Kerodon

Proof of Proposition 2.5.7.1. It follows immediately from the definitions that the shuffle product maps

depend functorially on $A_{\bullet }$ and $B_{\bullet }$ and satisfy $a \triangledown b = a \otimes b$ when $p=q=0$, and the Leibniz rule follows from Proposition 2.5.7.10. To complete the proof of Proposition 2.5.7.1, we will show that the shuffle product is the unique operation with these properties. To this end, suppose we are given another collection of bilinear maps

which depend functorially on $A_{\bullet }$ and $B_{\bullet }$ and satisfy the Leibniz rule. In the special case $A_{\bullet } = B_{\bullet } = \operatorname{\mathbf{Z}}[\Delta ^{0}]$, we can identify $\mathrm{N}_{0}(A)$, $\mathrm{N}_{0}(B)$, and $\mathrm{N}_{0}(A \otimes B)$ with the group $\operatorname{\mathbf{Z}}$ of integers, so that $1 \triangledown ' 1 = n$ for some integer $n$. We will complete the proof by showing that for every pair of simplicial abelian groups $A_{\bullet }$ and $B_{\bullet }$ and every pair of elements $a \in \mathrm{N}_{p}(A)$, $b \in \mathrm{N}_{q}(B)$, we have $a \triangledown ' b = n ( a \triangledown b)$ (in particular, if $a \triangledown ' b = a \otimes b$ whenever $p=q=0$, we must have $n=1$ and therefore $\triangledown ' = \triangledown $).

Without loss of generality, we may assume that $p,q \geq 0$. We will proceed by induction on $p+q$. Choose a lift of $a$ to an element of $\mathrm{C}_{p}(A) $, which we identify with a map of simplicial abelian groups $\operatorname{\mathbf{Z}}[ \Delta ^{p}] \rightarrow A_{\bullet }$. Invoking our assumption that $\triangledown '$ is functorial, we can assume without loss of generality that $A_{\bullet } = \operatorname{\mathbf{Z}}[ \Delta ^{p}]$ and that $a$ is the generator of $\mathrm{N}_{p}( \Delta ^ p; \operatorname{\mathbf{Z}})$ corresponding to the unique nondegenerate $p$-simplex of $\Delta ^ p$. Similarly, we may assume that $B_{\bullet } = \operatorname{\mathbf{Z}}[ \Delta ^{q}]$ and that $b \in \mathrm{N}_{q}( \Delta ^{q}; \operatorname{\mathbf{Z}})$ is the generator given by the unique nondegenerate $q$-simplex of $\Delta ^{q}$.

Let $\overline{a}$ and $\overline{b}$ denote the images of $a$ and $b$ in the relative chain complexes $\mathrm{N}_{\ast }( \Delta ^ p, \partial \Delta ^ p; \operatorname{\mathbf{Z}}) \simeq \operatorname{\mathbf{Z}}[p]$ and $\mathrm{N}_{\ast }( \Delta ^ q, \partial \Delta ^ q; \operatorname{\mathbf{Z}}) \simeq \operatorname{\mathbf{Z}}[q]$. Let $\partial ( \Delta ^ p \times \Delta ^ q ) \subseteq \Delta ^ p \times \Delta ^ q$ denote the union of the simplicial subsets $(\partial \Delta ^ p) \times \Delta ^ q$ and $\Delta ^{p} \times (\partial \Delta ^ q)$, so that we have an isomorphism of simplicial abelian groups

By virtue of Theorem 2.5.7.14, the Eilenberg-Zilber homomorphism

is a quasi-isomorphism. In particular, the $(p+q)$-cycles of the chain complex $\mathrm{N}_{\ast }( \Delta ^ p \times \Delta ^ q, \partial ( \Delta ^ p \times \Delta ^ q; \operatorname{\mathbf{Z}})$ form a cyclic group generated by the shuffle product $\overline{a} \triangledown \overline{b}$. Since the operation $\triangledown '$ satisfies the Leibniz rule, the chain $\overline{a} \triangledown ' \overline{b} \in \mathrm{N}_{p+q}( \Delta ^ p \times \Delta ^ q, \partial ( \Delta ^ p \times \Delta ^ q; \operatorname{\mathbf{Z}})$ is a cycle, and therefore satisfies $\overline{a} \triangledown ' \overline{b} = m (\overline{a} \triangledown \overline{b})$ for some integer $m$. Using the commutativity of the diagram

and the observation that the vertical maps are isomorphisms, we conclude that $a \triangledown ' b = m (a \triangledown b)$. We will complete the proof by showing that $m = n$. In the case $p=q=0$, this follows from the definition of the integer $n$. If $p + q > 0$, we invoke our inductive hypothesis to compute

Since $\partial (a \triangledown b)$ is a nonzero element of the free abelian group $\mathrm{N}_{p+q-1}( \Delta ^ p \times \Delta ^ q; \operatorname{\mathbf{Z}})$, we must have $m=n$ as desired. $\square$