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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