Schreier refinement theorem in an abelian category

For $0\le i<m$ and $0\le j\le n$, define $A_{i,j}:=A_i\vee (A_{i+1}\wedge B_j)\mathrm{.}$ Then $A_{i,0}=A_i$ and $A_{i,n}=A_{i+1}$, while $A_{i,j}\le A_{i,j+1}$ for every $j$. Concatenating the chains $A_i=A_{i,0}\le A_{i,1}\le \cdots \le A_{i,n}=A_{i+1}$ over $i=0,\dots ,m-1$ gives a refinement of the $A$-chain. Define $B_{j,i}:=B_j\vee (B_{j+1}\wedge A_i)$ symmetrically; concatenating those chains refines the $B$-chain.

For every cell $(i,j)$, apply [L1] to the pairs $A_i\le A_{i+1}$ and $B_j\le B_{j+1}$. It gives a canonical isomorphism $\frac{A_{i,j+1}}{A_{i,j}}\cong \frac{B_{j,i+1}}{B_{j,i}}\mathrm{.}$ So the successive quotients of the two refinements are paired by the same grid.

Some adjacent terms may coincide, producing zero successive quotients. By [F1], deleting those repetitions leaves equivalent refinements, and the quotient pairing from step 2.1 survives on every nonzero factor. Hence the original two chains admit equivalent refinements.