Serre classes are stable under finite filtrations

The group $F_{-1}A=0$ lies in $\mathcal{C}$. If $F_{i-1}A\in \mathcal{C}$, the filtration gives a short exact sequence $0⟶F_{i-1}A⟶F_{i}A⟶{\mathrm{gr}}_{i}A⟶0.$ The two end groups lie in $\mathcal{C}$, so extension closure in [F1] gives $F_{i}A\in \mathcal{C}$. Finite induction from $i=0$ through $i=m$ yields $A=F_{m}A\in \mathcal{C}$.

Write $f_i:F_{i}A\to F_{i}B$. For each $i$, the filtered map gives a commutative diagram whose rows are $0\to F_{i-1}A\to F_{i}A\to {\mathrm{gr}}_{i}A\to 0$ and the analogous sequence for $B$. Apply [F2]. If $f_{i-1}$ and ${\mathrm{gr}}_i(f)$ are $\mathcal{C}$-isomorphisms, all four of their kernel and cokernel groups lie in $\mathcal{C}$. Exactness of the snake sequence expresses $\ker f_i$ and $\mathrm{coker}{f}_i$ as extensions of subquotients of those four groups. By [F1], both belong to $\mathcal{C}$; hence $f_i$ is a $\mathcal{C}$-isomorphism.

The initial map $f_{-1}:0\to 0$ has zero kernel and cokernel. Starting there and applying step 1.2 finitely many times proves that $f_m=f$ is a $\mathcal{C}$-isomorphism. No simultaneous selection of lifts is made: the snake maps are homomorphisms supplied by [F2], and the argument only takes finitely many canonical kernels, images, quotients, and extensions.

If $m=-1$, then $A=B=0$ and both conclusions are immediate. Repeated filtration terms contribute zero graded pieces. A one-step filtration is exactly the defining extension closure, and a one-piece filtered map is the defining kernel-cokernel condition. Zero graded maps and zero source or target groups are included. The induction checks its initial and terminal endpoints, and step 1.2 checks both the kernel and cokernel sides of the snake sequence. Degenerate filtrations are handled by repeated terms. No AC is used and neither assertion is a biconditional.