Lemma 12.16.13. Let $\mathcal{A}$ be an abelian category. Let $f : A \to B$ be a morphism of finite filtered objects of $\mathcal{A}$. The following are equivalent

1. $f$ is strict,

2. the morphism $\mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f)$ is an isomorphism,

3. $\text{gr}(\mathop{\mathrm{Coim}}(f)) \to \text{gr}(\mathop{\mathrm{Im}}(f))$ is an isomorphism,

4. the sequence $\text{gr}(\mathop{\mathrm{Ker}}(f)) \to \text{gr}(A) \to \text{gr}(B)$ is exact,

5. the sequence $\text{gr}(A) \to \text{gr}(B) \to \text{gr}(\mathop{\mathrm{Coker}}(f))$ is exact, and

6. the sequence

$0 \to \text{gr}(\mathop{\mathrm{Ker}}(f)) \to \text{gr}(A) \to \text{gr}(B) \to \text{gr}(\mathop{\mathrm{Coker}}(f)) \to 0$

is exact.

Proof. The equivalence of (1) and (2) is Lemma 12.16.4. By Lemma 12.16.12 we see that (4), (5), (6) imply (3) and that (3) implies (4), (5), (6). Hence it suffices to show that (3) implies (2). Thus we have to show that if $f : A \to B$ is an injective and surjective map of finite filtered objects which induces and isomorphism $\text{gr}(A) \to \text{gr}(B)$, then $f$ induces an isomorphism of filtered objects. In other words, we have to show that $f(F^ pA) = F^ pB$ for all $p$. As the filtrations are finite we may prove this by descending induction on $p$. Suppose that $f(F^{p + 1}A) = F^{p + 1}B$. Then commutative diagram

$\xymatrix{ 0 \ar[r] & F^{p + 1}A \ar[r] \ar[d]^ f & F^ pA \ar[r] \ar[d]^ f & \text{gr}^ p(A) \ar[r] \ar[d]^{\text{gr}^ p(f)} & 0 \\ 0 \ar[r] & F^{p + 1}B \ar[r] & F^ pB \ar[r] & \text{gr}^ p(B) \ar[r] & 0 }$

and the five lemma imply that $f(F^ pA) = F^ pB$. $\square$

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