Lemma 52.12.1. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ be a short exact sequence of coherent modules on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Let $\mathcal{V}$ be the set of open subschemes $V \subset U$ containing $U \cap V(I)$ ordered by reverse inclusion. Consider the commutative diagram

\[ \xymatrix{ \mathop{\mathrm{colim}}\nolimits _\mathcal {V} H^0(V, \mathcal{F}') \ar[d] \ar[r] & \mathop{\mathrm{colim}}\nolimits _\mathcal {V} H^0(V, \mathcal{F}) \ar[d] \ar[r] & \mathop{\mathrm{colim}}\nolimits _\mathcal {V} H^0(V, \mathcal{F}'') \ar[d] \\ \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}') \ar[r] & \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}) \ar[r] & \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}'') } \]

If the left and right downarrows are isomorphisms so is the middle. If the middle and left downarrows are isomorphisms, so is the left.

**Proof.**
The sequences in the diagram are exact in the middle and the first arrow is injective. Thus the final statement follows from an easy diagram chase. For the rest of the proof we assume the left and right downward arrows are isomorphisms. A diagram chase shows that the middle downward arrow is injective. All that remains is to show that it is surjective.

We may choose finite $A$-modules $M$ and $M'$ such that $\mathcal{F}$ and $\mathcal{F}'$ are the restriction of $\widetilde{M}$ and $\widetilde{M}'$ to $U$, see Local Cohomology, Lemma 51.8.2. After replacing $M'$ by $\mathfrak a^ n M'$ for some $n \geq 0$ we may assume that $\mathcal{F}' \to \mathcal{F}$ corresponds to a module map $M' \to M$, see Cohomology of Schemes, Lemma 30.10.5. After replacing $M'$ by the image of $M' \to M$ and setting $M'' = M/M'$ we see that our short exact sequence corresponds to the restriction of the short exact sequence of coherent modules associated to the short exact sequence $0 \to M' \to M \to M'' \to 0$ of $A$-modules.

Let $\hat s \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ with image $\hat s'' \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}''/I^ n\mathcal{F}'')$. By assumption we find $V \in \mathcal{V}$ and a section $s'' \in \mathcal{F}''(V)$ mapping to $\hat s''$. Let $J \subset A$ be an ideal such that $V(J) = \mathop{\mathrm{Spec}}(A) \setminus V$. By Cohomology of Schemes, Lemma 30.10.5 after replacing $J$ by a power, we may assume there is an $A$-linear map $\varphi : J \to M''$ corresponding to $s''$. We fix this choice of $J$; in the rest of the proof we will replace $V$ by a smaller $V$ in $\mathcal{V}$, i.e, we will have $V \cap V(J) = \emptyset $.

Choose a presentation $A^{\oplus m} \to A^{\oplus n} \to J \to 0$. Denote $g_1, \ldots , g_ n \in J$ the images of the basis vectors of $A^{\oplus n}$, so that $J = (g_1, \ldots , g_ n)$. Let $A^{\oplus m} \to A^{\oplus n}$ be given by the matrix $(a_{ji})$ so that $\sum a_{ji} g_ i = 0$, $j = 1, \ldots , m$. Since $M \to M''$ is surjective, for each $i$ we can choose $m_ i \in M$ mapping to $\varphi (g_ i) \in M''$. Then the element $g_ i \hat s - m_ i$ of $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ lies in the submodule $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}')$. By assumption after shrinking $V$ we may assume there are $s'_ i \in \mathcal{F}'(V)$, $i = 1, \ldots , n$ with $s'_ i$ mapping to $g_ i \hat s - m_ i$. Set $s_ i = s'_ i + m_ i$ in $\mathcal{F}(V)$. Note that $\sum a_{ji} s_ i$ maps to $\sum a_{ji}g_ i\hat s = 0$ by the map

\[ \mathop{\mathrm{colim}}\nolimits _\mathcal {V} \mathcal{F}(V') \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

Since this map is injective (see above), we may after shrinking $V$ assume that $\sum a_{ji}s_ i = 0$ in $\mathcal{F}(V)$ for all $j = 1, \ldots , m$. Then it follows that we obtain an $A$-module map $J \to \mathcal{F}(V)$ sending $g_ i$ to $s_ i$. By the universal property of $\widetilde{J}$ this $A$-module map corresponds to an $\mathcal{O}_ V$-module map $\widetilde{J}|_ V \to \mathcal{F}$. However, since $V(J) \cap V = \emptyset $ we have $\widetilde{J}|_ V = \mathcal{O}_ V$. Thus we have produced a section $s \in \mathcal{F}(V)$. We omit the computation that shows that $s$ maps to $\hat s$ by the map displayed above.
$\square$

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