# The Stacks Project

## Tag 0D6M

Lemma 21.22.7. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $E \in D(\mathcal{O})$. Let $\mathcal{B} \subset \mathop{\rm Ob}\nolimits(\mathcal{C})$ be a subset. Assume

1. every object of $\mathcal{C}$ has a covering whose members are elements of $\mathcal{B}$, and
2. for every $V \in \mathcal{B}$ there exist a function $p(V, -) : \mathbf{Z} \to \mathbf{Z}$ and a cofinal system $\text{Cov}_V$ of coverings of $V$ such that $$H^p(V_i, H^{m - p}(E)) = 0$$ for all $\{V_i \to V\} \in \text{Cov}_V$ and all integers $p, m$ satisfying $p > p(V, m)$.

Then the canonical map $E \to R\mathop{\rm lim}\nolimits \tau_{\geq -n} E$ is an isomorphism in $D(\mathcal{O})$.

Proof. Set $K_n = \tau_{\geq -n}E$ and $K = R\mathop{\rm lim}\nolimits K_n$. The canonical map $E \to K$ comes from the canonical maps $E \to K_n = \tau_{\geq -n}E$. We have to show that $E \to K$ induces an isomorphism $H^m(E) \to H^m(K)$ of cohomology sheaves. In the rest of the proof we fix $m$. If $n \geq -m$, then the map $E \to \tau_{\geq -n}E = K_n$ induces an isomorphism $H^m(E) \to H^m(K_n)$. To finish the proof it suffices to show that for every $V \in \mathcal{B}$ there exists an integer $n(V) \geq -m$ such that the map $H^m(K)(V) \to H^m(K_{n(V)})(V)$ is injective. Namely, then the composition $$H^m(E)(V) \to H^m(K)(V) \to H^m(K_{n(V)})(V)$$ is a bijection and the second arrow is injective, hence the first arrow is bijective. By property (1) this will imply $H^m(E) \to H^m(K)$ is an isomorphism. Set $$n(V) = 1 + \max\{-m, p(V, m - 1) - m, -1 + p(V, m) - m, -2 + p(V, m + 1) - m\}.$$ so that in any case $n(V) \geq -m$. Claim: the maps $$H^{m - 1}(V_i, K_{n + 1}) \to H^{m - 1}(V_i, K_n) \quad\text{and}\quad H^m(V_i, K_{n + 1}) \to H^m(V_i, K_n)$$ are isomorphisms for $n \geq n(V)$ and $\{V_i \to V\} \in \text{Cov}_V$. The claim implies conditions (1) and (2) of Lemma 21.22.6 are satisfied and hence implies the desired injectivity. Recall (Derived Categories, Remark 13.12.4) that we have distinguished triangles $$H^{-n - 1}(E)[n + 1] \to K_{n + 1} \to K_n \to H^{-n - 1}(E)[n + 2]$$ Looking at the asssociated long exact cohomology sequence the claim follows if $$H^{m + n}(V_i, H^{-n - 1}(E)),\quad H^{m + n + 1}(V_i, H^{-n - 1}(E)),\quad H^{m + n + 2}(V_i, H^{-n - 1}(E))$$ are zero for $n \geq n(V)$ and $\{V_i \to V\} \in \text{Cov}_V$. This follows from our choice of $n(V)$ and the assumption in the lemma. $\square$

The code snippet corresponding to this tag is a part of the file sites-cohomology.tex and is located in lines 4236–4254 (see updates for more information).

\begin{lemma}
\label{lemma-is-limit-per-object}
Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $E \in D(\mathcal{O})$.
Let $\mathcal{B} \subset \Ob(\mathcal{C})$ be a subset. Assume
\begin{enumerate}
\item every object of $\mathcal{C}$ has a covering whose members
are elements of $\mathcal{B}$, and
\item for every $V \in \mathcal{B}$ there exist a function
$p(V, -) : \mathbf{Z} \to \mathbf{Z}$ and a cofinal system $\text{Cov}_V$
of coverings of $V$ such that
$$H^p(V_i, H^{m - p}(E)) = 0$$
for all $\{V_i \to V\} \in \text{Cov}_V$ and all integers $p, m$
satisfying $p > p(V, m)$.
\end{enumerate}
Then the canonical map $E \to R\lim \tau_{\geq -n} E$
is an isomorphism in $D(\mathcal{O})$.
\end{lemma}

\begin{proof}
Set $K_n = \tau_{\geq -n}E$ and $K = R\lim K_n$.
The canonical map $E \to K$
comes from the canonical maps $E \to K_n = \tau_{\geq -n}E$.
We have to show that $E \to K$ induces an isomorphism
$H^m(E) \to H^m(K)$ of cohomology sheaves. In the rest of the
proof we fix $m$. If $n \geq -m$, then
the map $E \to \tau_{\geq -n}E = K_n$ induces an isomorphism
$H^m(E) \to H^m(K_n)$.
To finish the proof it suffices to show that for every $V \in \mathcal{B}$
there exists an integer $n(V) \geq -m$ such that the map
$H^m(K)(V) \to H^m(K_{n(V)})(V)$ is injective. Namely, then
the composition
$$H^m(E)(V) \to H^m(K)(V) \to H^m(K_{n(V)})(V)$$
is a bijection and the second arrow is injective, hence the
first arrow is bijective. By property (1) this will imply
$H^m(E) \to H^m(K)$ is an isomorphism. Set
$$n(V) = 1 + \max\{-m, p(V, m - 1) - m, -1 + p(V, m) - m, -2 + p(V, m + 1) - m\}.$$
so that in any case $n(V) \geq -m$. Claim: the maps
$$H^{m - 1}(V_i, K_{n + 1}) \to H^{m - 1}(V_i, K_n) \quad\text{and}\quad H^m(V_i, K_{n + 1}) \to H^m(V_i, K_n)$$
are isomorphisms for $n \geq n(V)$ and $\{V_i \to V\} \in \text{Cov}_V$.
The claim implies conditions
(1) and (2) of Lemma \ref{lemma-cohomology-derived-limit-injective}
are satisfied and hence implies the desired injectivity.
Recall (Derived Categories, Remark
\ref{derived-remark-truncation-distinguished-triangle})
that we have distinguished triangles
$$H^{-n - 1}(E)[n + 1] \to K_{n + 1} \to K_n \to H^{-n - 1}(E)[n + 2]$$
Looking at the asssociated long exact cohomology sequence the claim follows if
$$H^{m + n}(V_i, H^{-n - 1}(E)),\quad H^{m + n + 1}(V_i, H^{-n - 1}(E)),\quad H^{m + n + 2}(V_i, H^{-n - 1}(E))$$
are zero for $n \geq n(V)$ and $\{V_i \to V\} \in \text{Cov}_V$.
This follows from our choice of $n(V)$
and the assumption in the lemma.
\end{proof}

Comment #2465 by anonymous on March 26, 2017 a 2:53 pm UTC

Two pedagogical suggestions:

(i) In item (2) of the lemma it would be good to add "for all integers $p$, $m$ satisfying".

(ii) The $m$ in the proof has a different role than the $m$ in the formulation of item (2) in the lemma. I found this a bit confusing. Could you use a different letter in item (2)?

Comment #2501 by Johan (site) on April 13, 2017 a 11:45 pm UTC

OK, I followed the first suggestion, but not the second. The reason is that actually roughly the $m = m_p$ in the proof is more or less the $m = m_s$ in statement (2). Namely, in the end we see that we are using condition in (2) for $m_s = m_p - 1$, $m_s = m_p$ and $m_s = m_p + 1$ (in the penultimate sentence of the proof). So they kind of play the same role. Alternatively, you could formulate this lemma one cohomology sheaf at a time and then you'd need the condition of (2) for that cohomological degree and the two adjacent ones. The change is here. Many thanks!

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