Suppose a complex w is not A-small. Concentrate on a simplex v in w. The preimages of the open sets of A cover the standard simplex δn. Thus a collection of open sets coverse the standard simplex, such that each open set maps entirely into one of the open sets of A. Since the simplex is compact, replace the open cover with a finite subcover. Then select a diameter d, such that every open ball of diameter d or less fits inside one of the open sets in our finite subcover. Apply the subdivision operator again and again, until each simplex has a diameter less than d. For some m, the mth order subdivision of v is A-small.
Do the same for each v in the collection w, and select the largest m. Thus the mth order subdivision of w is A-small.
In general, A could be any collection of sets that cover S, provided the interiors of the sets of A also cover S. This provides the aforementioned open cover for S. A complex is once again A-small if every simplex maps into one of the sets of A, and every complex can be subdivided into an A-small complex.
Let C be the chain complex of S, generated by the simplexes of S, without constraint. Let B be the A-small chains of S. Since B is a subchain of C, it is reasonable to talk about the homology of B, or the homology homomorphism as B embeds into C, or the relative homology of C/B.
In the previous section we mapped C onto E (using the subdivision operator), whereupon E embedded back into C. The composition of these two functions induced a homology isomorphism. Not just an isomorphism - but the identity map. Now we are ready to take another step.
C → E → B → C
The first arrow is the subdivision operator as usual, and the next two arrows are inclusions. The composition of those two arrows is an isomorphism, hence each arrow is injective. To show the third arrow is onto, start with y in C, which acts as a generator in h(C), move all the way to the left, and run y through all three arrows, giving y back again. This makes the last arrow surjective, whence the embedding of B into C induces a homology isomorphism. Apply an earlier theorem, and the relative homology h(C/B) = 0.
To show the second arrow is onto, start with z in B, map to y in C, move to the left, and run y through all three arrows, giving y. (Remember that y is a member of the homology, not a particular simplex.) Back up through the third and second arrows, which are injective, and find the preimage of z in E. Therefore, all three functions are homology isomorphisms.
Let U be a subspace of S. Every A-small complex that maps into U is an A-small complex of S. So we have for chain complexes, based on S, U, and the A-small chains of S and U. The relative homology of S/U is isomorphic to the relative homology of the A-small chains of S/U. The proof is essentially the same as the one presented in the previous section. Just use the A-small chains for the top row of the commutative diagram.