The Stacks project

@#9817 Replace "is fully faithful" by "is an equivalence." The functor $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ is essentially surjective since the hypothesis on $X\in\mathcal{C}$ gives an iso $Q(X)\to Q(X')$ in $S^{-1}\mathcal{C}$, with $X'\in\mathcal{C}'$.

Alternative proof: Refer to #9817. It is only left to check MS5 and MS6. This is justified in the four first sentences of the current proof.

One could add the following result to this section:

Lemma. Let $\mathcal{C}$ be a category. Let $S$ be a left multiplicative system in $\mathcal{C}$. Let $\mathcal{C}'\subset\mathcal{C}$ be a full subcategory. Assume that for every $X\in\mathcal{C}$ there is a morphism $X\to X'$ in $S$ with $X'\in\mathcal{C}'$. Then $S'=S\cap\operatorname{Mor}(\mathcal{C}')$ is a left multiplicative system in $\mathcal{C}'$ and the functor $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ coming from Lemma 4.27.8 is fully faithful.

Proof. LMS1 for $S'$ is clear. To check LMS2, suppose $Z\xleftarrow{t}X\xrightarrow{g}Y$ is a diagram in $\mathcal{C}'$ with $t\in S'$. Then there is a diagram $Z\xrightarrow{f_0}W_0\xleftarrow{s_0}Y$ with $s_0\in S$ and $W\in\mathcal{C}$ such that $s_0g=f_0t$. By hypothesis, there is $u:W_0\to W$ in $S$ with $W\in\mathcal{C}'$. Defining $s=us_0$ and $f=uf_0$ gives the result. Lastly, to check LMS3, suppose there is a diagram $W\xrightarrow{t} X\overset{f}{\underset{g}{\rightrightarrows}}Y$ in $\mathcal{C}'$ with $t\in S'$ and $ft=gt$. Then there is $s_0:Y\to Z_0$ in $S$ with $Z_0\in\mathcal{C}$ and $s_0 f=s_0 g$. By hypothesis, there is $v:Z_0\to Z$ in $S$ with $Z\in\mathcal{C}'$. Then $s=vs_0$ gives the result. We now prove $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ is fully faithful. Given $Y'\in\mathcal{C}'$, the hypothesis implies that the inclusion $Y'/S'\to Y'/S$ is a cofinal functor (Lemma 4.19.3). Thus, for $X'\in\mathcal{C}'$, the map $$\mathop{\mathrm{colim}}_{(Y'\to Y'')\in Y'/S'}\operatorname{Hom}_{\mathcal{C}'}(X',Y'') \to\mathop{\mathrm{colim}}_{(Y'\to Y'')\in Y'/S}\operatorname{Hom}_{\mathcal{C}}(X',Y'')$$ is bijective. That is, the map $$\operatorname{Hom}_{(S')^{-1}\mathcal{C}'}(X',Y') \to\operatorname{Hom}_{S^{-1}\mathcal{C}}(X',Y')$$ is bijective. $\square$

An alternative reference for this result is Kashiwara, Schapira, Categories and Sheaves, Corollary 14.1.8.

This is a minor point that would not confuse anyone who knows what is going on, but I am not certain what it would mean for a sequence of sets to be exact, considering that the category of sets does not have a zero object.

It seems to me that if you want your definition to make sense for sheaves of sets (in addition to abelian groups, etc.) it might be clearer to speak of the fragment of the diagram starting with $F(U)$ and ending with $\prod_{i,j}F(U_i\times_U U_j)$ being limiting or an equaliser diagram.

Small typo in the statement of the lemma: the derived tensor product on the right side should be over $f^{-1}\mathcal{O}'$, not $f^{-1}\mathcal{O}_Y$.

Sorry, the first formula in the previous comment is $\mathbb{\m z}$, and the second one is $k=\mathbb{\m q}$

I understand what you want to illustrate here, but your rings $R_n$ are not essentially of finite type over $\m z$ (unless maybe $k=\mathfrak{\m q}$)?

If you want, you can add in condition (3) that actually $S_{\lambda}$ is essentially of finite presentation over $R_{\lambda}$

Are you sure that assumption of $f$ being local is not needed?

I think it should be made clear that $f$ is required to be non-zero.

Minor correction: in the last sentence of the proof, $y$ should be replaced with $D(a)^{i+1}y$.

It should be $\varphi(I_n)\subseteq\varphi(I)^{2^{n-1}}R'$, but who cares...

There is a typo in the first displayed short exact sequence in the proof: the last term should be $H^0\otimes_B C$ instead of $N^0 \otimes_B C$.

typo: $J^\bullet$ and $I^\bullet$ are used for the same complex

A silly question: how can we deduce that the pro-objects $( H^ p(E_ n) )$ and $( H^ p(D_ n) )$ are isomorphic?

Typo: in the fourth line "to an object $M$ of $D(\textit{Ab}(\mathbf{N}))$".

Is $S_q$ intended in the statement of part (3) (in which case we indeed have it immediately from part (1)), instead of $S$?

I don't think it is necessarily the case that commutativity of the diagram $(FGA\to FGB\to FX\to (FGA)[1])\to(A\to B\to C\to A[1])$ implies commutativity of the diagram $(GA\to GB\to X\to (GA)[1])\to(GA\to GB\to GC\to (GA)[1])$. My worry is that when $G$ is regarded as a triangulated functor with the isomorphism $G\circ [1]\cong [1]\circ G$ as built in the proof it may happen that the unit $\eta:1\Rightarrow GF$ and counit $\varepsilon: FG\Rightarrow 1$ won't be trinatural (in the sense of #9797). Only if $\eta,\varepsilon$ are trinatural we will have an adjunction in the $2$-category of categories with translation (that contains (pre-)triangulated categories as a subcategory), which will induce an adjunction between categories of triangles via the $2$-functor $\mathrm{t}(-)$ (explained in #9797). From inspection of the current proof, the isomorphism $\xi':G\circ [1]\cong [1]\circ G$ can be seen to equal $$\xi'=T[GT^{-1}(\varepsilon_{T}\circ \xi_{T^{-1}GT}^{-1})\circ\eta_{T^{-1}GT}]$$ (where $\xi$ is the isomorphism $F\circ[1]\cong [1]\circ F$). However, it is not clear that $\eta,\varepsilon$ will be trinatural (i.e., compatible with $\xi,\xi'$). In [M, Proposition 47], Murfet chooses a different $\xi'$, namely (after dualizing his proof by assuming $F\dashv G$), $$(\xi')^{-1}=GT\varepsilon\cdot G\xi_G\cdot\eta_G.$$ With this definition for $\xi'$, Murfet shows that $\eta$ and $\varepsilon$ turn out to be trinatural.

References

[M] D. Murfet, Triangulated Categories Part I, http://therisingsea.org/notes/TriangulatedCategories.pdf

There is a typo at the last statement of the lemma: it should be "any $c$ elements of $I_{q'}$ which generate $(I/I^2)_q$..."