Lemma 12.19.15. Let $\mathcal{A}$ be an abelian category. Let $A \to B \to C$ be a complex of filtered objects of $\mathcal{A}$. Assume $A, B, C$ have finite filtrations and that $\text{gr}(A) \to \text{gr}(B) \to \text{gr}(C)$ is exact. Then
for each $p \in \mathbf{Z}$ the sequence $\text{gr}^ p(A) \to \text{gr}^ p(B) \to \text{gr}^ p(C)$ is exact,
for each $p \in \mathbf{Z}$ the sequence $F^ p(A) \to F^ p(B) \to F^ p(C)$ is exact,
for each $p \in \mathbf{Z}$ the sequence $A/F^ p(A) \to B/F^ p(B) \to C/F^ p(C)$ is exact,
the maps $A \to B$ and $B \to C$ are strict, and
$A \to B \to C$ is exact (as a sequence in $\mathcal{A}$).
Proof.
Part (1) is immediate from the definitions. We will prove (3) by induction on the length of the filtrations. If each of $A$, $B$, $C$ has only one nonzero graded part, then (3) holds as $\text{gr}(A) = A$, etc. Let $n$ be the largest integer such that at least one of $F^ nA, F^ nB, F^ nC$ is nonzero. Set $A' = A/F^ nA$, $B' = B/F^ nB$, $C' = C/F^ nC$ with induced filtrations. Note that $\text{gr}(A) = F^ nA \oplus \text{gr}(A')$ and similarly for $B$ and $C$. The induction hypothesis applies to $A' \to B' \to C'$, which implies that $A/F^ p(A) \to B/F^ p(B) \to C/F^ p(C)$ is exact for $p \geq n$. To conclude the same for $p = n + 1$, i.e., to prove that $A \to B \to C$ is exact we use the commutative diagram
\[ \xymatrix{ 0 \ar[r] & F^ nA \ar[r] \ar[d] & A \ar[r] \ar[d] & A' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & F^ nB \ar[r] \ar[d] & B \ar[r] \ar[d] & B' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & F^ nC \ar[r] & C \ar[r] & C' \ar[r] & 0 } \]
whose rows are short exact sequences of objects of $\mathcal{A}$. The proof of (2) is dual. Of course (5) follows from (2).
To prove (4) denote $f : A \to B$ and $g : B \to C$ the given morphisms. We know that $f(F^ p(A)) = \mathop{\mathrm{Ker}}(F^ p(B) \to F^ p(C))$ by (2) and $f(A) = \mathop{\mathrm{Ker}}(g)$ by (5). Hence $f(F^ p(A)) = \mathop{\mathrm{Ker}}(F^ p(B) \to F^ p(C)) = \mathop{\mathrm{Ker}}(g) \cap F^ p(B) = f(A) \cap F^ p(B)$ which proves that $f$ is strict. The proof that $g$ is strict is dual to this.
$\square$
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