Relative Kronecker evaluation is well defined, biadditive and natural

For a relative cocycle $\alpha \in Z^k(X,A;G)$ and a relative $k$-cycle $z$ define $E(\alpha ,z)=\alpha (z)\in G$; this is a $G$-valued function of the pair of representatives, and the relative chain condition $\mathrm{\partial }z\in C_{k-1}(A;\mathbb{Z})$ together with the vanishing of $\alpha$ on $A$-simplices is available by [L1] and [L2].

If $z^{\mathrm{\prime }}=z+\mathrm{\partial }b+c$ with $b\in C_{k+1}(X;\mathbb{Z})$ and $c\in C_k(A;\mathbb{Z})$ is another representative of the same relative class, then $E(\alpha ,z^{\mathrm{\prime }})=\alpha (z)+\alpha (\mathrm{\partial }b)+\alpha (c)=\alpha (z)+\delta \alpha (b)+0=\alpha (z)$ because $\delta \alpha =0$ and $\alpha$ vanishes on $A$-simplices, so $E$ is independent of the relative cycle representative.

If ${\alpha }^{\mathrm{\prime }}=\alpha +\delta \beta$ is another relative cocycle representative, then $E({\alpha }^{\mathrm{\prime }},z)=\alpha (z)+\beta (\mathrm{\partial }z)=\alpha (z)$ because $\mathrm{\partial }z\in C_{k-1}(A;\mathbb{Z})$ and $\beta$ vanishes on $A$-simplices, so $E$ is independent of the relative cocycle representative.

Steps 2.1 and 3.1 descend $E$ to a well-defined map $⟨-,-⟩:H^k(X,A;G)\times H_k(X,A;\mathbb{Z})\to G$, the relative form of the absolute descent in [L4]; for $k<0$ both groups are zero by [L1] and [L2] and the pairing is the zero map.

The descended pairing is biadditive: for relative cocycles $\alpha ,{\alpha }^{\mathrm{\prime }}$ and relative cycles $z,z^{\mathrm{\prime }}$ one has $E(\alpha +{\alpha }^{\mathrm{\prime }},z)=E(\alpha ,z)+E({\alpha }^{\mathrm{\prime }},z)$ and $E(\alpha ,z+z^{\mathrm{\prime }})=E(\alpha ,z)+E(\alpha ,z^{\mathrm{\prime }})$ because $C^k(X,A;G)$ consists of additive homomorphisms and evaluation is additive in the chain variable, and these identities pass to the quotients.

The pairing is coefficient-compatible: for a coefficient homomorphism $u:G\to G^{\mathrm{\prime }}$ the composite $u\circ \alpha$ again vanishes on $A$-simplices and satisfies $\delta (u\circ \alpha )=u\circ \delta \alpha =0$, and $E(u\circ \alpha ,z)=u(E(\alpha ,z))$; hence on classes $⟨u_{\ast }[\alpha ],[z]⟩=u⟨[\alpha ],[z]⟩$.

It is natural: a map of pairs $f:(X,A)\to (Y,B)$ has $f_{\mathrm{\#}}(C_{\bullet }(A;\mathbb{Z}))\subseteq C_{\bullet }(B;\mathbb{Z})$ by [L5], so $f_{\mathrm{\#}}$ descends to relative chains and $f^{\mathrm{\#}}$ carries cochains vanishing on $B$-simplices to cochains vanishing on $A$-simplices; on representatives $(f^{\mathrm{\#}}\phi )(z)=\phi (f_{\mathrm{\#}}z)$, which descends to $⟨f^{\ast }[\phi ],[z]⟩=⟨[\phi ],f_{\ast }[z]⟩$ by step 4.1.

If $R$ is a commutative unital ring and the cochains and chains are the $R$-linear ones, the same formulae with $R$-linear maps show that $E(r\alpha ,z)=rE(\alpha ,z)=E(\alpha ,rz)$, so the descended pairing is $R$-bilinear and the computation of steps 2.1, 3.1, 5.1 and 7.1 applies verbatim, giving the pairing $H^k(X,A;R)\times H_k(X,A;R)\to R$ with the same naturality; the abelian-group statement keeps integral chains on the homology side.

The special cases are consistent with the statement: $A=\mathrm{\varnothing }$ recovers the absolute pairing of [L3] and [L4]; $A=X$ or $X=\mathrm{\varnothing }$ or $G=0$ gives the zero pairing; and for $k<0$ both sides are zero.