Let $(X,A),(Y,B)$ be two pairs of topological spaces. Denote $\times$ as external cross product on cohomology level and $E:C_\star(X,A)\otimes C_\star(Y,B)\to C_\star((X,A)\times(Y\times B))$ as Eilenberg-Zilber map and assume appropriate hypothesis(i.e. excision pairs on $A,B$ and $A\times Y, X\times B$ excision pairs) s.t. $E$ induce homology level isomorphism. Set $<-,->:H^p(X;G_1)\otimes H_p(X,G_2)\to G_1\otimes_Z G_2$ as bilinear pairing where $G_i$ are abelian groups. Whenever abelian group is omitted in homology, it is assumed to be $Z$ coefficients.
$\textbf{Q:}$ I am quite confused about the following equation's sign $(-1)^{pq}$. "$a\in H_p(X,A),b\in H_q(Y,B),u\in H^p(X,A;G_1),v\in H^q(Y,B;G_2)$, then $(-1)^{pq}<u\times v,E(a\otimes b)>=<u,a>\otimes<v,b>$." Where is $(-1)^{pq}$ coming from?
The following is my reasoning. First $\times$ is supercommutative. Hence $u\times v=(-1)^{pq}v\times u$. Now I can use $H^q(Y,B,G_2)\otimes H^p(X,A,G_1)\otimes H_p(X,A)\otimes H_q(Y,B)\to H^q(Y,B,G_2)\otimes G_1\otimes H^q(Y,B,G_2)$ where I have used obvious $<-,->$ pairing once. Now I can flip $G_1$ by $A\otimes B\cong B\otimes A$ natural. And keep contraction. This gives the equation.
However, I do not see difference between $H^q(X,A,G_1)\times H^p(Y,B,G_2)\times H_q(X,A)\times H_p(Y,B)\to G_1\otimes G_2$ by directly contracting first slot and thirs slot and contracting second slot with the fourth slot. This reasoning does not give rise to $(-1)^{pq}$.
$\textbf{Q':}$ What is the loop hole in the above reasoning? Obviously, I did not use any cohomology ring supercommutativity property.
Say $X,Y$ are smooth orientable manifolds. Essentially I should think $<-,->$ as integration of forms on manifolds. Say I am looking at compact cohomology theory. Now $\int_{U\times V}\omega_1\wedge\omega_2=\int_U\omega_1\times\int_V\omega_2$ where I assume $\omega_i$ are parametrized by standard coordinates and using canonical orientation induced on $X\times Y$ with $U,V$ identified as homology cycle to be integrated. This does give rise to $(-1)^{pq}$ somehow.
$\textbf{Q'':}$ What is the loop hole in the reasoning here?