# The Stacks Project

## Tag: 012W

This tag has label homology-lemma-biregular-ss-converges and it points to

The corresponding content:

Lemma 11.18.9. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet, F)$ be a filtered complex of $\mathcal{A}$. Assume that the filtration on each $K^n$ is finite (see Definition 11.13.1). Then
1. the filtration on each $H^n(K^\bullet)$ is finite, and
2. the spectral sequence associated to $(K^\bullet, F)$ converges.

Proof. Part (1) is clear from Equation (11.18.6.1). We will use Lemma 11.18.8 to prove part (2). Fix $p, n \in \mathbf{Z}$. Look at the left hand side of Equation (11.18.6.3). The expression is equal to the right hand side since $F^mK^{n - 1} = 0$ for $m \ll 0$. Similarly, use $F^mK^{n + 1} = K^{n + 1}$ for $m \gg 0$ to prove equality in Equation (11.18.6.2). $\square$

\begin{lemma}
\label{lemma-biregular-ss-converges}
Let $\mathcal{A}$ be an abelian category.
Let $(K^\bullet, F)$ be a filtered complex of $\mathcal{A}$.
Assume that the filtration on each $K^n$ is finite
(see Definition \ref{definition-filtered}).
Then
\begin{enumerate}
\item the filtration on each $H^n(K^\bullet)$ is finite, and
\item the spectral sequence associated to $(K^\bullet, F)$ converges.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) is clear from Equation (\ref{equation-filtration-cohomology}).
We will use Lemma \ref{lemma-filtered-complex-ss-converges} to prove
part (2). Fix $p, n \in \mathbf{Z}$. Look at the left hand side of
equal to the right hand side since $F^mK^{n - 1} = 0$ for
$m \ll 0$. Similarly, use $F^mK^{n + 1} = K^{n + 1}$ for
$m \gg 0$ to prove equality in Equation (\ref{equation-on-top-bigraded}).
\end{proof}


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