Finite and complete filtered isomorphism lifting

Consider a commutative diagram of short exact sequences $0\to U\to X\to V\to 0$ and $0\to U^{\mathrm{\prime }}\to X^{\mathrm{\prime }}\to V^{\mathrm{\prime }}\to 0$, whose maps on $U$ and $V$ are isomorphisms. If a morphism into $X$ is killed by the middle map, its image in $V$ is killed by the isomorphism to $V^{\mathrm{\prime }}$, hence zero. It factors through $U$, where the isomorphism to $U^{\mathrm{\prime }}$ and the monic inclusion force it to be zero. Thus the middle map is monic. To lift a morphism into $X^{\mathrm{\prime }}$, first project to $V^{\mathrm{\prime }}$, use the inverse on $V$, and lift into $X$ after epic pullback. Its difference from the prescribed map lies in $U^{\mathrm{\prime }}$, so use the inverse on $U$ to correct the lift. Thus the middle map is epic by epic cancellation. A monic epic in an abelian category is invertible by its coimage-image factorization. The local lifts and their cancellation have precisely the meaning of [F1]; no global representatives are chosen.

Under the complete module hypotheses, completeness identifies $F_{p}A$ with . Indeed a compatible tuple in these subquotients is a tuple in $A\mathrm{/}{F}_{k}A$ for $k<p$. Its component at $p$ and at larger indices is zero, since each tuple entry has a representative in $F_{p}A$. This extends it uniquely to a compatible tuple in the full quotient system. Completeness supplies a unique $a\in A$ with those residues, and its zero residue at $p$ says $a\in F_{p}A$. Conversely an element of $F_{p}A$ gives that tuple, and its uniqueness follows from separatedness (also from the injective completion map). The formulas preserve addition and scalar multiplication. The same argument applies to $B$.

For finite filtrations take common integer bounds $a<b$ such that both $F_a$ pieces are zero and both $F_b$ pieces are the whole objects. At $a$ the restriction of $f$ is an isomorphism of zero objects. Apply step 1.1 to the sequences $0\to F_{p-1}\to F_p\to F_p\mathrm{/}{F}_{p-1}\to 0$ for $a<p\le b$. Finite induction proves every restriction $F_{p}A\to F_{p}B$ invertible, including $f$ at $b$. Below $a$ and above $b$ the restrictions are respectively the zero and whole-object maps. Their inverses are the restrictions of $f^{-1}$ by uniqueness, so the inverse is filtered. Empty graded pieces and repeated filtration terms cause no change to the argument.

Now assume the complete module hypotheses. For any fixed $k<p$, filter $F_{p}A\mathrm{/}{F}_{k}A$ and $F_{p}B\mathrm{/}{F}_{k}B$ by the images of the intermediate $F_j$ for $k\le j\le p$. The successive quotients are the original graded pieces by the nested-quotient comparison. The finite argument therefore gives an isomorphism $f_{p,k}:F_{p}A\mathrm{/}{F}_{k}A\to F_{p}B\mathrm{/}{F}_{k}B$. For $k^{\mathrm{\prime }}<k$ its quotient-transition squares commute; applying inverses on both sides proves the inverse squares commute as well.

The compatible inverse maps in step 3.1 send a compatible tuple on the $B$ side to one on the $A$ side. By step 1.2 they give an inverse to $f:F_{p}A\to F_{p}B$ for every $p$. This constructs the inverse without selecting representatives: all residue inverses and their limits are unique. Every $b\in B$ lies in some $F_{p}B$ by exhaustiveness and therefore has a preimage in $F_{p}A$. Every kernel element in $A$ lies in some $F_{p}A$ and is zero by injectivity there. Thus $f$ is bijective and linear, and its inverse sends $F_{p}B$ into $F_{p}A$. The zero module and any one-step finite filtration satisfy the same formulas. Infinite index sets enter only through unique compatible tuples, so no AC is used.