Good pairs and quotient reduced homology

Since $V$ deformation retracts onto $A$, $H_{\ast }(V,A;G)=0$. The quotient homotopy gives the same conclusion for $(V\mathrm{/}A,A\mathrm{/}A)$. The long exact sequences of the short exact chain-complex sequences $0\to C_{\ast }(V,A;G)\to C_{\ast }(X,A;G)\to C_{\ast }(X,V;G)\to 0$ and its quotient analogue therefore give $H_{\ast }(X,A;G)\cong H_{\ast }(X,V;G),H_{\ast }(X\mathrm{/}A,A\mathrm{/}A;G)\cong H_{\ast }(X\mathrm{/}A,V\mathrm{/}A;G)\mathrm{.}$

Because $A$ is closed and lies in the open set $V$, excision applies to $(X,V)$ and to $(X\mathrm{/}A,V\mathrm{/}A)$ after removing respectively $A$ and the quotient point. The resulting removed pairs are homeomorphic, so the quotient map induces $H_{\ast }(X,V;G)\cong H_{\ast }(X\mathrm{/}A,V\mathrm{/}A;G)$. Combining with step 1.1 gives $H_{\ast }(X,A;G)\cong H_{\ast }(X\mathrm{/}A,A\mathrm{/}A;G)$. Since $A\mathrm{\ne }\mathrm{\varnothing }$, $A\mathrm{/}A$ is one point, and the augmented singular complex identifies the latter groups with ${\tilde{H}}_{\ast }(X\mathrm{/}A;G)$, including degree zero.