Is every filtration on an abelian category strict? It's well-known that the category of filtered objects in an abelian category is generally not an abelian category. One must generally add extra structure to ensure that all morphisms are strict for the filtration. This issue is discussed in Deligne's Hodge Theory II
Is there any kind of converse to this? In other words, if I have an abelian category and a functorial filtration on every object, then must every morphism be strict for that filtration?
 A: This is not true.
Example. Let $\mathscr A = \mathbf{Ab}$ (you may restrict to finitely generated abelian groups if you like), and consider the functorial two-step filtration $F^0 \supseteq F^1 \supseteq F^2 = 0$ given by $F^0(A) = A$ and $F^1(A) = A_{\text{tors}}$. This is functorial as a torsion element is mapped to a torsion element under any group homomorphism.
But the quotient map $\mathbf Z \twoheadrightarrow \mathbf Z/n\mathbf Z$ is not strict for this functorial filtration, as $\operatorname{Gr}_F(\mathbf Z) \to \operatorname{Gr}_F(\mathbf Z/n\mathbf Z)$ is the zero map (the first is $\mathbf Z \oplus 0$ and the second $0 \oplus \mathbf Z/n\mathbf Z$). For instance, use Prop. 1.1.11(1) of Hodge II, or the explicit description of strict morphisms of filtered modules immediately after that proposition.
In fact, we have the following criterion:
Lemma. For a functorial filtration $F^\bullet \colon \mathscr A \to \operatorname{Fil}(\mathscr A)$, consider the following statements:

*

*For each morphism $f \colon A \to B$ in $\mathscr A$, the morphism $F^\bullet f \colon F^\bullet A \to F^\bullet B$ is strict,

*The functor $F^n \colon \mathscr A \to \mathscr A$ is exact for every $n$.

*The functor $\operatorname{gr}^n \colon \mathscr A \to \mathscr A$ is exact for every $n$.
Then the implications (1) $\Leftrightarrow$ (2) $\Rightarrow$ (3) hold, and the converse (3) $\Rightarrow$ (2) holds if there exists $a \ll 0$ with $F^a \cong \operatorname{id}$ or $b \gg 0$ with $F^b = 0$.
In the example above, as well as the variations in the comments, the functor $F^1$ is only left exact.
Proof. (1) $\Leftrightarrow$ (2): Let $A \stackrel f\to B \stackrel g\to C$ be an exact sequence in $\mathscr A$, and consider the commutative diagram
$$\begin{array}{ccccc} F^nA & \stackrel{F^nf}\longrightarrow & F^nB & \stackrel{F^ng}\longrightarrow & F^nC \\ \cap & & \cap & & \cap \\ A & \underset f\longrightarrow & B & \underset g\longrightarrow & C.\!\end{array}$$
Strictness of $f$ means that $\operatorname{im}(F^nf) = \operatorname{im}(f) \cap F^nB$, i.e. the top sequence is exact.
(2) $\Rightarrow$ (3): Consider the short exact sequences
$$0 \to F^{n+1} \to F^n \to \operatorname{gr}^n \to 0\label{1}\tag{1}.$$
By the lemma below, if all $F^n$ are exact, then so are all $\operatorname{gr}^n$.
(3) $\Rightarrow$ (2): If $F^a \cong \operatorname{id}$ for $a \ll 0$ (resp. $F^b = 0$ for $b \gg 0$), then induction (resp. descending induction) on \eqref{1} shows that all $F^n$ are exact if all $\operatorname{gr}^n$ are. $\square$
Lemma. Let $0 \to F \to G \to H \to 0$ be a short exact sequence of functors $\mathscr A \to \mathscr B$ of abelian categories. If two out of the three functors are exact, then so is the third.
Proof. Note that the assumptions imply $F(0) = G(0) = H(0) = 0$. Indeed, this holds for two out of three since an exact functor satisfies this condition, hence for the third by the short exact sequence $0 \to F(0) \to G(0) \to H(0) \to 0$.
Thus if $0 \to A \to B \to C \to 0$ is a short exact sequence in $\mathscr A$, viewed as an acyclic complex $C^\bullet$ in $\mathscr A$, then $0 \to F(C^\bullet) \to G(C^\bullet) \to H(C^\bullet) \to 0$ is a short exact sequence of chain complexes in $\mathscr B$. Indeed, since $A \to B \to C$ is the zero map, we conclude that the same holds after applying each of the functors as $F(0) = G(0) = H(0) = 0$. Then the long exact cohomology sequence shows that if two of them are acyclic, then so is the third. $\square$
(For deducing exactness of $F$ or $H$, this is just the nine lemma, and for $G$ you can also prove it directly with a diagram chase.)
