Lemma 20.34.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)$ be an inverse system in $D(\mathcal{O}_ X)$. Let $x \in X$ and $m \in \mathbf{Z}$. Assume there exist an integer $n(x)$ and a fundamental system $\mathfrak {U}_ x$ of open neighbourhoods of $x$ such that for $U \in \mathfrak {U}_ x$

$R^1\mathop{\mathrm{lim}}\nolimits H^{m - 1}(U, K_ n) = 0$, and

$H^ m(U, K_ n) \to H^ m(U, K_{n(x)})$ is injective for $n \geq n(x)$.

Then the map on stalks $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)_ x \to H^ m(K_{n(x)})_ x$ is injective.

**Proof.**
Let $\gamma $ be an element of $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)_ x$ which maps to zero in $H^ m(K_{n(x)})_ x$. Since $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)$ is the sheafification of $U \mapsto H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ (by Lemma 20.32.3) we can choose $U \in \mathfrak {U}_ x$ and an element $\tilde\gamma \in H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ mapping to $\gamma $. Then $\tilde\gamma $ maps to $\tilde\gamma _{n(x)} \in H^ m(U, K_{n(x)})$. Using that $H^ m(K_{n(x)})$ is the sheafification of $U \mapsto H^ m(U, K_{n(x)})$ (by Lemma 20.32.3 again) we see that after shrinking $U$ we may assume that $\tilde\gamma _{n(x)} = 0$. For this $U$ we consider the short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{m - 1}(U, K_ n) \to H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits H^ m(U, K_ n) \to 0 \]

of Lemma 20.34.1. By assumption (1) the group on the left is zero and by assumption (2) the group on the right maps injectively into $H^ m(U, K_{n(x)})$. We conclude $\tilde\gamma = 0$ and hence $\gamma = 0$ as desired.
$\square$

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