Kerodon

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

Proof. We first show that the unnormalized Alexander-Whitney homomorphisms determine a nonunital colax monoidal structure on the functor $\mathrm{C}_{\ast }$ (Variant 2.1.4.16). By construction, the homomorphism $\overline{\mathrm{AW}}: \mathrm{C}_{\ast }(A \otimes B) \rightarrow \mathrm{C}_{\ast }(A) \boxtimes \mathrm{C}_{\ast }(B)$ is natural in $A_{\bullet }$ and $B_{\bullet }$. It will therefore suffice to show that, for every triple of simplicial abelian groups $A_{\bullet }$, $B_{\bullet }$, and $C_{\bullet }$, the diagram of chain complexes

commutes, where the horizontal maps are given by the associativity constraints of the monoidal categories $\operatorname{ Ab }_{\Delta }$ and $\operatorname{Ch}(\operatorname{\mathbf{Z}})$, respectively. Unwinding the definitions, we see that both the clockwise and counterclockwise composition are given by the construction

for $a \in A_ n$, $b \in B_ n$, and $c \in C_ n$, where $\rho $ denotes the nondecreasing map $[q-p] \hookrightarrow [n]$ given by $\rho (i) = i + p$.

Note that the unit object of the category of simplicial abelian groups is the constant functor $\operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{ Ab }$ taking the value $\operatorname{\mathbf{Z}}$, which we can identify with the free simplicial abelian group $\operatorname{\mathbf{Z}}[ \Delta ^0]$ generated by the simplicial set $\Delta ^0$. The image of this object under the functor $\overline{\mathrm{AW}}$ is the unnormalized chain complex $\mathrm{C}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}})$. On the other hand, the unit object of $\operatorname{Ch}(\operatorname{\mathbf{Z}})$ is the chain complex $\operatorname{\mathbf{Z}}[0]$, which we will identify with the normalized chain complex $\mathrm{N}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}})$. We will complete the proof of Proposition 2.5.8.4 by showing that the quotient map $\epsilon : \mathrm{C}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}}) \twoheadrightarrow \mathrm{N}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}})$ is a counit for the nonunital colax monoidal structure constructed above (in the sense of Variant 2.1.5.11). To prove this, we must show that for every simplicial abelian group $A_{\bullet }$, both of the composite maps

are equal to the identity. This follows immediately from the construction (using the fact that $\epsilon $ vanishes on every element of $\mathrm{C}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}})$ of positive degree). $\square$

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

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 $p \leq i$. $\square$

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 $\mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B)$). $\square$

Proof. By virtue of Proposition 2.5.8.9, the Alexander-Whitney homomorphism is a left inverse to the Eilenberg-Zilber map $\mathrm{EZ}: \mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B) \rightarrow \mathrm{N}_{\ast }(A \otimes B)$, which is a quasi-isomorphism by virtue of Theorem 2.5.7.14. $\square$