Lemma 12.19.13. 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 12.19.4. By Lemma 12.19.12 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 $f(F^ pA) = F^ pB$. $\square$

Comment #6228 by 57Jimmy on

Have I got this right? If $\mathcal{A}$ is abelian and comes endowed with a functorial filtration on $\mathrm{Id}_\mathcal{A}$ (in particular, all morphisms are compatible with the filtration), then condition (2) is automatically satisfied (as $\mathcal{A}$ is abelian), hence all morphisms are actually strictly compatible with the filtration (by (1)), and moreover the functor $\mathrm{Grad}_*(-)$ is exact (by (6)).

Comment #6229 by 57Jimmy on

(Sorry for the double comment, I don't think I can delete one). My aim was just to point out that from the lemma mentioned here one can easily deduce "global" results about abelian categories that come with a functorial filtration. It might be relevant to point this out, as there are many cases of interest (Hodge structures, motives).

Comment #6230 by 57Jimmy on

I think my first comment is flawed ((2) is not automatically satisfied), but it would still be interesting to write somewhere that $\mathrm{gr}_*(-)$ is exact iff all morphisms are strict

Comment #6231 by 57Jimmy on

Actually, I do think that, if the filtration on $\mathcal{A}$ is assumed to be bounded from below, then isomorphisms are automatically strictly compatible with the filtration (by induction on filtration steps), and so exactness of $\mathrm{gr}_*(-)$ also follows.

Comment #6232 by on

What is a counterexample where you have the filtration on $\text{id}_\mathcal{A}$ but morphisms aren't automatically strict? OK, maybe the category of finite modules $M$ over the polynomial ring $k[x]$ with filtration on $M$ given by $M_{tors} \subset M$ where $M_{tors}$ is the torsion submodule?

Comment #6233 by 57Jimmy on

@Johan Sorry for making a bit of a mess with my comments. I agree with you, my point is precisely that with $\mathcal{A}$ abelian and endowed with a functorial filtration (possibly boundedness needed, not sure), many properties follow automatically: all morphisms are strict, $\mathrm{gr}_*(-)$ is exact, $F^k$ is exact for all $k$. So it might be worth pointing this out somewhere. (I am totally unfamiliar with how this project works, so I don't know who should do it. I just thought it could be worth doing it. But if it doesn't fit with the philosophy of the site or of this specific page, just ignore all my comments :) ).

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