The Stacks project

Proof. Let $p$ be the largest integer such that $F^ pI \not= 0$. In particular $\text{gr}^ p(I) = F^ pI$. Let $I'$ be the object of $\text{Fil}^ f(\mathcal{A})$ whose underlying object of $\mathcal{A}$ is $F^ pI$ and with filtration given by $F^ nI' = 0$ for $n > p$ and $F^ nI' = I' = F^ pI$ for $n \leq p$. Note that $I' \to I$ is a strict monomorphism too. The fact that $u$ is a strict monomorphism implies that $F^ pI \to A/F^{p + 1}(A)$ is injective, see Homology, Lemma 12.19.13. Choose a splitting $s : A/F^{p + 1}A \to F^ pI$ in $\mathcal{A}$. The induced morphism $s' : A \to I'$ is a strict morphism of filtered objects splitting the composition $I' \to I \to A$. Hence we can write $A = I' \oplus \mathop{\mathrm{Ker}}(s')$ and $I = I' \oplus \mathop{\mathrm{Ker}}(s'|_ I)$. Note that $\mathop{\mathrm{Ker}}(s'|_ I) \to \mathop{\mathrm{Ker}}(s')$ is a strict monomorphism and that $\mathop{\mathrm{Ker}}(s'|_ I)$ is a filtered injective object. By induction on the length of the filtration on $I$ the map $\mathop{\mathrm{Ker}}(s'|_ I) \to \mathop{\mathrm{Ker}}(s')$ is a split injection. Thus we win. $\square$