**Proof.**
Let $B$ a vector space over a field $k$ with basis $e_1, e_2$, with the filtration $F^ nB = B$ for $n < 0$, with $F^0B = ke_1$, and $F^ nB = 0$ for $n > 0$. Now take $A = k(e_1 + e_2)$ and $C = B/ke_2$ with filtrations induced by $B$, i.e., such that $A \to B$ and $B \to C$ are strict (Lemma 12.16.7). Then $F^ n(A) = A$ for $n < 0$ and $F^ n(A) = 0$ for $n \geq 0$. Also $F^ n(C) = C$ for $n \leq 0$ and $F^ n(C) = 0$ for $n > 0$. So the (nonzero) composition $A \to C$ is not strict.

Assume $g$ is injective. Then

\begin{align*} g(f(F^ pA)) & = g(f(A) \cap F^ pB) \\ & = g(f(A)) \cap g(F^ p(B)) \\ & = (g \circ f)(A) \cap (g(B) \cap F^ pC) \\ & = (g \circ f)(A) \cap F^ pC. \end{align*}

The first equality as $f$ is strict, the second because $g$ is injective, the third because $g$ is strict, and the fourth because $(g \circ f)(A) \subset g(B)$.

Assume $f$ is surjective. Then

\begin{align*} (g \circ f)^{-1}(F^ iC) & = f^{-1}(F^ iB + \mathop{\mathrm{Ker}}(g)) \\ & = f^{-1}(F^ iB) + f^{-1}(\mathop{\mathrm{Ker}}(g)) \\ & = F^ iA + \mathop{\mathrm{Ker}}(f) + \mathop{\mathrm{Ker}}(g \circ f) \\ & = F^ iA + \mathop{\mathrm{Ker}}(g \circ f) \end{align*}

The first equality because $g$ is strict, the second because $f$ is surjective, the third because $f$ is strict, and the last because $\mathop{\mathrm{Ker}}(f) \subset \mathop{\mathrm{Ker}}(g \circ f)$.
$\square$

## Comments (2)

Comment #1011 by JuanPablo on

Comment #1029 by Johan on

There are also: