Let $k$ be a field, and let $C$ be the category whose objects are finite dimensional $k$-vector spaces endowed with two finite filtrations $W$ and $F$, the former being ascending and the latter descending. A morphism in $C$ is a morphism of the underlying vector spaces preserving both filtrations $W$ and $F$.
Although $C$ admits direct sums, kernels and cokernels, it is not abelian: given an arbitrary morphism $f:A\to B$ in $C$ we do not have for granted that the natural map $\iota(f):\operatorname{coim} f \to \operatorname{im} f$ is an isomorphism in $C$ (while this is an isomorphism of the underlying vector spaces, the inverse map might not preserve $W$ and $F$). We call $f$ strict whenever $\iota(f)$ is an isomorphism in the category $C$.
Let $f:A\to B$ be a morphism in $C$. The following two assertions should be equivalent (e.g. A. Huber Mixed Motives and Their Realizations in Derived Categories, Lemma 3.1.3):
- $f$ is strict;
- For all integers $m$ and $n$, we have $f(W_nA\cap F^m A)=W_nB\cap F^m B\cap f(A)$ as subspaces of $B$.
In particular, for $f$ to be strict it is not enough to ask for $f$ to be strictly compatible with both $W$ and $F$ (that is $f(W_n A)=W_n B\cap f(A)$ and $f(F^n A)=F^n B\cap f(A)$ for all $n$).
But why is that so? Are 1. and 2. equivalent?
Here is my attempt so far. We know that $\operatorname{coim}f$ is the space $A/\ker f$ equipped with the induced filtrations, that is $W_n\operatorname{coim}f=(W_nA+\ker f)/\ker f$ (and idem for $F$). On the other-hand, the induced filtration on $\operatorname{im} f$ is given by $W_n\operatorname{im} f=W_n B\cap f(A)$. Therefore, if $f$ is strict, we obtain $f(W_n A)=W_n B\cap f(A)$ and the analogous equality for $F$. But to me, it already means that $f^{-1}$ preserves both $W$ and $F$, so what am I missing here?
Many thanks in advance!