Proposition 2.5.8.3. Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. Then the unnormalized Alexander-Whitney homomorphism $\overline{\mathrm{AW}}: \mathrm{C}_{\ast }(A \otimes B) \rightarrow \mathrm{C}_{\ast }(A) \boxtimes \mathrm{C}_{\ast }(B)$ is a chain map.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Let $x$ be an element of the abelian group $\mathrm{C}_{n}(A \otimes B) = A_{n} \otimes B_{n}$; we wish to show that $\partial ( \overline{\mathrm{AW}}(x) ) = \overline{\mathrm{AW}}( \partial x)$. Without loss of generality, we may assume that $n>0$ and that $x$ has the form $a \otimes b$, for some elements $a \in A_ n$ and $b \in B_ n$. In this case, we compute
\begin{eqnarray*} \overline{\mathrm{AW}}( \partial (a \otimes b) ) & = & \sum _{i=0}^{n} (-1)^{i} \overline{\mathrm{AW}}( d^{n}_ i a \otimes d^{n}_ i b ) \\ & = & \sum _{i=0}^{n} \sum _{p=0}^{n-1} (-1)^{i} \iota _{\leq p}^{\ast }( d^{n}_ i a) \boxtimes \iota _{\geq p}^{\ast }( d^{n}_ i b ) \\ & = & \sum _{i=0}^{n} \sum _{p=0}^{i-1} (-1)^{i} \iota _{\leq p}^{\ast }( d^{n}_ i a) \boxtimes \iota _{\geq p}^{\ast }( d^{n}_ i b ) + \sum _{i=0}^{n} \sum _{p=i}^{n-1} (-1)^{i} \iota _{\leq p}^{\ast }( d^{n}_ i a) \boxtimes \iota _{\geq p}^{\ast }( d^{n}_ i b ) \\ & = & \sum _{i=0}^{n} \sum _{p=0}^{i-1} (-1)^{i} \iota _{\leq p}^{\ast }(a) \boxtimes d^{n-p}_{i-p} \iota _{\geq p}^{\ast }(b) + \sum _{i=0}^{n} \sum _{q=i+1}^{n} (-1)^{i} d^{q}_{i} \iota _{\leq q}^{\ast }(a) \boxtimes \iota _{\geq q}^{\ast }(b) \\ & = & \sum _{i=0}^{n} \sum _{p=0}^{i} (-1)^{i} \iota _{\leq p}^{\ast }(a) \boxtimes d^{n-p}_{i-p} \iota _{\geq p}^{\ast }(b) + \sum _{i=0}^{n} \sum _{q=i}^{n} (-1)^{i} d^{q}_{i} \iota _{\leq q}^{\ast }(a) \boxtimes \iota _{\geq q}^{\ast }(b) \\ & = & \sum _{p=0}^{n} (-1)^{p} \iota _{\leq p}^{\ast }(a) \boxtimes (\sum _{j=0}^{n-p} (-1)^{j} d^{n-p}_ j \iota _{\geq p}^{\ast }(b) ) + \sum _{q=0}^{n} ( \sum _{i=0}^{q} (-1)^{i} d^{q}_ i \iota _{\leq q}^{\ast }(a) ) \boxtimes \iota _{\geq q}^{\ast }(b) \\ & = & \sum _{p=0}^{n} (-1)^{p} \iota _{\leq p}^{\ast }(a) \boxtimes \partial \iota _{\geq p}^{\ast }(b) + \sum _{q=0}^{n} \partial \iota _{\leq q}^{\ast }(a) \boxtimes \iota _{\geq q}^{\ast }(b) \\ & = & \partial ( \sum _{p=0}^{n} \iota _{\leq p}^{\ast }(a) \boxtimes \iota _{\geq p}^{\ast }(b) ) \\ & = & \partial ( \overline{\mathrm{AW}}( a \otimes b)). \end{eqnarray*}
$\square$