Relative singular cohomology group

Fix an integer $n\ge0$ , using zero chain and cochain groups in negative degrees. Let $A\subseteq X$ be a subspace of a topological space $X$ , and let $G$ be an abelian group. The inclusion gives a subcomplex $C_\bullet(A;\mathbb Z)\subseteq C_\bullet(X;\mathbb Z)$ . The relative singular chain group is

C_n(X,A;\mathbb Z):=C_n(X;\mathbb Z)/C_n(A;\mathbb Z),

and the boundary on $C_\bullet(X;\mathbb Z)$ induces a boundary on these quotient groups because $C_\bullet(A;\mathbb Z)$ is a subcomplex. The relative cochains with coefficients in $G$ are

C^n(X,A;G):=\operatorname{Hom}\!\bigl(C_n(X,A;\mathbb Z),G\bigr).

Their coboundary is $\delta\varphi=\varphi\circ\partial$ . The $n$ th relative singular cohomology group is

H^n(X,A;G):= \ker(\delta:C^n(X,A;G)\to C^{n+1}(X,A;G)) /\operatorname{im}(\delta:C^{n-1}(X,A;G)\to C^n(X,A;G)).

Thus relative cohomology is the cohomology of the cochain complex dual to the quotient chain complex of the pair. When $A=\varnothing$ , this recovers ordinary singular cohomology $H^n(X;G)$ .

Maps of pairs

A continuous map of pairs $f:(X,A)\to(Y,B)$ , meaning $f:X\to Y$ with $f(A)\subseteq B$ , induces a pullback

f^*:H^n(Y,B;G)\longrightarrow H^n(X,A;G).

The short exact sequence of chain complexes $0\to C_\bullet(A;\mathbb Z)\to C_\bullet(X;\mathbb Z)\to C_\bullet(X,A;\mathbb Z)\to0$ yields the long exact sequence of the pair.

Cup product with an absolute class

For coefficients in a commutative ring $R$ , the usual cochain cup product restricts to

H^p(X;R)\times H^q(X,A;R)\longrightarrow H^{p+q}(X,A;R).

Indeed, a relative cochain is an absolute cochain vanishing on chains in $A$ ; its product with an absolute cochain still vanishes on such chains. This gives relative cohomology a module structure over the absolute cohomology ring.