Kerodon

Proof. Fix element $a \in \mathrm{N}_{p}(A)$ and $b \in \mathrm{N}_{q}(B)$ having shuffle product $a \triangledown b \in \mathrm{N}_{p+q}(A \otimes B)$. We wish to show that the Alexander-Whitney homomorphism $\mathrm{AW}$ satisfies $\mathrm{AW}( a \triangledown b) = a \boxtimes b$. Lift $a$ and $b$ to elements $\overline{a} \in \mathrm{C}_{p}(A) = A_ p$ and $\overline{b} \in \mathrm{C}_ q(B) = B_ q$, respectively. Unwinding the definitions, we see that $\mathrm{AW}( a \triangledown b)$ is given by the image of

under the quotient map $\mathrm{C}_{\ast }(A) \boxtimes \mathrm{C}_{\ast }(B) \twoheadrightarrow \mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B)$; here the sum is taken over all $(p,q)$-shuffles $\sigma = (\sigma _{-}, \sigma _{+})$ (see Notation 2.5.7.2). Note that the simplex $(\iota _{\leq r}^{\ast } \sigma _{-}^{\ast })(\overline{a}) \in A_{r}$ is degenerate unless $\sigma _{-}(r) = r$ (which implies that $r \leq p$). Similarly, the simplex $( \iota _{\geq r}^{\ast } \sigma _{+}^{\ast })(\overline{b}) \in B_{n-r}$ is degenerate unless $\sigma _{+}(r) = r-p$ (which guarantees that $r \geq p$). We may therefore ignore every term in the sum except for the one with $r = p$ and $\sigma (i) = \begin{cases} (i,0) & \text{ if } i \leq p \\ (p, i-p) & \text{ if } i \geq p, \end{cases}$ for which the corresponding summand is equal to $\overline{a} \boxtimes \overline{b}$ (and therefore has image $a \boxtimes b$ in $\mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B)$). $\square$