The Thom quotient identifies relative and reduced cohomology

For a based space $Y$, its point cochain complex with coefficients $G$ is $G\stackrel{0}{\to }G\stackrel{1}{\to }G\stackrel{0}{\to }\cdots$: there is one simplex in each degree, and the alternating boundary sum is zero or identity. Thus its cohomology is $G$ in degree0 and zero otherwise. Restriction $H^0(Y;G)\to H^0(\{\ast \};G)$ is split onto by constant degree-zero cocycles. By [F3], $H^0(Y,\{\ast \};G)$ is its kernel and the relative groups equal the absolute ones in positive degrees; negative groups vanish. In degree0, subtracting the constant value at the basepoint canonically identifies the quotient by constant cocycles with that kernel; in positive degrees these are the ordinary reduced groups. Thus we obtain the canonical relative-basepoint identification in every degree.

Suppose $S\mathrm{\ne }\mathrm{\varnothing }$. It is closed in $D$ by continuity of the norm. The open neighbourhood $V=\{v\in D:\mathrm{\parallel }v\mathrm{\parallel }>1\mathrm{/}2\}$ strongly deformation retracts onto $S$ by $v\mapsto ((1-s)+s\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel })v$, $0\le s\le 1$. This stays in $V$, fixes $S$, and reaches the sphere. Its quotient homotopy contracts $V\mathrm{/}S$ onto the quotient point. It is continuous after passage to the quotient because product with the compact interval preserves the quotient construction. Also $q(V)$ is open in $Y$ and is $V\mathrm{/}S$, since $V$ is open and saturated.

Apply [F3] to the inclusion of pairs $(D,S)\to (D,V)$. The maps on $D$ are identities and those from $V$ to $S$ are cohomology isomorphisms by the retraction and [F4]. In the five-term window $H^{k-1}(D)\to H^{k-1}(V)\to H^k(D,V)\to H^k(D)\to H^k(V)$ and its $S$ analogue, [F4] therefore makes $H^k(D,V)\to H^k(D,S)$ an isomorphism. Applying the same argument to $(Y,\{\ast \})\to (Y,q(V))$, using the contraction in step 1.2, makes $H^k(Y,q(V))\to H^k(Y,\{\ast \})$ an isomorphism. The windows include their zero negative-degree groups, so no degree0 endpoint is omitted.

If $S=\mathrm{\varnothing }$, [F1] gives $Y=D_{+}$ and $q$ is the inclusion of the clopen component $D$, not an onto quotient map. Every singular simplex has connected domain and hence lies wholly in $D$ or wholly at the added point. The relative chain complex $C_{\ast }(D_{+},\{\ast \};\mathbb{Z})$ is therefore exactly $C_{\ast }(D;\mathbb{Z})$, by deleting the point-simplex summand; dualizing gives an explicit cochain isomorphism induced by $q$, for arbitrary $G$. Consequently $H^k(D_{+},\{\ast \};G)=H^k(D;G)=H^k(D,\mathrm{\varnothing };G)$. This proves rank zero; if $B$ is empty, $D$ is empty and $Y$ a point, so both complexes and groups are zero.

Excision [F5] removes $S$ from $(D,V)$, since $S$ is closed and contained in open $V$, and removes the closed quotient point from $(Y,q(V))$. The resulting pairs $(D\setminus S,V\setminus S)$ and $(Y\setminus \{\ast \},q(V)\setminus \{\ast \})$ are homeomorphic under $q$. Thus their cohomology groups are isomorphic, and the two excision isomorphisms identify $q^{\ast }:H^k(Y,q(V);G)\to H^k(D,V;G)$ as an isomorphism. Combine with step 2.1 and naturality [F3] to obtain the asserted $q^{\ast }:H^k(Y,\{\ast \};G)\to H^k(D,S;G)$.

In the compactly generated convention, kification leaves continuous maps from compact Hausdorff domains unchanged by its defining final topology. Singular simplices and their homotopies have such domains, so the singular chain and relative cochain complexes used above are unchanged. Hence the same isomorphisms apply. Every map in the proof is induced by the actual quotient map and commutes with maps of disk/sphere pairs and coefficient maps by [F3]; the temporary radial neighbourhood proves invertibility, not an extra choice of isomorphism. No choice of representatives, Hom-exactness assumption, global trivializing cover or base compactness was used.