The Stacks project

After Lemma 12.25.1, in “these spectral sequences define two filtrations on $H^n(\operatorname{Tot}(K^{\bullet,\bullet}))$. We will denote these $F_I$ and $F_{II}$,” perhaps one could spell out $F_J^pH^n(\operatorname{Tot}(K^{\bullet,\bullet}))=\operatorname{Im}\left(H^n(F_J^p\operatorname{Tot}(K^{\bullet,\bullet}))\to H^n(\operatorname{Tot}(K^{\bullet,\bullet}))\right)$, for $J=I,II$. That is, we are plugging $(\operatorname{Tot}(K^{\bullet,\bullet}),F_J)$ into Definition 12.24.5.

Analogously as in Comment #9487:

1. In the statement, instead of “the limit $E_{\infty}$ exists” shouldn't we specify “the limit $E_{\infty}$ of the spectral sequence of Lemma 12.24.2 exists”?
2. In the proof, second sentence, instead of $Z_r=\bigoplus Z_r^{p,q}$, $B_r=\bigoplus B_r^{p,q}$, shouldn't one write $Z_r=\left(Z_r^{p,q}\right)$, $B_r=\left(Z_r^{p,q}\right)$? (We are working over an abelian category $\mathcal{A}$ that might not satisfy AB3.)

1. In the statement, instead of “the limit $E_\infty$ exists” shouldn't we specify “the limit $E_\infty$ of the spectral sequence of Lemma 12.23.2 exists”?
2. In the proof, second sentence, instead of $Z_r=\bigoplus Z_r^p$, $B_r=\bigoplus B_r^p$, shouldn't one write $Z_r=(Z_r^p)$, $B_r=(Z_r^p)$? (We are working over an abelian category $\mathcal{A}$ that might not satisfy AB3.)

Typo: In the statement, I think the very last equality should be $E_1^p=H(\operatorname{gr}^pK,\operatorname{gr}^pd)$.

Can you please explain why in the proof of lemma 26.14.1 the commutative diagram imply the isomorphism of sheaves $\mathscr O_{U_i}|_{U_i\cap U_j}\rightarrow \mathscr O_{U_j}|_{U_i\cap U_j}$. What I think is all three morphisms in the commutative diagram are isomorphisms and hence induce a sequence of isomorphisms $\mathscr O_{U_i}|_{U_i\cap U_j}\cong \varphi _i|{U_{ij}}^*\mathscr O_i|_{U_{ij}}\cong \varphi _j{U_{ji}}^*\mathscr O_{j}|_{U_{ji}}\cong \mathscr O_{U_{j}}|_{U_i\cap U_j}$. But I could be wrong.

Maybe a easy question but why $\varphi_i$ in Lemma 6.33.2 is isomorphism. Let $W\subset U_i$ be open for fixed $i$. Let $(s_j)_{j\in I}\in \mathscr F|_{U_i}$ then for any other index $j\in I$, $s_j$ is uniquely determined by the image of $s_i$. In detail, we construct $\alpha: \mathscr F|_{U_i}\rightarrow \mathscr F_i: (s_j)_{j\in I}\rightarrow s_i$ and $\beta: \mathscr F_i\rightarrow \mathscr F_{U_i}: s\rightarrow (s_j)_{j\in I}$, where in the second map $s_j=s$ if $j=i$, otherwise set $s_j=\varphi(s|{W\cap U_j})$. And they are mutually inverses.

Should Lemma 17.22.2.(1) be correct only for $F$ being coherent(or finite presentation), then it follows from the stalk isomorphism of intern Hom, while the general case the proof would not go through since surjective of sheaves only implies surjective on the stalk level?

The diagram should compile like this. I don't know why Mathjax doesn't compile it well here in the Stacks Project (same happened to me in #6829).

How one can describe exactly how $g'$ is induced? What I thought is to define $g'$ as the connecting homomorphism arisen from the snake lemma applied to the commutative diagram with exact rows $$\require{AMScd} \begin{CD} @.TE@>{Tf}>> TA@>{T\alpha}>>A'@>>> 0\\ @.@V{Td}VV@V{Tg}VV@VV{0}V@.\\ 0@>>>\operatorname{Ker}TS d@>>> TSE@>{\smash{ TSd}}>> TS^2E \end{CD}$$ Note that this also gives a description of $g'$ in Lemma 12.21.2 if $T=S=\operatorname{id}_\mathcal{A}$. I don't know if there is another way, though. The footnote seems to be using something else, instead of the snake lemma.

Maybe in the list of properties (1) – (3) to be found before Definition 12.22.5 one could mention Lemma 12.21.4? It seems it is what one needs.

(See also [1, Sect. 2.6, Proposition 6.1].)

I disagree with the “obviously” before Remark 12.21.5 in “note that in the situation of the lemma we obviously have $B_\infty = g\left(\bigcup\nolimits_r \operatorname{Ker}(\alpha^r)\right) \subset Z_\infty = f^{-1}\left(\bigcap\nolimits_r \operatorname{Im}(\alpha^r)\right)$.” The non-obvious thing to me is the fact that taking inverse of subobjects commutes with unions (more generally, it will commute with subobject sums) and that taking image of subobjects commutes with intersections.

It is not a difficult matter, though: Given an abelian category $\mathcal{A}$ and an object $X$ of $\mathcal{A}$, denote $S(X)$ to the class of isomorphism classes of subobjects of $X$. Then $S(X)$ is partially ordered. Given a morphism $f:X\to Y$ in $\mathcal{A}$, one can define maps $f:S(X)\rightleftarrows S(Y):f^{-1}$ which are taking image of a subobject and taking inverse image of a subobject along $f$, see [1, p. 40]. Moreover, the former is left adjoint to the latter [1, Sect. 2.6, Proposition 6.8]. Hence, taking direct images commutes with arbitrary colimits and taking inverse images with arbitrary limits.

References

1. N. Popescu, Abelian categories with applications to rings and modules

I have two small comments on this lemma:

In order to apply Lemma 03Q6 at the end of the proof of (1) we need $f:X\to Y$ to be quasi-compact and quasi-separated. \

The last equation in the proof of (1) says: By Lemma 099S we have $\Gamma(X\times_Y V, \epsilon^* \mathcal{F}) = \Gamma(X\times_Y V, \mathcal{F})$

To be precise this should be $\Gamma(X\times_Y V, \epsilon^* \mathcal{F}) = \Gamma(X\times_Y V, g^{-1}\mathcal{F})$ where $g:X\times_Y V \rightarrow X$ is the canonical map. Moreover, the used lemma is not 099S but Lemma 0GLZ.

Alternative suggested proof: By Lemma 10.97.2 it is flat. Thus, if $M$ is some non-zero $R$-module, to see that $M\otimes_R\hat{R}\neq 0$ it suffices to see that $N\otimes_R\hat{R}\neq 0$, for finitely generated submodules $N\subset M$. By Lemma 10.97.1, $N\otimes_R\hat{R}\cong\hat{N}$. Since $\hat{N}=\varprojlim_n N/I^nN$, the map $N\to \hat{N}$ is injective by Krull's intersection theorem (Lemma 10.51.4), so $N\neq 0$ implies $\hat{N}\neq 0$.

Typo : in the proof of Lemma 0F2N, $Y$ should be replaced by $S$.

Typo : in the proof of Lemma 0F2N, $Y$ should be replaced by $S$.

Thank you! I think I just confused myself. I'm curious: why do you say this section is "a bit obsolete"?

Regarding the last sentence of the proof: I think there are currently no “remarks following Definition 13.23.2.” One could instead substitute this last sentence by “This is true by the commutative square in proof of Lemma 13.23.3. In this square, if $\alpha$ is a quasi-isomorphism, then $i_{L^\bullet}\circ\alpha=j(\alpha)\circ i_{K^\bullet}$ is a quasi-isomorphism too, hence so is $j(\alpha)$; by Proposition 13.23.1, $j(\alpha)$ is an iso in $K^+(\mathcal{I})$.”

In the statement, one could sharpen:

1. “then $j=j'\circ Q$ for a unique exact functor...,” since this is what we actually get from the proof (say, from Lemma 13.5.7).

2. “which is quasi-inverse in the $2$-category of triangulated categories to the canonical functor...” (I wrote the proof below).

Finally, at the end of the proof, shouldn't we say something like “we leave to the reader to check that $j'$ is quasi-inverse to $K^+(\mathcal{I})\to K^+(\mathcal{A})$” at least? Here's the proof anyway:

First we note that the components $i_{K^\bullet}$ assemble into a natural transformation $i:\operatorname{id}_{K^+(\mathcal{A})}\to\iota\circ j$, where $\iota:K^+(\mathcal{I})\to K^+(\mathcal{A})$ is the inclusion. This is witnessed by the commutative square in the proof of Lemma 13.23.3. Now, on the one hand, $\operatorname{id}_{K^+(\mathcal{I})}$ is isomorphic to the composite $K^+(\mathcal{I})\to D^+(\mathcal{A})\xrightarrow{j'} K^+(\mathcal{I})$ since $i$ becomes a natural isomorphism when restricted to $K^+(\mathcal{I})$, i.e., $i_{I^\bullet}$ is an isomorphism if $I^\bullet\in K^+(\mathcal{I})$ (use Proposition 13.23.1). By Comment #9471, this natural isomorphism is in the $2$-category of triangulated categories. On the other hand, leveraging again naturality of $i$ it is not difficult to see that $\operatorname{id}_{D^+(\mathcal{A})}$ is isomorphic to the composite $D^+(\mathcal{A})\xrightarrow{j'}K^+(\mathcal{I})\to D^+(\mathcal{A})$. Again, Comment #9471 implies this natural isomorphism lives in the $2$-category of triangulated categories. $\square$

One could add to the statement something like “Moreover, the morphisms $i_{K^\bullet}:K^\bullet\to j(K^\bullet)$ assemble into a $2$-morphism in the $2$-category of triangulated categories (see explanation before Definition 13.3.4).” Indeed, the proof already gives $\xi_{K^\bullet}\circ i_{K^\bullet[1]}=i_{K^\bullet}[1]$ for free.

In the proof a spectral sequence 15.67.0.1 is used. But why we can use it? There is an assumption that $K$ is bounded below in Section 15.67...