Defintion of relative cohomology cross product

Question

This is a passage from my notes which lacks some details:

Let $X,Y$ be topological spaces, and $A\subseteq X$ and $B\subseteq Y$ be open subsets. Then there is a natural map $K\colon S_{\ast}(X,A)\otimes S_{\ast}(Y,B)\rightarrow S_{\ast}(X\times Y, (A\times Y) \cup (X \times B))$ which induces in isomorphism in homology. This map is obtained as the composition of two maps $$L\colon S_{\ast}(X,A)\otimes S_{\ast}(Y,B)\rightarrow S_{\ast}(X\times Y)/ (S_{\ast}(A\times Y) + S_{\ast}(X \times B))$$ $$M\colon S_{\ast}(X\times Y)/ (S_{\ast}(A\times Y) + S_{\ast}(X \times B)) \rightarrow S_{\ast}(X\times Y, (A\times Y) \cup (X \times B))$$ Since $K$ induces an isomorphism in homology it is a chain homotopy equivalence. Pick a homotopy inverse $EZ$ of $K$.

1. What are the maps $L$ and $K$ explicitly?
2. Why do they induce an isomorphism in homology?
3. In what sense is $K$ natural? When I hear natural I am thinking natural transformation. Is this meant in this way here?
4. Why does the fact that $K$ is a quasi-isomorphism imply that it is a chain homotopy equivalence? In general, only the converse is true.
5. Given maps of pairs $f\colon (X,A)\rightarrow (X',A')$, $g\colon (Y,B)\rightarrow (Y',B')$ and $p,q,n\in \mathbb{N}$ with $p+q=n$, are the following maps equal and why? $$S_n(X\times Y; X\times B \cup A\times Y)\xrightarrow{S_n(f,g)} S_n(X'\times Y';X'\times B' \cup A'\times Y')\xrightarrow{EZ} (S_{\ast}(X',A')\otimes S_{\ast}(Y',B'))_n\xrightarrow{\pi_{p,q}}S_p(X',A')\otimes S_q(Y',B')$$ and $$S_n(X\times Y;X\times B \cup A\times Y)\xrightarrow{EZ} (S_{\ast}(X,A)\otimes S_{\ast}(Y,B))_n \xrightarrow{\pi_{p,q}} S_{p}(X,A)\otimes S_{q}(Y,B)\xrightarrow{S_p(f)\otimes S_q(g)}S_p(X',A')\otimes S_q(Y',B')$$