## 26.7 Quasi-coherent sheaves on affines

Recall that we have defined the abstract notion of a quasi-coherent sheaf in Modules, Definition 17.10.1. In this section we show that any quasi-coherent sheaf on an affine scheme $\mathop{\mathrm{Spec}}(R)$ corresponds to the sheaf $\widetilde M$ associated to an $R$-module $M$.

Lemma 26.7.1. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. Let $M$ be an $R$-module. There exists a canonical isomorphism between the sheaf $\widetilde M$ associated to the $R$-module $M$ (Definition 26.5.3) and the sheaf $\mathcal{F}_ M$ associated to the $R$-module $M$ (Modules, Definition 17.10.6). This isomorphism is functorial in $M$. In particular, the sheaves $\widetilde M$ are quasi-coherent. Moreover, they are characterized by the following mapping property

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _ R(M, \Gamma (X, \mathcal{F}))$

for any sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$. Here a map $\alpha : \widetilde M \to \mathcal{F}$ corresponds to its effect on global sections.

Proof. By Modules, Lemma 17.10.5 we have a morphism $\mathcal{F}_ M \to \widetilde M$ corresponding to the map $M \to \Gamma (X, \widetilde M) = M$. Let $x \in X$ correspond to the prime $\mathfrak p \subset R$. The induced map on stalks are the maps $\mathcal{O}_{X, x} \otimes _ R M \to M_{\mathfrak p}$ which are isomorphisms because $R_{\mathfrak p} \otimes _ R M = M_{\mathfrak p}$. Hence the map $\mathcal{F}_ M \to \widetilde M$ is an isomorphism. The mapping property follows from the mapping property of the sheaves $\mathcal{F}_ M$. $\square$

Lemma 26.7.2. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. There are canonical isomorphisms

1. $\widetilde{M \otimes _ R N} \cong \widetilde M \otimes _{\mathcal{O}_ X} \widetilde N$, see Modules, Section 17.15.

2. $\widetilde{\text{T}^ n(M)} \cong \text{T}^ n(\widetilde M)$, $\widetilde{\text{Sym}^ n(M)} \cong \text{Sym}^ n(\widetilde M)$, and $\widetilde{\wedge ^ n(M)} \cong \wedge ^ n(\widetilde M)$, see Modules, Section 17.20.

3. if $M$ is a finitely presented $R$-module, then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \widetilde N) \cong \widetilde{\mathop{\mathrm{Hom}}\nolimits _ R(M, N)}$, see Modules, Section 17.21.

First proof. Using Lemma 26.7.1 and Modules, Lemma 17.10.5 we see that the functor $M \mapsto \widetilde M$ can be viewed as $\pi ^*$ for a morphism $\pi$ of ringed spaces. And pulling back modules commutes with tensor constructions by Modules, Lemmas 17.15.4 and 17.20.3. The morphism $\pi : (X, \mathcal{O}_ X) \to (\{ *\} , R)$ is flat for example because the stalks of $\mathcal{O}_ X$ are localizations of $R$ (Lemma 26.5.4) and hence flat over $R$. Thus pullback by $\pi$ commutes with internal hom if the first module is finitely presented by Modules, Lemma 17.21.4. $\square$

Second proof. Proof of (1). By Lemma 26.7.1 to give a map $\widetilde{M \otimes _ R N}$ into $\widetilde M \otimes _{\mathcal{O}_ X} \widetilde N$ we have to give a map on global sections $M \otimes _ R N \to \Gamma (X, \widetilde M \otimes _{\mathcal{O}_ X} \widetilde N)$ which exists by definition of the tensor product of sheaves of modules. To see that this map is an isomorphism it suffices to check that it is an isomorphism on stalks. And this follows from the description of the stalks of $\widetilde{M}$ (either in Lemma 26.5.4 or in Modules, Lemma 17.10.5), the fact that tensor product commutes with localization (Algebra, Lemma 10.11.16) and Modules, Lemma 17.15.1.

Proof of (2). This is similar to the proof of (1), using Algebra, Lemma 10.12.5 and Modules, Lemma 17.20.2.

Proof of (3). Since the construction $M \mapsto \widetilde{M}$ is functorial there is an $R$-linear map $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M}, \widetilde{N})$. The target of this map is the global sections of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \widetilde N)$. Hence by Lemma 26.7.1 we obtain a map of $\mathcal{O}_ X$-modules $\widetilde{\mathop{\mathrm{Hom}}\nolimits _ R(M, N)} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \widetilde N)$. We check that this is an isomorphism by comparing stalks. If $M$ is finitely presented as an $R$-module then $\widetilde M$ has a global finite presentation as an $\mathcal{O}_ X$-module. Hence we conclude using Algebra, Lemma 10.10.2 and Modules, Lemma 17.21.3. $\square$

Third proof of part (1). For any $\mathcal{O}_ X$-module $\mathcal{F}$ we have the following isomorphisms functorial in $M$, $N$, and $\mathcal{F}$

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M} \otimes _{\mathcal{O} _ X} \widetilde{N}, \mathcal{F}) & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O} _ X} (\widetilde{N}, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M, \Gamma (X, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde{N}, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{N}, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, \Gamma (X,\mathcal{F}))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, \Gamma (X, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M \otimes _ R N}, \mathcal{F}) \end{align*}

The first equality is Modules, Lemma 17.21.1. The second equality is the universal property of $\widetilde{M}$, see Lemma 26.7.1. The third equality holds by definition of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits$. The fourth equality is the universal property of $\widetilde{N}$. Then fifth equality is Algebra, Lemma 10.11.8. The final equality is the universal property of $\widetilde{M \otimes _ R N}$. By the Yoneda lemma (Categories, Lemma 4.3.5) we obtain (1). $\square$

Lemma 26.7.3. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(S), \mathcal{O}_{\mathop{\mathrm{Spec}}(S)})$, $(Y, \mathcal{O}_ Y) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be affine schemes. Let $\psi : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of affine schemes, corresponding to the ring map $\psi ^\sharp : R \to S$ (see Lemma 26.6.5).

1. We have $\psi ^* \widetilde M = \widetilde{S \otimes _ R M}$ functorially in the $R$-module $M$.

2. We have $\psi _* \widetilde N = \widetilde{N_ R}$ functorially in the $S$-module $N$.

Proof. The first assertion follows from the identification in Lemma 26.7.1 and the result of Modules, Lemma 17.10.7. The second assertion follows from the fact that $\psi ^{-1}(D(f)) = D(\psi ^\sharp (f))$ and hence

$\psi _* \widetilde N(D(f)) = \widetilde N(D(\psi ^\sharp (f))) = N_{\psi ^\sharp (f)} = (N_ R)_ f = \widetilde{N_ R}(D(f))$

as desired. $\square$

Lemma 26.7.3 above says in particular that if you restrict the sheaf $\widetilde M$ to a standard affine open subspace $D(f)$, then you get $\widetilde{M_ f}$. We will use this from now on without further mention.

Lemma 26.7.4. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then $\mathcal{F}$ is isomorphic to the sheaf associated to the $R$-module $\Gamma (X, \mathcal{F})$.

Proof. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Since every standard open $D(f)$ is quasi-compact we see that $X$ is a locally quasi-compact, i.e., every point has a fundamental system of quasi-compact neighbourhoods, see Topology, Definition 5.13.1. Hence by Modules, Lemma 17.10.8 for every prime $\mathfrak p \subset R$ corresponding to $x \in X$ there exists an open neighbourhood $x \in U \subset X$ such that $\mathcal{F}|_ U$ is isomorphic to the quasi-coherent sheaf associated to some $\mathcal{O}_ X(U)$-module $M$. In other words, we get an open covering by $U$'s with this property. By Lemma 26.5.1 for example we can refine this covering to a standard open covering. Thus we get a covering $\mathop{\mathrm{Spec}}(R) = \bigcup D(f_ i)$ and $R_{f_ i}$-modules $M_ i$ and isomorphisms $\varphi _ i : \mathcal{F}|_{D(f_ i)} \to \mathcal{F}_{M_ i}$ for some $R_{f_ i}$-module $M_ i$. On the overlaps we get isomorphisms

$\xymatrix{ \mathcal{F}_{M_ i}|_{D(f_ if_ j)} \ar[rr]^{\varphi _ i^{-1}|_{D(f_ if_ j)}} & & \mathcal{F}|_{D(f_ if_ j)} \ar[rr]^{\varphi _ j|_{D(f_ if_ j)}} & & \mathcal{F}_{M_ j}|_{D(f_ if_ j)}. }$

Let us denote these $\psi _{ij}$. It is clear that we have the cocycle condition

$\psi _{jk}|_{D(f_ if_ jf_ k)} \circ \psi _{ij}|_{D(f_ if_ jf_ k)} = \psi _{ik}|_{D(f_ if_ jf_ k)}$

on triple overlaps.

Recall that each of the open subspaces $D(f_ i)$, $D(f_ if_ j)$, $D(f_ if_ jf_ k)$ is an affine scheme. Hence the sheaves $\mathcal{F}_{M_ i}$ are isomorphic to the sheaves $\widetilde M_ i$ by Lemma 26.7.1 above. In particular we see that $\mathcal{F}_{M_ i}(D(f_ if_ j)) = (M_ i)_{f_ j}$, etc. Also by Lemma 26.7.1 above we see that $\psi _{ij}$ corresponds to a unique $R_{f_ if_ j}$-module isomorphism

$\psi _{ij} : (M_ i)_{f_ j} \longrightarrow (M_ j)_{f_ i}$

namely, the effect of $\psi _{ij}$ on sections over $D(f_ if_ j)$. Moreover these then satisfy the cocycle condition that

$\xymatrix{ (M_ i)_{f_ jf_ k} \ar[rd]_{\psi _{ij}} \ar[rr]^{\psi _{ik}} & & (M_ k)_{f_ if_ j} \\ & (M_ j)_{f_ if_ k} \ar[ru]_{\psi _{jk}} }$

commutes (for any triple $i, j, k$).

Now Algebra, Lemma 10.23.5 shows that there exist an $R$-module $M$ such that $M_ i = M_{f_ i}$ compatible with the morphisms $\psi _{ij}$. Consider $\mathcal{F}_ M = \widetilde M$. At this point it is a formality to show that $\widetilde M$ is isomorphic to the quasi-coherent sheaf $\mathcal{F}$ we started out with. Namely, the sheaves $\mathcal{F}$ and $\widetilde M$ give rise to isomorphic sets of glueing data of sheaves of $\mathcal{O}_ X$-modules with respect to the covering $X = \bigcup D(f_ i)$, see Sheaves, Section 6.33 and in particular Lemma 6.33.4. Explicitly, in the current situation, this boils down to the following argument: Let us construct an $R$-module map

$M \longrightarrow \Gamma (X, \mathcal{F}).$

Namely, given $m \in M$ we get $m_ i = m/1 \in M_{f_ i} = M_ i$ by construction of $M$. By construction of $M_ i$ this corresponds to a section $s_ i \in \mathcal{F}(U_ i)$. (Namely, $\varphi ^{-1}_ i(m_ i)$.) We claim that $s_ i|_{D(f_ if_ j)} = s_ j|_{D(f_ if_ j)}$. This is true because, by construction of $M$, we have $\psi _{ij}(m_ i) = m_ j$, and by the construction of the $\psi _{ij}$. By the sheaf condition of $\mathcal{F}$ this collection of sections gives rise to a unique section $s$ of $\mathcal{F}$ over $X$. We leave it to the reader to show that $m \mapsto s$ is a $R$-module map. By Lemma 26.7.1 we obtain an associated $\mathcal{O}_ X$-module map

$\widetilde M \longrightarrow \mathcal{F}.$

By construction this map reduces to the isomorphisms $\varphi _ i^{-1}$ on each $D(f_ i)$ and hence is an isomorphism. $\square$

Lemma 26.7.5. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. The functors $M \mapsto \widetilde M$ and $\mathcal{F} \mapsto \Gamma (X, \mathcal{F})$ define quasi-inverse equivalences of categories

$\xymatrix{ \mathit{QCoh}(\mathcal{O}_ X) \ar@<1ex>[r] & \text{Mod-}R \ar@<1ex>[l] }$

between the category of quasi-coherent $\mathcal{O}_ X$-modules and the category of $R$-modules.

From now on we will not distinguish between quasi-coherent sheaves on affine schemes and sheaves of the form $\widetilde M$.

Lemma 26.7.6. Let $X = \mathop{\mathrm{Spec}}(R)$ be an affine scheme. Kernels and cokernels of maps of quasi-coherent $\mathcal{O}_ X$-modules are quasi-coherent.

Proof. This follows from the exactness of the functor $\widetilde{\ }$ since by Lemma 26.7.1 we know that any map $\psi : \widetilde{M} \to \widetilde{N}$ comes from an $R$-module map $\varphi : M \to N$. (So we have $\mathop{\mathrm{Ker}}(\psi ) = \widetilde{\mathop{\mathrm{Ker}}(\varphi )}$ and $\mathop{\mathrm{Coker}}(\psi ) = \widetilde{\mathop{\mathrm{Coker}}(\varphi )}$.) $\square$

Lemma 26.7.7. Let $X = \mathop{\mathrm{Spec}}(R)$ be an affine scheme. The direct sum of an arbitrary collection of quasi-coherent sheaves on $X$ is quasi-coherent. The same holds for colimits.

Proof. Suppose $\mathcal{F}_ i$, $i \in I$ is a collection of quasi-coherent sheaves on $X$. By Lemma 26.7.5 above we can write $\mathcal{F}_ i = \widetilde{M_ i}$ for some $R$-module $M_ i$. Set $M = \bigoplus M_ i$. Consider the sheaf $\widetilde{M}$. For each standard open $D(f)$ we have

$\widetilde{M}(D(f)) = M_ f = \left(\bigoplus M_ i\right)_ f = \bigoplus M_{i, f}.$

Hence we see that the quasi-coherent $\mathcal{O}_ X$-module $\widetilde{M}$ is the direct sum of the sheaves $\mathcal{F}_ i$. A similar argument works for general colimits. $\square$

Lemma 26.7.8. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. Suppose that

$0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$

is a short exact sequence of sheaves of $\mathcal{O}_ X$-modules. If two out of three are quasi-coherent then so is the third.

Proof. This is clear in case both $\mathcal{F}_1$ and $\mathcal{F}_2$ are quasi-coherent because the functor $M \mapsto \widetilde M$ is exact, see Lemma 26.5.4. Similarly in case both $\mathcal{F}_2$ and $\mathcal{F}_3$ are quasi-coherent. Now, suppose that $\mathcal{F}_1 = \widetilde M_1$ and $\mathcal{F}_3 = \widetilde M_3$ are quasi-coherent. Set $M_2 = \Gamma (X, \mathcal{F}_2)$. We claim it suffices to show that the sequence

$0 \to M_1 \to M_2 \to M_3 \to 0$

is exact. Namely, if this is the case, then (by using the mapping property of Lemma 26.7.1) we get a commutative diagram

$\xymatrix{ 0 \ar[r] & \widetilde M_1 \ar[r] \ar[d] & \widetilde M_2 \ar[r] \ar[d] & \widetilde M_3 \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_1 \ar[r] & \mathcal{F}_2 \ar[r] & \mathcal{F}_3 \ar[r] & 0 }$

and we win by the snake lemma.

The “correct” argument here would be to show first that $H^1(X, \mathcal{F}) = 0$ for any quasi-coherent sheaf $\mathcal{F}$. This is actually not all that hard, but it is perhaps better to postpone this till later. Instead we use a small trick.

Pick $m \in M_3 = \Gamma (X, \mathcal{F}_3)$. Consider the following set

$I = \{ f \in R \mid \text{the element }fm\text{ comes from }M_2\} .$

Clearly this is an ideal. It suffices to show $1 \in I$. Hence it suffices to show that for any prime $\mathfrak p$ there exists an $f \in I$, $f \not\in \mathfrak p$. Let $x \in X$ be the point corresponding to $\mathfrak p$. Because surjectivity can be checked on stalks there exists an open neighbourhood $U$ of $x$ such that $m|_ U$ comes from a local section $s \in \mathcal{F}_2(U)$. In fact we may assume that $U = D(f)$ is a standard open, i.e., $f \in R$, $f \not\in \mathfrak p$. We will show that for some $N \gg 0$ we have $f^ N \in I$, which will finish the proof.

Take any point $z \in V(f)$, say corresponding to the prime $\mathfrak q \subset R$. We can also find a $g \in R$, $g \not\in \mathfrak q$ such that $m|_{D(g)}$ lifts to some $s' \in \mathcal{F}_2(D(g))$. Consider the difference $s|_{D(fg)} - s'|_{D(fg)}$. This is an element $m'$ of $\mathcal{F}_1(D(fg)) = (M_1)_{fg}$. For some integer $n = n(z)$ the element $f^ n m'$ comes from some $m'_1 \in (M_1)_ g$. We see that $f^ n s$ extends to a section $\sigma$ of $\mathcal{F}_2$ on $D(f) \cup D(g)$ because it agrees with the restriction of $f^ n s' + m'_1$ on $D(f) \cap D(g) = D(fg)$. Moreover, $\sigma$ maps to the restriction of $f^ n m$ to $D(f) \cup D(g)$.

Since $V(f)$ is quasi-compact, there exists a finite list of elements $g_1, \ldots , g_ m \in R$ such that $V(f) \subset \bigcup D(g_ j)$, an integer $n > 0$ and sections $\sigma _ j \in \mathcal{F}_2(D(f) \cup D(g_ j))$ such that $\sigma _ j|_{D(f)} = f^ n s$ and $\sigma _ j$ maps to the section $f^ nm|_{D(f) \cup D(g_ j)}$ of $\mathcal{F}_3$. Consider the differences

$\sigma _ j|_{D(f) \cup D(g_ jg_ k)} - \sigma _ k|_{D(f) \cup D(g_ jg_ k)}.$

These correspond to sections of $\mathcal{F}_1$ over $D(f) \cup D(g_ jg_ k)$ which are zero on $D(f)$. In particular their images in $\mathcal{F}_1(D(g_ jg_ k)) = (M_1)_{g_ jg_ k}$ are zero in $(M_1)_{g_ jg_ kf}$. Thus some high power of $f$ kills each and every one of these. In other words, the elements $f^ N \sigma _ j$, for some $N \gg 0$ satisfy the glueing condition of the sheaf property and give rise to a section $\sigma$ of $\mathcal{F}_2$ over $\bigcup (D(f) \cup D(g_ j)) = X$ as desired. $\square$

Comment #5526 by Kang Taeyeoup on

I think the first proof of the lemma 01I8 a) is not clear. The proof shows that there is the isomorhpism $M\otimes_R N \to \Gamma(X,\widetilde{M}\otimes_{\mathscr{O}_X} \widetilde{N})$, and this gives the corresponding isomorphism $\widetilde{M\otimes_R N} \to \widetilde{M}\otimes_{\mathscr{O}_X}\widetilde{N}$

However, an adjunction does not guarantee that the isomorphisms in the one part goes to isomorphisms in the other part.

Suppose $F : C \to D$ and $G : D \to C$ are given and $F$ is left adjoint to $G$.

Let $\alpha : c \to Gd$ be an isomorphism, then the corresponding morphism in $Hom_D(Fc,d)$ is given by the composite $Gd \xrightarrow{F\alpha} FGc \xrightarrow{\epsilon_c} c$. Here $\epsilon : FG \to id$ is a counit. In general there is noreason that $\epsilon_c\circ F\alpha$ should be an isomorphism.

Comment #5527 by Kang Taeyeoup on

Alternatively, I believe that the proof can be done by much easier and elegant way.

We have a Hom-Tensor adjuntion in the category of $\mathscr{O}_X$-modules.

Using this with the lemma 26.7.1, we get the following bunch of natural isomorphisms.

$\mathrm{Hom} _{\mathcal{O} _X}(\widetilde{M}\otimes _{\mathcal{O} _X} \widetilde{N},F) \cong \mathrm{Hom} _{\mathcal{O} _X}(\widetilde{M},\mathrm{Hom} _{\mathcal{O} _X} (\widetilde{N},F))$ $\cong \mathrm{Hom} _R(M,\Gamma(X,\mathrm{Hom} _{\mathcal{O} _X}(\widetilde{N},F)) = \mathrm{Hom} _R(M,\mathrm{Hom} _R(N,\Gamma(X,F)))$ $\cong \mathrm{Hom} _R(M\otimes _R N,\Gamma(X,F)) \cong \mathrm{Hom} _{\mathcal{O}_X}(\widetilde{M\otimes _R N},F)$

Since this hold for any $F$, we have that $\widetilde{M}\otimes _{\mathcal{O}_X} \widetilde{N} \cong \widetilde{M\otimes _R N}$.

Comment #5631 by Felipe Monteiro on

I believe there are some typos on the proof of Lemma 01IE. , in the paragraph starting with "Take any point $z$ ...". The element $m' \in \mathcal{F}_2( D(fg) ) = (M_2)_{fg}$, since it's a difference of sections of $\mathcal{F}_2$ over $D(fg)$, and $m' \in (M_2)_g$, for the same reason.

Comment #5719 by on

@Kang Taeyeoup: You comment #5526 does not accurately portray the first proof of Lemma 26.7.2 part (1). What is says is to use the universal property of $\widetilde{M \otimes_R N}$ to construct a map $\widetilde{M \otimes_R N} \to \widetilde{M} \otimes_{\mathcal{O}_X} \widetilde{N}$ and then check it is an isomorphism by looking at stalks! So we aren't claiming at all that this follows from any kind of abstract nonsense argument.

OTOH your comment #5527 is wonderful. I have added your argument in this commit. Thanks very much.

Comment #5759 by on

@#5631: OK but $\mathcal{F}_1$ is the subsheaf of $\mathcal{F}_2$ of sections mapping to zero in $\mathcal{F}_3$ and the section we are looking at does map to zero.

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