The Stacks project

Suggested slogan: the kernel of an exact functor of (pre-)triangulated categories is a (pre-)triangulated subcategory.

Very minor comment: Since $R$ is Noetherian, it suffices to let $M$ be finite. $N$ being finite is then automatic.

Typo: the '(4)' at the end of the proof should be '(7)'.

Typos: in the subscripts of the colimits, instead of $S$ and $S'$, it should be $(S/X')^\text{opp}$ and $(S'/X')^\text{opp}$, respectively.

Also, to justify equality of the two colimits maybe it would be nice to explicitly invoke 4.17.2 plus the following observation: if $F:\mathcal{I}\to\mathcal{J}$ is a fully faithful functor such that $\mathcal{J}$ is filtered and condition (1) from Definition 4.17.1 is satisfied, then $F$ is cofinal and $\mathcal{I}$ is filtered. Here we apply the observation to the inclusion $S'/X'\to S/X'$.

Update: I tried to get the uniqueness right in the pull request I just did on the GitHub project.

@#8344 The version you mention can only be used if $F$ is from $\mathcal{A}$ into the category of abelian groups or modules or something like that.

It is actually enough if for every $n > 0$, $A \in \mathrm{Ob}(\mathcal{A})$ and $s \in F^n(A)$ there exists an injective morphism $u : A \to B$ (depending on $n,A,s$) such that $s$ maps to zero under $H^n(u)$. The proof mostly doesn't change.

A reference is Prop. 4.2 in Buchsbaum's paper "On satellites and universal functors" (https://people.brandeis.edu/~buchsbau/miscpapers/027.pdf). This refined effaceability criterion is very useful in cases that there don't exist uniform $F^n(A)$-killing injections $A \to B$.

A non-example for this situation would be cohomology with respect to Zariski (hyper-)coverings, where there exists a single hypercovering computing the correct cohomology, and an example would be étale cohomology, where we truly need to take the colimit over all hypercoverings.

I think the translations $[n]:S^{-1}\mathcal{D}\to S^{-1}\mathcal{D}$ are not uniquely determined, and uniqueness is only guaranteed if we ask that $Q(X)[1]=Q(X[1])$ holds (as actual equality, not natural isomorphism). Maybe the result should be restated to “there is a pre-triangulated structure on $S^{-1}\mathcal{D}$ that is unique up to (unique) exact isomorphism that turns $Q$ into a triangulated functor.”

I'm not sure if the following is completely right, but here the non-uniqueness argument I came up with: Suppose $F:S^{-1}\mathcal{D}\to S^{-1}\mathcal{D}$ is an additive isomorphism for which there are natural isomorphisms $\eta: F\cong \text{id}_{S^{-1}\mathcal{D}}$ and $\varepsilon: F^{-1}\cong \text{id}_{S^{-1}\mathcal{D}}$. Then, if we define $[n]'=F\circ [n]\circ F^{-1}$ as a new translation in $S^{-1}\mathcal{D}$, we get an isomorphism of functors $\eta_{[n]}\star\varepsilon :[n]'\cong [n]$ (notation from 4.28). For a triangle $\Delta=(X,Y,Z,f,g,h)$ in $S^{-1}\mathcal{D}$, define $\Delta'=(X,Y,Z,f,g,(\eta_{[1]}\star\varepsilon)_X\circ h)$. If $\mathcal{T}$ is the class of distinguished triangles in $S^{-1}\mathcal{D}$, define $\mathcal{T}'=\{\Delta'\mid\Delta\in\mathcal{T}\}$. Then $Q:\mathcal{D}\to S^{-1}\mathcal{D}$ is also an exact functor of pre-triangulated categories when the codomain is the pre-triangulated category $(S^{-1}\mathcal{D},[\ ]',\mathcal{T}')$.

In proof of 38.30.2, in the last paragraph, it seems we want Fit_{r-1} instead of Fit_{k-1} because we didn't define k.

Should this lemma maybe be stated for $I=\mathcal{I}$ any filtered (small) category? If the result is true for a directed index set, then it must be true for a filtered index category, for the skeleton (https://en.wikipedia.org/wiki/Skeleton_(category_theory)#Definition) of a filtered category is a directed set. On the other hand, the inclusion of the skeleton into the ambient category is an equivalence and, thus, a cofinal functor. So Lemma 4.17.2 applies.

$\DeclareMathOperator{\Mor}{Mor}$A minor suggestion, the argument for the equality $u^s g_{\ast}(h_U^{\sharp})=f_{\ast}(h_U^{\sharp})$ actually works for any $\mathcal{F}\in Sh(\mathcal{C})$. In fact, let $\mathcal{F}\in Sh(\mathcal{C})$ and $V$ be an object of $\mathcal{D}$. Then $$\begin{equation} \begin{split} (u^sg_{\ast}\mathcal{F})(V) &= (g_{\ast}\mathcal{F})(f^{-1}h_V^{\sharp})\\ &=\Mor_{Sh(\mathcal{C}')}(h_{f^{-1}h_V^{\sharp}},g_{\ast}\mathcal{F})\\ &=\Mor_{Sh(\mathcal{C})}(g^{-1}h_{f^{-1}h_V^{\sharp}},\mathcal{F})\\ &=\Mor_{Sh(\mathcal{C})}(f^{-1}h_V^{\sharp},\mathcal{F})\\ &=\Mor_{Sh(\mathcal{D})}(h_V^{\sharp},f_{\ast}\mathcal{F})\\ &=f_{\ast}\mathcal{F}(V) \end{split} \end{equation}$$ Here in the fourth equality is by Lemma 7.29.4 (4).

Thank you, this will be very helpful! I needed the fact that if $E$ is the injective hull of the residue field of a noetherian local ring $(R,\mathfrak m)$ (or for that matter any Artinian $R$-module), then $E \otimes_R \widehat{R}^{\mathfrak m} \cong E$, but could not find a suitable reference for the last isomorphism.

Typo in the last paragraph: "is equal to σ^n−1 modulo J^n" should be "is equal to σ^n−1 modulo J^n-1"

(And dually also for Remark 4.27.15.)

I think we can give a more user-readable derivation of $t^{-1}(h\circ f)=t^{-1}(f''\circ g)$ (notation as explained after Definition 4.27.4). After making the observation that this notation satisfies $u^{-1}k=Q(u)^{-1}\circ Q(k)$ (for some diagram $\bullet\xrightarrow{k}\bullet\xleftarrow{u}\bullet$ in $\mathcal{C}$, with $u\in S$), we have

$$\begin{aligned} t^{-1}(h\circ f) &=Q(t)^{-1}\circ Q(h\circ f)\\ &=Q(t)^{-1}\circ Q(h)\circ Q(f)\\ &=t^{-1}h\circ Q(f)\\ &=b\circ Q(f)\\ &=Q(f')\circ a\\ &=Q(f')\circ s^{-1}g\\ &=Q(f')\circ Q(s)^{-1}\circ Q(g)\\ &=Q(t)^{-1}\circ Q(f'')\circ Q(g)\\ &=Q(t)^{-1}\circ Q(f''\circ g)\\ &=t^{-1}(f''\circ g). \end{aligned}$$

Regarding "now the combined results of Lemmas 4.27.5 and 4.27.6 tell us formula 4.27.7.1". I was thinking that maybe it is more direct to argue:

"4.27.7.1 follows at once from unwrapping the definition of the equivalence relation for a filtered colimit explained before Categories, Lemma 4.19.2. This yields exactly point (3) in the definition of $S^{-1}\mathcal{C}$ in the left calculus of fractions."

I'm not sure whether equality $\operatorname{id}_{Y''}^{-1}(a\circ s)\circ s^{-1}f=\operatorname{id}_{Y''}^{-1}(a\circ f)$ is formally implied by "the rules discussed in the text following Definition 4.27.4". I'm not saying this is a non-trivial identity (it follows at once from the definition of the composition in $S^{-1}\mathcal{C}$), but maybe it is worth mentioning it in the comments after Definition 4.27.4.

Indeed, this is true. We could say that the first part works if $M$ is killed by a power of $I$ and $R \to S$ is an isomorphism modulo any power of $I$ without assuming $I$ is finitely generated. The conclusion about the $I$-adic completion only (as far as I know) works when $I$ is finitely generated by the reference you pointed out. I will make these changes the next time I go through all the comments and I will move this lemma into Section 15.88 because it is more suitable.

@#8327. Yes, very good for finding this mistake! Thanks! I will fix this the next time I go through all the comments. I checked all the places in the stacks project where this gets used and in the places where we use the vanishing of the tor it comes from projectivity of $\Omega$.

I think you could cite lemma 10.131.10 to finish the proof, right?