Relative homology Mayer–Vietoris for closed supports

The singular simplex generators common to $D$ and $E$ are exactly the maps with image in $U\cap V$. Hence $D\cap E=C_{\ast }(U\cap V;R)$, including for the zero ring. The chain maps $0⟶C\mathrm{/}(D\cap E)\stackrel{j}{\to }(C\mathrm{/}D)\oplus (C\mathrm{/}E)\stackrel{q}{\to }C\mathrm{/}(D+E)⟶0,$ given by $j[c]=([c],[c])$ and $q([a],[b])=[a-b]$, are well-defined and commute with boundary because $D,E$ are subcomplexes.

The map $j$ is injective since a representative mapping to zero lies in both $D$ and $E$. The map $q$ is onto since $q([a],0)=[a]$, and $qj=0$. If $q([a],[b])=0$, write $a-b=d+e$ with $d\in D,e\in E$. Then $c=a-d=b+e$ has residues $[a]$ in $C\mathrm{/}D$ and $[b]$ in $C\mathrm{/}E$, so $j[c]=([a],[b])$. This proves exactness in every degree; the argument uses only the existence of a decomposition for one element of $D+E$.

By De Morgan's laws, $U\cap V=X\setminus (A\cup B)$ and $U\cup V=X\setminus (A\cap B)$. The first two nonzero complexes in step 1.1 therefore have exactly the relative homology groups displayed in the statement. Since $U,V$ are open, [F2] identifies the homology of the final quotient with $H_{\ast }(X\mathrm{\parallel }A\cap B;R)$. Applying [F3] to step 2.1 yields the asserted sequence. Composition of $q$ with this canonical quotient map is the difference of the two relative quotient maps, so the printed sign is precisely $a{\mathrm{\mid }}_{A\cap B}-b{\mathrm{\mid }}_{A\cap B}$.

If $A=\mathrm{\varnothing }$, then $U=X$, $D=C$, and the sequence reduces to identity maps on the groups supported in $B$, with zero groups for empty support; the other empty case is symmetric. If $A=B$, the diagonal and difference sequence has the stated exactness. For $X=\mathrm{\varnothing }$, $R=0$ or negative chain degrees all complexes concerned are zero. Degree zero follows from the same degreewise short exact sequence, with no reduced-group substitution. All singular generators, including degenerate ones, were retained in step 1.1. The quotient equivalence in [F2] uses prescribed small-chain operators; no AC or choice of a splitting is required.