The Stacks project

Isn't it a little bit weird to talk about sheafification in this lemma, since you introduce it only after this subsection?

Typos in the proof, second paragraph: in “let $x\in\lim_{\beta < \alpha} K_\alpha^0$” it should be $K_\beta^0$ in the limit, and in “we will find a $y\in K^{-1}_\alpha$”, instead of $K^{-1}_\alpha$ it should be $\lim_{\beta < \alpha} K_\beta^{-1}$.

The previous comments #9501 and #9502 should be inside Tag 19.11.7. I thought I was commenting inside the latter result, my bad.

In the proof, I got a little bit confused because $n$ means different things in “let $n\in\mathbb{Z}$ and $\psi:U\to M^n$ be a morphism” and “$|N^m|\leq c^{\max\{n-m,0\}}(|U|)$ for some $k\geq n$.” Maybe in the former case one could write “let $n_0\in\mathbb{Z}$ and $\psi:U\to M^{n_0}$ be a morphism” instead? If I got it right, the base case of the downward induction is $k=n$ and one fixes some $n\geq n_0+2$.

In the proof of (1), the definition of $S_{i'}$ should be replaced by its complement.

Oh you mean the local values in 03RU by "the values"

I thought (2) did not imply (1). For example, let $x$ be a geometric point, $X = \bigsqcup_{i=0}^\infty x$ and $U = X\sqcup X$.

Proof for finite order thickenings, second paragraph, line 2: should be ``the map $A' \to A$ is surjective''.

(Since the previous doesn't currently compile well on my browser, I assume it won't either in other people's browsers, so you may look at the comment's plain text here.)

Typo: In “note that $M\to\mathbf{M}(N)$ is a functor” I think it should be $M\mapsto\mathbf{M}(M)$ instead. This was also pointed out in #7169.

In the proof, after “pick any ordinal $\alpha$ whose cofinality is greater than $|U|$,” I think we later use that $\alpha$ is actually a limit ordinal when we say “then we see that $\varphi$ factors through $M_{\alpha'}(M)$ for some $\alpha'<\alpha$ by Proposition 19.11.5.” Maybe one should mention some of this? ($\alpha$ will always be a limit ordinal since $|U|\geq 1$ and by Sets, Comment #9498),

Also, in the transfinite recursion, we are not only defining $\mathbf{M}_\alpha(M)$ for each $\alpha$ but also a natural transformation $\mu_{\alpha,\beta}:\mathbf{M}_\alpha\to \mathbf{M}_\beta$ for each $\alpha<\beta$, right? And such that $\mu_{\alpha,\beta,M}$ is injective for all $M\in\operatorname{ob}\mathcal{A}$ and $\mu_{\beta,\gamma}\circ\mu_{\alpha,\beta}=\mu_{\alpha,\gamma}$ for $\alpha<\beta<\gamma$. In other words, we are getting a functor $\operatorname{Ord}\to\operatorname{Fun}_\mathrm{inj}(\mathcal{A},\mathcal{A})$, where $\operatorname{Ord}$ is the totally ordered class of ordinal numbers and $\operatorname{Fun}_\mathrm{inj}(\mathcal{A},\mathcal{A})$ is the wide subcategory of $\operatorname{Fun}(\mathcal{A},\mathcal{A})$ where the morphisms are the natural transformations whose components are all injective maps. After thinking for a while, the following is what I came up with.

For each ordinal number $\alpha$, denote $S(\alpha)$ to the set of ordinal numbers $\leq\alpha$ (this is $S(\alpha)=\alpha\cup\{\alpha\}$, the successor of $\alpha$). For each ordinal number $\alpha$, we want to define a functor $$\begin{aligned} \mathbf{M}^{\leq\alpha}:S(\alpha) &\to\operatorname{Fun}_\mathrm{inj}(\mathcal{A},\mathcal{A})\\ \beta&\mapsto{}\mathbf{M}^{\leq\alpha}_\beta\\ \gamma\leq\beta &\mapsto\mu^{\leq\alpha}_{\gamma,\beta} \end{aligned}$$ such that $\mathbf{M}^{\leq\alpha}|_{S(\beta)}=\mathbf{M}^{\leq\beta}$ for all $\beta\leq\alpha$. The desired functor $\operatorname{Ord}\to\operatorname{Fun}_\mathrm{inj}(\mathcal{A},\mathcal{A})$ will be obtained by taking the union of the functors $\mathbf{M}^{\leq\alpha}$.

We do so by transfinite recursion in $\alpha$.

• Base step. For $\alpha=0$, define $\mathbf{M}^{\leq 0}:S(0)\to\operatorname{Fun}_\mathrm{inj}(\mathcal{A},\mathcal{A})$ as $\mathbf{M}^{\leq 0}_0=\operatorname{id}_\mathcal{A}$.

• $\mathbf{+1}$ case. Suppose such a functor $\mathbf{M}^{\leq\alpha}$ is defined. For an ordinal $\alpha$, define $\mathbf{M}^{\leq\alpha+1}:S(\alpha+1)\to\operatorname{Fun}_\mathrm{inj}(\mathcal{A},\mathcal{A})$ as $$\begin{align*} \mathbf{M}^{\leq\alpha+1}|_{S(\alpha)} &=\mathbf{M}^{\leq\alpha}\\ \mathbf{M}^{\leq\alpha+1}_{\alpha+1} &=\mathbf{M}\circ\mathbf{M}^{\leq\alpha+1}_\alpha\\ \mu^{\leq\alpha+1}_{\alpha,\alpha+1} &=\mu_{\mathbf{M}^{\leq\alpha}_\alpha}\\ \mu^{\leq\alpha+1}_{\beta,\alpha+1} &=\mu^{\leq\alpha+1}_{\alpha,\alpha+1} \circ \mu^{\leq\alpha+1}_{\beta,\alpha}&\text{for }\beta<\alpha, \end{align*}$$ where $\mu:\operatorname{id}_\mathcal{A}\to\mathbf{M}$ is the natural transformation whose component at $M$ is the morphism $M\to\mathbf{M}(M)$ in the cocartesian square that defines $\mathbf{M}(M)$, and in the second-to-last equality we are using the whiskering notation from Categories, Section 4.28. Since $\mu$ is injective, so is the morphism in the second-to-last equality; hence also the morphism in the last equality is injective.

• Limit case. Suppose $\alpha\neq 0$ is a limit ordinal and that $\mathbf{M}^{\leq\beta}$ is defined for all $\beta<\alpha$. We define $\mathbf{M}^{\leq\alpha}$, on the one hand, by setting $\mathbf{M}^{\leq\alpha}|_{\alpha}$ to be the union of the functors $\mathbf{M}^{\leq\beta}$ for $\beta<\alpha$, and on other hand, by $$\label{colim}\tag{1} \mathbf{M}^{\leq\alpha}_\alpha=\underset{\beta<\alpha}{\operatorname{colim}}\mathbf{M}_\beta^{\leq\alpha}.$$ For $\beta<\alpha$, the map $\mu_{\beta,\alpha}^{\leq\alpha}$ is defined to be the leg at $\beta$ of the limiting cocone \eqref{colim}. Since $\mathcal{A}$ is AB5, by the Lemma in Comment #9497, $\mu_{\beta,\alpha}^{\leq\alpha}$ is injective.

no

Is there an implied 0 of the left in the exact sequence that appears in the definition of a finitely presented module?

In case it's any interesting, one could add to the statement “moreover, if $\kappa>0$, then such an ordinal is necessarily a limit ordinal.” Indeed, for an ordinal $\alpha\neq 0$, $\alpha$ is a sucessor ordinal if and only if $\operatorname{cf}\alpha=1$.

$\def\C{\mathcal{C}}$I think the phrase “so suppose to the contrary that all of the $f^{-1}(B_\beta)$ were proper subobjects of $M$” may be deleted. The argument does not do a proof by contradiction.

For the slow-thinkers out there (like me myself) the justification of the sentence “we only need to show that the map (19.2.0.1) is a surjection” is because $\operatorname{Hom}_\mathcal{A}(M,-)$ is left-exact (Homology, Lemma 12.5.8) and because of the following

Lemma. Suppose $\C$ is an AB5 abelian category. If $C=\{C_i\mid i\in I\}$ is a direct system in $\C$ such that $C_i\to C_j$ is injective for all $i\leq j$, then $C_i\to\varinjlim_{j\in I}C_j$ is injective.

Proof. We may assume that $i$ is initial in $I$ (replace $I$ by the directed subset $\{j\in\C\mid i\leq j\}$, which is cofinal, and apply Categories, Lemma 4.17.2). The components of the morphism of direct systems $C_i\Rightarrow C$ are all injective. Hence $\varinjlim(C_i\Rightarrow C)=C_i\to\varinjlim_{j\in I}C_j$ is injective. $\square$

I was a little confused by the way (2) is phrased. At the beginning I understood that what was being asserted is that for each cardinal $\kappa$ there is an object $M$ in $\mathcal{A}$ with $|M|\leq\kappa$. Maybe it's me alone, but I would phrase it as "the collection of isomorphisms classes of objects $M$ with $|M|\leq\kappa$ is a set" or something like that.

Definition is not clear. Maybe it is better to define 'standard smooth presentation' or use terms like 'there exists a presentation such that ...' in the definition. As Peter Johnson mentioned.

After we know $Q$ is additive, I think one has a direct proof by Homology, Lemma 12.3.7.

Is it argued somewhere that $K(\mathcal{A})$ has direct sums? (That is part of the requirements for a category to be triangulated.) This follows directly by application of Homology, Lemma 12.3.7 to the functor $\operatorname{Comp}(\mathcal{A})\to K(\mathcal{A})$.

Typo: In the proof, third-to-last sentence, I think it should be ${}'d_2^{p,q}={}'d_3^{p,q}=\dots=0$.

I don't understand “hence we conclude that $H^n(\operatorname{Tot}(A^{\bullet,\bullet}))=H^n(K^{\bullet})$.” I asked about this and expressed my concerns in MOs; by the time of publication of this comment no one has answered yet.

In Kashiwara, Schapira, Categories and Sheaves, a stronger result (Theorem 12.5.4) is proven without resorting to spectral sequences. It says that if $f:X\to Y$ is a morphism of double complexes that are diagonally bounded (i.e., with a finite amount of nonzero terms in each diagonal $p+q=n$) and such that $H_{II}(H_I(X))\to H_{II}(H_I(Y))$ is an isomorphism, then $\operatorname{Tot}f:\operatorname{Tot}X\to\operatorname{Tot}Y$ is a quasi-isomorphism.

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.