The Stacks project

I believe that in \tag{034R}, the divisions are the wrong way around. It should be $\cdots x_3 \mid x_2 \mid x_1$, since $x_n = x_{n+1}r_n$ for some $r_n \in R$, thus $x_{n+1} \mid x_n$. But of course the inclusions are the right way around, $x_{n+1} \mid x_n$ is equivalent to $(x_n) \subset (x_{n+1})$.

I am a little confused. We know that $x\circ s$ and $x\circ t$ are connected by a $2$-isomorphism defining $\alpha$. Hence the two "equalities" $(x\circ s)^{\ast}F| _ {R_{\acute{e}tale}}=(x\circ t)^{\ast}F| _ {R_{\acute{e}tale}}$ and $(x\circ s)^{\ast}H| _ {R_{\acute{e}tale}}=(x\circ t)^{\ast}H| _ {R_{\acute{e}tale}}$ in the commutative diagramme are in fact the canonical isomorphisms induced from $x\circ s\rightarrow x\circ t$ ? If I understand correctly, the upper horizontal morphism (after inverting $s_ {small}^{\ast} (x^{\ast} \mathcal{F}| _ {U_ {\acute{e}tale}}) \rightarrow (x \circ s) ^* \mathcal{F}| _ {R_ {\acute{e}tale}}$) is exactly the $\alpha: t^{\ast}_ {small}( x^{\ast} \mathcal{F}| _ {U_ {\acute{e}tale}})\rightarrow s^{\ast}_ {small} (x^{\ast} \mathcal{F}| _ {U_ {\acute{e}tale}})$ ?

Small typo at the end: we've shown $\mathcal{F}_x$ is free, not $\mathcal{F}$.

If I understand correctly, the topology on $F(X)$ is discrete. When you say $F$ preserves decomposition of connected components, do you in fact mean that $F$ sends connected components of an object $X$ to the $G$-orbits of $F(X)$?

I believe there is a typo in the formulation of (1). It should be map $X\rightarrow Gal(E/F)$ instead of $X\rightarrow Aut(E/F)$ in the first row, maybe?

Typo : The paragraph before Lemma 58.3.8, "we see that $F(g)s=s$ if and only if $g=\mathrm{id}_X$"

In the proof: Is the fact that $B \subset B'$ is a finite free extension used anywhere?

The freeness of the map $R \rightarrow S$ of example 00SR, is explicitly used in the proof of Lemma 03GD. Thus I suggest replacing "locally free" with "free" in the statement of the current Lemma.

End of proof: the claim is "obvious" to the same extent as the axiom of choice; in fact it is exactly the "dependent choice axiom".

The hypothesis that $M$ be a Mittag-Leffler $S$-module is redundant: by 10.88.8, this follows from from $M$ being a flat $S$-module.

I hope it is a suitable place to ask questions. In the definition of proper morphism, you require finite type but not finite presentation. Does it mean that there are proper morphisms which are not of finite presentation?

I hope it is a suitable place to ask questions. In the definition of proper morphism, you require finite type but not finite presentation. Does it mean that there are proper morphisms which are not of finite presentation?

The first displayed equation in the proof, $X_K$ should be $X_k$?

(Original code here.) For reference, this is the actual result:

Lemma. Let $(F,\xi),(G,\zeta):\mathcal{D}\rightrightarrows\mathcal{D}'$ be triangulated functors between pre-triangulated categories. For each object $A\in\mathcal{D}$, suppose given a morphism $\eta_A:F(A)\to G(A)$ and denote $\eta=\{\eta_A\}_{A\in\mathcal{D}}$. The following are equivalent:

1. $\eta:F\Rightarrow G$ is a natural transformation and we have a commutative triangle of natural transformations

$$\require{AMScd} \begin{CD} F\circ [1]@>{\xi}>>[1]\circ F\\ @V\eta_T VV@VVT'\eta V\\ G\circ [1]@>>\zeta> [1]\circ G \end{CD}$$

1. For every distinguished triangle $A\to B\to C\xrightarrow{h} A[1]$ in $\mathcal{D}$, the diagram $$\require{AMScd} \begin{CD} F(A)@>>> F(B) @>>> F(C)@>>> F(A)[1]\\ @VVV@VVV@VVV@VVV\\ G(A)@>>> G(B) @>>> G(C)@>>> G(A)[1] \end{CD}$$ commutes, where the vertical maps are given by $\eta_A,\eta_B,\eta_C$, and the last top and bottom arrows are $\xi_A\circ F(h)$ and $\zeta_A\circ G(h)$, respectively.

Succintly, $\eta$ is a transnatural transformation (a 2-morphism in the category of categories with translation) if and only if it is a trinatural transformation (a 2-morphism in the category of pre-triangulated categories). I wrote the proof in this question (see the Lemma at the end).

I am wondering where the hypothesis $\mathcal{F}$ is an étale sheaf is used. It seems to me that this Lemma holds true even for presheaves on $\mathcal{X}$. ((1)(3)(4) are true for presheaves since the comparison morphisms are constructed for presheaves and (2) can be verified by a direct computation.) Maybe I missed some point?

I think in the proof of lemma 27.10.4 in second paragraph third last line it should be $g\in (S_h)_{d\text{deg(g)}}$ instead of $(S_h)_{-d\text{deg(g)}}$

Sorry the formatting of my previous comment was messed up. $A$ is supposed to be the union of the $Y_i$ that don't contain $x$.

Lemma 5.9.6 (04MF): At the end of the proof, one can exhibit the open connected neighborhood of $x$ directly without having to mention $E''$. Let $A=\bigcup\\{Y_i:x\notin Y_i\\}$, which is closed in $E\cap E'$. So $V=(E\cap E')\setminus A$ is an open neighborhood of $x$. Claim: $V$ is connected. Reason: $V$ is the union of the $Y_j\setminus A$ for all the $Y_j$ containing $x$. Every open subset of an irreducible set is irreducible. So for $Y_j$ containing $x$, the set $Y_j\setminus A$ is also irreducible, hence connected. And so is their union since they all contain the common point $x$.

(math free comment) Since they are consecutives, it would be nice to put lemmas 10.50.4 and 10.50.5 together as two equivalent conditions. Also, with 10.50.15 and 10.50.16 one could produce a list of equivalent definitions like many others on the site.

In (2), it is better to replace (05Y9) by a link to this tag.