$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proposition 2.5.8.5. Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups, and let $K_{\ast } \subseteq \mathrm{C}_{\ast }(A) \boxtimes \mathrm{C}_{\ast }(B)$ be the subcomplex generated by $\mathrm{C}_{\ast }(A) \boxtimes \mathrm{D}_{\ast }(B)$ and $\mathrm{D}_{\ast }(A) \boxtimes \mathrm{C}_{\ast }(B)$. Then $K_{\ast }$ contains the image of the composite map
\[ \mathrm{D}_{\ast }(A \otimes B) \hookrightarrow \mathrm{C}_{\ast }(A \otimes B) \xrightarrow { \overline{\mathrm{AW}} } \mathrm{C}_{\ast }(A) \boxtimes \mathrm{C}_{\ast }(B). \]
Proof.
Let $x$ be an $n$-simplex of the tensor product $A_{\bullet } \otimes B_{\bullet }$, let $0 \leq i \leq n$, and let $s^{n}_ i(x)$ denote the associated degenerate $(n+1)$-simplex of $A_{\bullet } \otimes B_{\bullet }$. We wish to show that $\overline{\mathrm{AW}}( s^{n}_ i(x) )$ belongs to $K_{\ast }$. Without loss of generality, we may assume that $x = a \otimes b$ for $n$-simplices $a \in A_ n$ and $b \in B_ n$. In this case, we have
\[ \overline{\mathrm{AW}}( s^{n}_ i(x) ) = \overline{\mathrm{AW}}( s^{n}_ i(a) \otimes s^{n}_ i(b) ) = \sum _{p=0}^{n+1} \iota _{\leq p}^{\ast }( s^{n}_ i(a) ) \boxtimes \iota _{\geq p}^{\ast }( s^{n}_ i(b) ). \]
It will therefore suffice to show that each summand $\iota _{\leq p}^{\ast }( s^{n}_ i(a) ) \boxtimes \iota _{\geq p}^{\ast }( s^{n}_ i(b) )$ belongs to $K_{\ast }$. This is clear: the simplex $\iota _{\leq p}^{\ast }( s^{n}_ i(a) )$ is degenerate if $p > i$, and the simplex $\iota _{\geq p}^{\ast }( s^{n}_ i(b) )$ is degenerate for $p \leq i$.
$\square$