Consider pairs $(X,A),(Y,B)$ of $CW$-complexes, i.e. $X$ is a CW-complex and $A\subseteq X$ a CW-subcomplex and the same for $(Y,B)$.
I want to know the following: How to prove in singular (relative) homology for all $n\in\mathbb{N}_0$: $$H_n(X\times B\cup A\times Y, X\times B)\cong H_n(A\times Y, A\times B)?$$
This should be excision with $Z=(X\setminus A)\times B$... but I need $\overline{Z}=\overline{(X\setminus A)}\times\overline{ B}\subseteq int(X\times B)=int(X)\times int(B)=X\times int(B)$ (Int=interior) for this, or am I wrong? How to apply exicision here and how to prove that the homology-groups above are isomorphic?
Tools (Also the following might help, since I guess it should be used that we consider CW-complexes (but I don't know how to use it):
In lecture we discussed that if $A$ is a $CW$-subcomplex of $X$, then $A$ is a strong neighbourhood-deformation retract (SNDR, for short) in $X$- this means:
Let $X$ be a topological space. A subspace $A\subset X$ is said to be a strong neighbourhood-deformation retract of $X$, if there is an open neighbourhood $U$ of $A$ and a continuous map $h:U\times [0,1]\to U$ such that
$h(u,0)=u$, $\;h(u,1)\in A$ and $h(a,t)=a$ for all $u\in U$ and $(a,t)\in A\times [0,1]$.
Furthermore, if $(X,A)$ is a pair of topological spaces such that $A$ is a SNDR in $X$, then $H_n(X,A)\cong \tilde{H}_n(X/A)$ for all $n\in\mathbb{N}_0$ ($H_*$ is the singular homology and $\tilde{H}_*$ the reduced singular homology).
I appreciate your help.
Edit: I know the two (equivalent) statements of excision:
1)Let $Z\subseteq A\subseteq X$ be subspaces with $\overline{Z}\subseteq int(A)$. Then the inclusion $i:(X\setminus Z, A\setminus Z)\to (X,A)$ induces an isomorphism $H_n(i):H_n(X\setminus Z, A\setminus Z)\to H_n(X,A)$ for all $n$.
2)Let $A, B\subseteq X$ subspaces such that $int(A)\cup int(B)=X$. Then the inclusion $i:(B,A\cap B)\to (X,A)$ induces an isomorphism $H_n(i):H_n(B,A\cap B)\to H_n(X,A)$ for all $n$.