Excision for singular cohomology

The interiors of $A$ and $Y$ cover $X$: a point outside $\mathrm{int}A$ lies outside $\bar{Z}$, hence in the open set $X\setminus \bar{Z}\subseteq Y$. Use this two-member family in [F2]. Its small subcomplex is $M=C_A+C_{\ast }(Y;\mathbb{Z})$, since its generators are precisely simplices whose image lies in one of the two members. The simplex subsets intersect exactly in those with image in $A\cap Y=A\setminus Z$, so the finite formal-chain model gives $C_A\cap C_{\ast }(Y)=C_{\ast }(A\setminus Z)$.

Use [F1]'s constructed maps $R:C\to M$ and $D:C_n\to C_{n+1}$. Here every simplex in $A$, and every simplex in $Y$, is already small. For such a simplex the least-depth recursion gives $m=0$, so $D\sigma =D_0\sigma =0$. Its faces are also small; hence $R\sigma =\sigma$ from the explicit formula. Consequently $D{\mathrm{\mid }}_M=0$, $R{\mathrm{\mid }}_M=1$, and in particular both maps preserve $C_A$. With inclusion $i:M\to C$ the identities are $Ri=1$ and $1-iR=\mathrm{\partial }D+D\mathrm{\partial }$. These are the actual operators of the supplier proof, not an inference that an arbitrary homotopy inverse preserves subspaces.

Quotienting the identities in step 2.1 by $C_A$ yields maps $\bar{i}:M\mathrm{/}{C}_A\to C\mathrm{/}{C}_A$ and $\bar{R}$ in reverse, with $\bar{R}\bar{i}=1$ and $1-\bar{i}\bar{R}=\bar{\mathrm{\partial }}\bar{D}+\bar{D}\bar{\mathrm{\partial }}$. Step 1.1 identifies $M\mathrm{/}{C}_A$ with $C_{\ast }(Y)\mathrm{/}{C}_{\ast }(A\setminus Z)$: send a $Y$ chain to its class modulo $C_A$. This is onto since every element of $M$ is a sum of an $A$ chain and a $Y$ chain, and its kernel is exactly the displayed intersection. The isomorphism commutes with boundaries, since all maps are induced by chain inclusions and quotients. Thus $\bar{i}$ is the actual relative-chain inclusion for excision.

Precompose with $\bar{i}$ and $\bar{R}$ to obtain cochain maps ${\bar{i}}^{\mathrm{♯}}$ and ${\bar{R}}^{\mathrm{♯}}$. The identity $\bar{R}\bar{i}=1$ gives ${\bar{i}}^{\mathrm{♯}}{\bar{R}}^{\mathrm{♯}}=1$. On a degree-$n$ cochain $\phi$ on $C\mathrm{/}{C}_A$, put $K^n\phi =\phi {\bar{D}}_{n-1}$, with $K^0=0$. The other identity in step 3.1 gives $\phi -{\bar{R}}^{\mathrm{♯}}{\bar{i}}^{\mathrm{♯}}\phi =\delta K^n\phi +K^{n+1}\delta \phi \mathrm{.}$ For a cocycle the right side is a coboundary; therefore both induced composites are identities on cohomology. This proves bijectivity, with inverse induced by ${\bar{R}}^{\mathrm{♯}}$, for arbitrary $G$; no exactness property of the Hom functor has been assumed.

By [F4] and step 3.1 the map just proved invertible is precisely the claimed restriction map, proving the theorem. If $Z=\mathrm{\varnothing }$ it is the identity; if $A=X$, both relative complexes are zero; if $A=\mathrm{\varnothing }$, the hypothesis forces $Z=\mathrm{\varnothing }$. Empty spaces, a point, zero coefficients and negative degrees are included. Degree zero in step 4.1 uses $D_{-1}=0$, so the homotopy formula has its correct zero first summand. The least-depth construction in [F1] and all quotient/dual formulas are specified without any choice of bases or representatives; no AC, flatness or coefficient injectivity is used.