G} \coprod_{p, \mathcal{F}} \mathcal{G}$ is a separated presheaf (using that $\mathcal{F}$ is a subsheaf of $\mathcal{G}$), and hence maps injectively to its sheafification.

We can refine the Warning before Lemma 15.86.9 in the following way: Denote $\omega$ to the category where $\operatorname{ob}\omega=\mathbb{N}$ and where for $n,m\in\omega$ there is one morphism from $n$ to $m$ if $n\leq m$ and none otherwise. Write $\overline{\omega}=\omega^\mathrm{op}$. The category of inverse systems over a category $\mathcal{C}$ is the functor category $\mathcal{C}^{\overline{\omega}}$. Let $\mathcal{A}$ be an abelian category. There is a natural isomorphism $\operatorname{Ch}(\mathcal{A}^{\overline{\omega}})\cong\operatorname{Ch}(\mathcal{A})^{\overline{\omega}}$, which identifies the quasi-isomorphisms in $\operatorname{Ch}(\mathcal{A}^{\overline{\omega}})$ with the natural transformations $\alpha:F\Rightarrow G$ between functors $F,G:{\overline{\omega}}\rightrightarrows\operatorname{Ch}(\mathcal{A})$ whose components are qis, i.e., $\alpha_n$ is a qis for every $n\in\omega$ [1, Lemma 3]. Thus, by the universal property of the category localization (it holds that $D(\mathcal{B})=\operatorname{Ch}(\mathcal{B})[\mathrm{qis}^{-1}]$ for any abelian category $\mathcal{B}$), we have a unique functor $$\tag{1}\label{fun} D(\mathcal{A}^{\overline{\omega}})\to D(\mathcal{A})^{\overline{\omega}}$$ fitting into a commutative square as this one. An alternative way to build \eqref{fun} is first to build a functor $K(\mathcal{A}^{\overline{\omega}})\to K(\mathcal{A})^{\overline{\omega}}$ (which is possible thanks to [1, Remark 4], beware that this functor is not faithful and possible neither full [2]), and then using the universal property of the localization $D(\mathcal{B})=K(\mathcal{B})[\mathrm{qis}^{-1}]$ to get a obtain a unique \eqref{fun} making this square commute.

Lemma 15.86.9 says that \eqref{fun} is essentially surjective. However, \eqref{fun} is not faithful and might not be full [3].

References

1. A characterization of AB5 and AB4 categories. Does this result show up in the literature?, Mathematics Stack Exchange.

2. Derivators - diagrams in homotopy category of chain complexes, MathOverflow.

3. Is the derived category of inverse systems the inverse systems of the derived category?, MathOverflow.

Typo: last sentence "We conclude by Lemma 10.5.3 (2) that $N_ 2$ is finitely presented. "

Wait! Sorry, now I realize you are not taking union over all associated primes, but just over some of them...

I don't think the proof works. What if you take $R$ to be normal of dimension $n>2$ and take any nonzero $a\in R$. Then $R_f$ is clearly normal, but the equality that you are claiming at the end doesn't hold since the right-hand-side is nonempty. In fact you have only proven inclusion $(\subseteq)$.

The last diagram on the morphism of distinguished triangles involving $\beta$ seems not to be commutative on the squared adjacent to $\beta$. At least not in the level of the category of complexes of $R$-modules. How can we see the diagram being commutative in $\mathrm{D}(R)$? (A diagram chasing still can show that $H^{n-1}(F^{n-1}\to \ldots)$ surjects to $H^{n-1}(K)$ and $H^{n}(F^{n-1}\to \ldots)$ is isomorphic to $H^{n}(K)$, however.)

I think this lemma also holds if $\mathcal{S}$ and $\mathcal{S}'$ are arbitrary fibred categories over $\mathcal{C}$ (not necessarily fibred in groupoids) and if $G$ is moreover assumed to be a morphism of fibred categories (automatic when $\mathcal{S}'$ is fibred in groupoids).

Sorry, I guess you do want just the Leibniz rule since then automatically $D(b^p)=0$ for all $b\in R$. However, still $D:R\rightarrow R$ is the statement is not correct I think

I don't see how Lemma 10.9.9 can be used in the second part of the proof. However, after giving up on it I found that there is an elementary alternative: Since $x_i/1 = 0/1 \in M_\mathfrak{p}$ there must exist an $f_i\notin\mathfrak{p}$ such that $f_ix_i = 0$. Now choose $f = f_1\cdots f_r$, and note that on one hand $f\notin\mathfrak{p}$ since $\mathfrak{p}$ is prime, and on the other hand $f\in \mathrm{Ann}(M)$. Thus $\mathrm{Ann}(M)\not\subseteq\mathfrak{p}$, which is equivalent to the desired result $\mathfrak{p}\notin V(\mathrm{Ann}(M))$.

I am not sure what you mean by derivative here? If you mean "$A$-derivative", then you would automatically have $D(a)=0$, so I guess not. If you mean that only the Lebniz rule should be satisfied, then I don't see why you can replace $a$ with $f^pa$. My guess is that you want this to be $A^p$-derivative, where $A^p\sseq A$ is the subring of all $p$-powers, but should be in the statement. Also, in the statement it should be $D:K\rightarrow K$ instead of $D:A\rightarrow A$.

In the statement of the Lemma it seems that the first occurence of $X_{pro-étale}$ should be replaced by $Y_{pro-étale}$.