The Stacks project

Trivial typo in first line of statement: 'finite morphism OF algebraic".

SS: The canonical map between a presheaf of sets on a site and its corresponding zeroth Čech cohomology admits, for every object and every section of the zeroth Čech cohomology over that object, a covering such that each restriction of the section corresponding to that covering is in the image of this canonical map when applied over the object generating such restriction.

SS: A pair of covergins such that each refines the other induces a canonical identification of the corresponding zeroth Čech cohomologies of any presheaf of sets on the site with respect to these two coverings.

SS: Two morphisms between the same pair of covergins of a site inducing (covariantly) the same morphism on the objects of the site covered by these coverings induce (contravariantly) the same map on the zeroth Čech cohomologies of any presheaf of sets on the site with respect to these two coverings.

SS: For any given pair of coverings of an object of a site, there exists a covering which is a common refinement of these two coverings.

SS: The association sending a presheaf to the map of presheaves sending this presheaf to its zeroth Čech cohomology is a functor.

SS: The colimit, when the coverings vary over the opposite of the category of all coverings of a fixed object on a site, of the zeroth Čech cohomology of a fixed presheaf of sets with respect to these coverings of the fixed object on the site defines a presheaf (the zeroth Čech cohomology of the fixed presheaf of sets over the fixed object on the site) and a canonical map of presheaves from the original sheaf to the one produced here via the colimit.

SS: Limit of diagram of sheaves exists and coincides with limit as presheaves.

A minor typo: The third line from bottom "is a coequalizer diagram in $\mathop{\mathit{Sh}}\nolimits (F_{\acute{e}tale})$" should be "is a coequalizer diagram in $\mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{\acute{e}tale})$"

Also a small question: The conclusion $\pi _ F \circ i_ F = \text{id}$ does it include that the composition of structure ring maps $\mathcal{O}_F \rightarrow \pi _{F,\ast}\mathcal{O} _{\mathcal{X}}\rightarrow \pi _{F,\ast}i _{F,\ast}\mathcal{O} _F$ is isomorphic to identity?

Instead of this lemma, one could prove the following more general result and obtain Lemma 17.28.11 as a Corollary:

Lemma. Let $X$ be a topological space, let $\mathcal{O}$ be a sheaf of rings over $X$ and let $\mathcal{A},\mathcal{B}$ be sheaves of $\mathcal{O}$-algebras. Suppose $\mathcal{I}\subset\mathcal{B}$ is an ideal sheaf with $\mathcal{I}^2=0$. Then $\mathcal{I}$ is naturally a $\mathcal{B}/\mathcal{I}$-module. Let $\varphi:\mathcal{A}\to\mathcal{B}/\mathcal{I}$ be a morphism of $\mathcal{O}$-algebras ($\mathcal{I}$ becomes an $\mathcal{A}$-module via $\varphi$). Denote $\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$ to the set of $\mathcal{O}$-algebra homomorphisms $\mathcal{A}\to\mathcal{B}$ that lift $\varphi$ to $\mathcal{B}$. Then $\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$ is a $\operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})$-torsor via $$\begin{align*} \operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})\times\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})&\to\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})\\ (D,\psi)&\mapsto D+\psi. \end{align*}$$

Proof. The map is well-defined: It is clear that $D+\psi$ is $\mathcal{O}$-linear, that it preserves the multiplicative unit and that as a map of sheaves of abelian groups, it's a lifting of $\varphi$ to $\mathcal{B}$. It is left to show that $D+\psi$ is multiplicative. For this, we need first to prove the following: Let $\psi\in\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$, and let $h,\tau$ be local sections respectively of $\mathcal{A}$ and of $\mathcal{I}$, over the same open subset $U\subset X$. Then we claim that $h\tau=\psi(h)\tau$, i.e., that $\varphi(h)\tau=\psi(h)\tau$. It suffices to verify the equality on germs at $x\in U$, and we leave this as an exercise to the reader (use the natural $\mathcal{B}_x/\mathcal{I}_x$-module structure on $\mathcal{I}_x$). Now, one has $$\begin{align*} (Df+\psi f)(Dg+\psi g) &=\psi(fg)+(\psi f)Dg+(\psi g)Df,&\text{for }\mathcal{I}^2=0,\\ &=\psi(fg)+fDg+gDf,&\text{by the claim,}\\ &=\psi(fg)+D(fg),&\text{by Leibniz's rule.} \end{align*}$$ Thus, $D+\psi\in \operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$ and the map from the statement constitutes a group action of $\operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})$ on $\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$. It is clear that the action is free. To show transitivity, let $\psi,\psi'\in\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$. We have to show that $D=\psi'-\psi\in\operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})$. On the one hand, it is clear that $D$ is $\mathcal{O}$-linear. On the other hand, using the same facts as in the last computation, we get $$\begin{align*} \psi'(fg)=\psi'(f)\psi'(g)&=(\psi(f)+D(f))(\psi(g)+D(g))\\ &=\psi(fg)+fD(g)+gD(f), \end{align*}$$ which shows Leibniz's rule for $D$. $\square$

In the statement of the theorem, I think one could write "then we have an isomorphism $\Omega_{\mathcal{O}_2/\mathcal{O}_1,x}\cong\Omega_{\mathcal{O}_{2,x}/\mathcal{O}_{1,x}}$ compatible with universal derivations." (I.e., that $\varphi\circ \mathrm{d}_{\mathcal{O}_2/\mathcal{O}_1,x}=\mathrm{d}_{\mathcal{O}_{2,x}/\mathcal{O}_{1,x}}$, where $\varphi$ is the isomorphism.) The proof doesn't change, for Lemma 17.28.6 guarantees this on its statement.

In case anyone needs it, here's a very detailed proof of the last three equalities:

Along the proof we will use the adjunction $$\tag{1}\label{1} \operatorname{Hom}_{\mathcal{O}}(\mathcal{O}[\mathcal{F}],\mathcal{G})\cong\operatorname{Hom}(\mathcal{F},\mathcal{G}),$$ pointed out in the second paragraph of this comment.

Lemma 1. Let $\mathcal{O}$, $\mathcal{F}$ be sheaves over $X$, respectively, of rings and of sets, and let $x\in X$ be a point. There is an isomorphism of $\mathcal{O}_x$-modules $\mathcal{O}[\mathcal{F}]_x\cong\mathcal{O}_x[\mathcal{F}_x]$, that is natural on $\mathcal{F}$.

Proof. Consider the unit of the adjunction $\mathcal{F}\to\mathcal{O}[\mathcal{F}]$ and take map on stalks $\mathcal{F}_x\to\mathcal{O}[\mathcal{F}]_x$. This induces an $\mathcal{O}_x$-linear map $\mathcal{O}_x[\mathcal{F}_x]\to\mathcal{O}[\mathcal{F}]_x$ (natural on $\mathcal{F}$, by naturality of the unit). Namely, it maps $[s_x]$ to $[s]_x$. Conversely, for each open neighborhood $U\subset X$ of $x$, there is an obvious map $\mathcal{F}(U)\to\mathcal{O}_x[\mathcal{F}_x]$. Since $\mathcal{O}_x[\mathcal{F}_x]$ is an $\mathcal{O}(U)$-module, this induces a map $\mathcal{O}(U)[\mathcal{F}(U)]\to\mathcal{O}_x[\mathcal{F}_x]$. Since the latter is compatible with restrictions to open neighborhoods of $x$, we obtain a canonical map $\mathcal{O}[\mathcal{F}]_x\to\mathcal{O}_x[\mathcal{F}_x]$. But it maps $[s]_x$ to $[s_x]$, so the two maps are mutually inverse. $\square$

Lemma 2. Same notation as in Lemma 1, and suppose that $\mathcal{G}$ is a sheaf of $\mathcal{O}$-modules. Then the following square commutes: $$\require{AMScd} \begin{CD} \operatorname{Hom}_{\mathcal{O}}(\mathcal{O}[\mathcal{F}],\mathcal{G}) @>{\cong}>>\operatorname{Hom}(\mathcal{F},\mathcal{G})\\ @VVV@VVV\\ \operatorname{Hom}_{\mathcal{O}_x}(\mathcal{O}_x[\mathcal{F}_x],\mathcal{G}_x) @>>{\cong}>\operatorname{Hom}(\mathcal{F}_x,\mathcal{G}_x) \end{CD}$$ where the left arrow is obtained from the induced map on stalks plus the isomorphism of Lemma 1.

We leave the proof to the reader. In the diagram, the horizontal maps are \eqref{1} and the universal property of the free module over a set.

Lemma 3. Notations as in Lemma 1. Let $f:Y\to X$ be a continuous map of topological spaces. Then $f^{-1}(\mathcal{O}[\mathcal{F}])\cong f^{-1}\mathcal{O}[f^{-1}\mathcal{F}]$.

Proof. Pulling back the unit of the adjunction $\eta:\mathcal{F}\to\mathcal{O}[\mathcal{F}]$ gives a map $f^{-1}\mathcal{F}\to f^{-1}(\mathcal{O}[\mathcal{F}])$. Note that $f^{-1}(\mathcal{O}[\mathcal{F}])$ is a $f^{-1}\mathcal{O}$-module. By the adjunction \eqref{1}, we get a map $f^{-1}\mathcal{O}[f^{-1}\mathcal{F}]\to f^{-1}(\mathcal{O}[\mathcal{F}])$. On stalks, this map equals $$\label{2}\tag{2} (f^{-1}\mathcal{O})_x[(f^{-1}\mathcal{F})_x]\to f^{-1}(\mathcal{O}[\mathcal{F}])_x,$$ where we have used Lemma 1. By Lemma 2, \eqref{2} is adjoint to $(f^{-1}\mathcal{F})_x\to f^{-1}(\mathcal{O}[\mathcal{F}])_x$, i.e., to $\eta_x:\mathcal{F}_x\to \mathcal{O}[\mathcal{F}]_x$. In other words, the map \eqref{2} equals the canonical morphism $\mathcal{O}_x[\mathcal{F}_x]\to \mathcal{O}[\mathcal{F}]_x$ of Lemma 1, which is an isomorphism. $\square$

A few comments:

1. $\tau$ is used both in the proof and the statement in different roles.
2. Statement (3): Strictly speaking, a map $A \rightarrow S$ mod $q$ becomes $A/qA \rightarrow S/qS$ which is not of the form $\kappa(q) \rightarrow S/m_S$. So, perhaps, make this statement more precise.
3. Because this statement is an important fact, I suggest making the proof a bit more transparent. For instance, more explanation can be added in the proof as to why $A \rightarrow A \otimes_R S \rightarrow S$ reduces to $\tau$, possibly highlighting the role of the uniqueness of the splitting.

Statement (8): As far as I understand, the section $S \rightarrow R$ is unique. This follows from Lemma 06RR.

How does one argue uniqueness of $\alpha_D$? I guess it is because $D=\alpha_D\circ d$ plus facts (i) $\alpha_D$ is $\mathcal{O}_2$-linear and (ii) the image (pre)sheaf of $d$ generates $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ as an $\mathcal{O}_2$-module (in the sense of Definition 17.4.5). Does this sound right?

Also, I think it might be interesting to point out that $\mathcal{F}\mapsto\mathcal{O}[\mathcal{F}]$ is the left adjoint of the forgetful functor $Mod(\mathcal{O})\to\operatorname{Sh}_{Set}(X)$, as it is easy to show (this is actually how one obtains $\mathcal{O}_2[\mathcal{O}_2]\to\mathcal{F}$ from $D:\mathcal{O}_2\to\mathcal{F}$ in the third to last sentence of the proof).

The readers might find interesting to know the following: In Matsumura's Commutative Algebra (not to be confused with Commutative Ring Theory), Ch. 10, (26.I), in Theorem 58 it is stated a sufficient and equivalent condition for the map $I/I^2\to\Omega_{S/R}\otimes_S S'$ from Lemma 10.131.9 to have an $S'$-linear retraction (namely, that there is an $R$-algebra map $S'\to S/I^2$ which is a right inverse to the canonical map $S/I^2\to S'$). It is immediate to verify that "there exists an $R$-algebra map $S'\to S$ which is a right inverse to $S\to S'$" implies the condition between parentheses.

In the proof of the Lemma 28.26.14, where can I find ( associated ) proof of $X'_{s'} = i^{-1}(X_s)$ ?

Two very minor comments now:

1. Perhaps write $d: k' \cdot \bar{f} \rightarrow k' \cdot dx$ to indicate the class of $f$ mod $(f^2)$, instead of $d: k' \cdot f \rightarrow k' \cdot dx$.

2. Last sentence of second paragraph of the proof: "extensions" instead of "extension".

+1 for the example suggestion #8512 above

In the statement, one could change "canonical" by "unique." Here's the proof of the uniqueness:

Suppose $Q$ is an additive functor of preadditive categories. Let $X, Y$ be objects of $\mathcal{C}$ and $\alpha, \beta : X \to Y$ be morphisms in $S^{-1}\mathcal{C}$. According to Categories, Lemma 4.27.5, we may represent these by pairs $s^{-1}f, s^{-1}g$ with common denominator $s$. Then $$\begin{align*} s^{-1}f+s^{-1}g &=Q(s)^{-1}\circ Q(f)+Q(s)^{-1}\circ Q(g)\\ &=Q(s)^{-1}\circ(Q(f)+Q(g))\\ &=Q(s)^{-1}\circ Q(f+g)\\ &=s^{-1}(f+g). \end{align*}$$ Hence, the preadditive structure in $S^{-1}\mathcal{C}$ is determined by that of $\mathcal{C}$.

The reason I find interesting to add the uniqueness to this lemma is to guarantee the uniqueness in the statement of Derived Categories, Lemma 13.5.6.