# Kerodon

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Proposition 2.5.7.1. There exists a collection of maps

$\triangledown : \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B) \quad \quad (a,b) \mapsto a \triangledown b,$

defined for every pair of simplicial abelian groups $A_{\bullet }$ and $B_{\bullet }$ and every pair of integers $p,q \in \operatorname{\mathbf{Z}}$, and uniquely determined by the following properties:

• Each of the maps $\triangledown : \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B)$ is bilinear and satisfies the Leibniz rule $\partial ( a \triangledown b) = (\partial a) \triangledown b + (-1)^{p} a \triangledown (\partial b)$ (and therefore induces a chain map $\mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B) \rightarrow \mathrm{N}_{\ast }(A \otimes B)$; see Exercise 2.5.1.15).

• The operation $\triangledown$ depends functorially on $A_{\bullet }$ and $B_{\bullet }$. That is, if $f: A_{\bullet } \rightarrow A'_{\bullet }$ and $g: B_{\bullet } \rightarrow B'_{\bullet }$ are homomorphisms of simplicial abelian groups, then the diagram

$\xymatrix@R =50pt@C=50pt{ \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \ar [r]^-{ \triangledown } \ar [d]^{ \mathrm{N}_{p}(f) \times \mathrm{N}_{q}(g)} & \mathrm{N}_{p+q}( A \otimes B) \ar [d]^{ \mathrm{N}_{p+q}(f \otimes g) } \\ \mathrm{N}_{p}(A') \times \mathrm{N}_{q}(B') \ar [r]^-{ \triangledown } & \mathrm{N}_{p+q}( A' \otimes B') }$

commutes.

• For $a \in A_0$ and $b \in B_0$, we have $a \triangledown b = a \otimes b$ (where we identify $a$, $b$, and $a \otimes b$ with the corresponding elements of $\mathrm{N}_0(A)$, $\mathrm{N}_0(B)$, and $\mathrm{N}_{0}(A \otimes B)$, respectively).

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

$\triangledown : \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B)$

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

$\triangledown ': \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B)$

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

$(\operatorname{\mathbf{Z}}[ \Delta ^ p ] / \operatorname{\mathbf{Z}}[ \partial \Delta ^ p]) \otimes ( \operatorname{\mathbf{Z}}[ \Delta ^ q] / \operatorname{\mathbf{Z}}[ \partial \Delta ^ q] ) \simeq \operatorname{\mathbf{Z}}[ \Delta ^{p} \times \Delta ^ q ] / \operatorname{\mathbf{Z}}[ \partial ( \Delta ^ p \times \Delta ^ q)].$

By virtue of Theorem 2.5.7.14, the Eilenberg-Zilber homomorphism

$\mathrm{EZ}: \mathrm{N}_{\ast }( \Delta ^ p, \partial \Delta ^ p; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }( \Delta ^ q, \partial \Delta ^ q; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{\ast }( \Delta ^ p \times \Delta ^ q, \partial ( \Delta ^ p \times \Delta ^ q; \operatorname{\mathbf{Z}})$

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

$\xymatrix@R =50pt@C=50pt{ \mathrm{N}_{p}( \Delta ^ p; \operatorname{\mathbf{Z}}) \times \mathrm{N}_{q}( \Delta ^ q; \operatorname{\mathbf{Z}}) \ar [r]^-{ \triangledown '} \ar [d]^{\sim } & \mathrm{N}_{p+q}( \Delta ^ p \times \Delta ^ q; \operatorname{\mathbf{Z}}) \ar [d]^{\sim } \\ \mathrm{N}_{p}( \Delta ^ p, \partial \Delta ^ p; \operatorname{\mathbf{Z}}) \times \mathrm{N}_{q}( \Delta ^ q, \partial \Delta ^ q; \operatorname{\mathbf{Z}}) \ar [r]^-{ \triangledown } & \mathrm{N}_{p+q}( \Delta ^ p \times \Delta ^ q, \partial ( \Delta ^ p \times \Delta ^ q); \operatorname{\mathbf{Z}}) }$

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

\begin{eqnarray*} m \partial (a \triangledown b) & = & \partial ( a \triangledown ' b) \\ & = & (\partial a) \triangledown ' b + (-1)^{p} a \triangledown ' (\partial b) \\ & = & n( (\partial a) \triangledown b + (-1)^{p} a \triangledown (\partial b) ) \\ & = & n \partial ( a \triangledown b). \end{eqnarray*}

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. $\square$