Kerodon

Proof. Without loss of generality, we may assume that $(p,q) \neq (0,0)$ and that the simplicial abelian groups $A_{\bullet } \simeq \operatorname{\mathbf{Z}}[ \Delta ^{p} ]$ and $B_{\bullet } \simeq \operatorname{\mathbf{Z}}[ \Delta ^{q} ]$ are freely generated by $a$ and $b$, respectively. In this case, we can identify $(A \otimes B)_{p+q-1}$ with the free abelian group generated by the set of $(p+q-1)$-simplices of $\Delta ^{p} \times \Delta ^{q}$, which we view as nondecreasing functions $\tau : [p+q-1] \rightarrow [p] \times [q]$. For every such simplex $\tau $, let $c$, $c_{-}$, and $c_{+}$ denote the coefficients of $\tau $ appearing in $\partial ( a \bar{\triangledown } b)$, $(\partial a) \bar{\triangledown } b$, and $a \bar{\triangledown } (\partial b)$, respectively. We wish to prove that $c = c_{-} + (-1)^{p} c_{+}$. We may assume without loss of generality that the map $\tau $ is injective (otherwise, we have $c= c_{-} = c_{+} = 0$). Let us identify $\tau $ with a pair $(\tau _{-}, \tau _{+} )$, where $\tau _{-}: [p+q-1] \rightarrow [p]$ and $\tau _{+}: [p+q-1] \rightarrow [q]$ are nondecreasing functions. We now distinguish three cases:

$(1)$

Suppose that the map $\tau _{-}: [p+q-1] \rightarrow [p]$ is not surjective (that is, $\tau $ belongs to the simplicial subset $(\partial \Delta ^ p) \times \Delta ^ q \subseteq \Delta ^{p} \times \Delta ^{q}$). Then $p > 0$ and there exists a unique integer $0 \leq i \leq p$ which does not belong to the image of $\tau _{-}$. We proceed under the assumption that $i < p$ (the case $i > 0$ follows by a similar argument, with minor changes in notation). We then make the following observations:

• There is a unique $(p,q)$-shuffle $\sigma $ and integer $0 \leq j \leq p+q$ satisfying $\tau = d^{p+q}_ j(\sigma )$. Here $j$ is the smallest integer satisfying $\tau _{-}(j) = i+1$, and $\sigma $ is given by the formula

  \[ \sigma (k) = \begin{cases} ( \tau _{-}(k), \tau _{+}(k) ) & \text{ if } k < j \\ (i, \tau _{+}(j) ) & \text{ if } k = j \\ (\tau _{-}(k-1), \tau _{+}(k-1) ) & \text{ if } k > j. \end{cases} \]

  It follows that $c = (-1)^{j} \cdot (-1)^{\sigma }$.

• There is a unique $(p-1, q)$-shuffle $\sigma '$ and integer $0 \leq a \leq p$ such that $\tau $ is given by the composition

  \[ [p+q-1] \xrightarrow { \sigma ' } [p-1] \times [q] \xrightarrow { \delta ^{a}_{p} \times \operatorname{id}} [p] \times [q]; \]

  here $\delta ^{a}_{p}: [p-1] \hookrightarrow [p]$ denotes the unique monomorphism whose image does not contain $a$ (Construction 1.1.1.4). These conditions guarantee that $a = i$ and that $\sigma '$ is given by the formula

  \[ \sigma '(k) = \begin{cases} (\tau _{-}(k), \tau _{+}(k) ) & \text{ if } k < j \\ ( \tau _{-}(k)-1, \tau _{+}(k) ) & \text{ if } k \geq j. \end{cases} \]

  Consequently, we have $c_{-} = (-1)^{i} \cdot (-1)^{\sigma '}$.

• There does not exist a $(p,q-1)$-shuffle $\sigma ''$ and an integer $0 \leq b \leq q$ for which $\tau $ is equal to the composition

  \[ [p+q-1] \xrightarrow { \sigma '' } [p] \times [q-1] \xrightarrow {\operatorname{id}\times \delta ^{b}_{q} } [p] \times [q]. \]

  Consequently, the coefficient $c_{+}$ vanishes.

We are therefore reduced to verifying the identity $(-1)^{j} \cdot (-1)^{\sigma } = (-1)^{i} \cdot (-1)^{\sigma '}$, which is an immediate consequence of the definitions.

$(2)$

Suppose that the map $\tau _{+}: [p+q-1] \rightarrow [q]$ is not surjective (that is, $\tau $ belongs to the simplicial subset $\Delta ^{q} \times (\partial \Delta ^{q}) \subseteq \Delta ^{p} \times \Delta ^{q}$). The argument in this case proceeds as in $(1)$, with minor adjustments in notation.

$(3)$

The functions $\tau _{-}$ and $\tau _{+}$ are both surjective. In this case, we have $c_{-} = c_{+} = 0$. Note that there is a unique integer $1 \leq j \leq p+q-1$ satisfying $\tau _{-}(j-1) < \tau _{-}(j)$ and $\tau _{+}(j-1) < \tau _{+}(j)$. From this, it is easy to see that if $\sigma $ is a $(p,q)$-shuffle satisfying $d^{p+q}_ k(\sigma ) = \tau $ for some $0 \leq k \leq p+q$, then we must have $k=j$. Moreover, there are exactly two $(p,q)$-shuffles $\sigma $ satisfying $d^{p+q}_ j(\sigma ) = \tau $, given by the formulae

\[ \sigma (i) = \begin{cases} \tau (i) & \text{ if } i < j \\ (\tau _{-}(j-1), \tau _{+}(j)) & \text{ if } i = j \\ \tau (i-1) & \text{ if } i > j \end{cases} \quad \quad \sigma (i) = \begin{cases} \tau (i) & \text{ if } i < j \\ (\tau _{-}(j), \tau _{+}(j-1)) & \text{ if } i = j \\ \tau (i-1) & \text{ if } i > j. \end{cases} \]

Since these $(p,q)$-shuffles have opposite sign, we conclude that $c = 0 = c_{-} + (-1)^{p} c_{+}$, as desired.

Proof. Let $a \in A_ p$ and $b \in B_{q}$ be simplices of $A_{\bullet }$ and $B_{\bullet }$, respectively. We wish to show that if either $a$ belongs to $\mathrm{D}_{p}(A)$ or $b$ belongs to $\mathrm{D}_{q}(B)$, then the unnormalized shuffle product $a \bar{\triangledown } b$ belongs to $\mathrm{D}_{p+q}(A \otimes B)$. Without loss of generality, we may assume that $a$ belongs to $\mathrm{D}_{p}(A)$. Decomposing $a$ into summands, we can further assume that $a = s^{p-1}_ i(a')$ for some $0 \leq i \leq p-1$ and some $a' \in A_{p-1}$. Let $\sigma = (\sigma _{-}, \sigma _{+})$ be a $(p,q)$-shuffle. Then there exists a unique integer $0 \leq j < p+q$ satisfying $\sigma _{-}( j) = i$ and $\sigma _{-}(j+1) = i+1$. It then follows that both $\sigma _{-}^{\ast }(a)$ and $\sigma _{+}^{\ast }(b)$ are fixed points of the composite maps

so that $\sigma _{-}^{\ast }(a) \otimes \sigma _{+}^{\ast }(b)$ is a degenerate simplex of $(A \otimes B)_{\bullet }$. Allowing $\sigma $ to vary, we deduce that the shuffle product

belongs to $\mathrm{D}_{p+q}( A \otimes B)$. $\square$

Proof of Theorem 2.5.7.14. Let us first regard the simplicial abelian group $A_{\bullet }$ as fixed. Let $M_{\ast } \in \operatorname{Ch}(\operatorname{\mathbf{Z}})_{\geq 0}$ be a chain complex of abelian groups which is concentrated in degrees $\geq 0$, and let $\mathrm{K}(M_{\ast })$ be the associated Eilenberg-MacLane space (Construction 2.5.6.9). We will say that $M_{\ast }$ is good if the Eilenberg-Zilber map

is a quasi-isomorphism. By virtue of Theorem 2.5.6.1, it will suffice to show that every object $M_{\ast } \in \operatorname{Ch}(\operatorname{\mathbf{Z}})_{\geq 0}$ is good. Writing $M_{\ast }$ as a filtered direct limit of bounded subcomplexes, we may assume that $M_{\ast }$ is concentrated in degrees $\leq n$ for some integer $n \geq 0$. We proceed by induction on $n$. Let $T$ denote the abelian group $M_ n$, so that we have a short exact sequence of chain complexes

where $M'_{\ast }$ is concentrated in degrees $\leq n-1$. Note that this sequence is degreewise split, so that the associated exact sequence of simplicial abelian groups

is also degreewise split (see Remark 2.5.6.18). We therefore have a commutative diagram of short exact sequences

where the left vertical map is a quasi-isomorphism by virtue of our inductive hypothesis. Invoking Remark 2.5.1.7, we see that $M_{\ast }$ is good if and only if the chain complex $T[n]$ is good. In particular, the condition that $M_{\ast }$ is good depends only the abelian group $T = M_ n$. We may therefore assume without loss of generality that $M_{\ast }$ factors as a tensor product $\mathrm{N}_{\ast }( \Delta ^ n; \operatorname{\mathbf{Z}}) \boxtimes T[0]$. We are therefore reduced to proving Theorem 2.5.7.14 in the special case where $B_{\bullet }$ factors as a tensor product of $\operatorname{\mathbf{Z}}[ \Delta ^ n ]$ with the abelian group $T$.

Applying the same argument with the roles of $A_{\bullet }$ and $B_{\bullet }$ reversed, we can also assume that $A_{\bullet }$ factors as the tensor product of $\operatorname{\mathbf{Z}}[ \Delta ^ m ]$ with another abelian group $T'$. In this case, we are reduced to proving that the Eilenberg-Zilber map

becomes a quasi-isomorphism after tensoring both sides with the abelian group $T' \otimes T$. In fact, we claim that this map is chain homotopy equivalence. To prove this, let $u$ and $v$ denote the initial vertices of $\Delta ^ m$ and $\Delta ^ n$, respectively, and write $[u]$ and $[v]$ for the corresponding generators of $\mathrm{N}_{0}( \Delta ^ m; \operatorname{\mathbf{Z}})$ and $\mathrm{N}_{0}( \Delta ^ n; \operatorname{\mathbf{Z}})$. Then the shuffle product $[u] \triangledown [v]$ is given by $[w]$, where $w = (u,v)$ is the vertex of $\Delta ^ m \times \Delta ^ n$ corresponding to the least element of the partially ordered set $[m] \times [n]$. We have a commutative diagram of chain complexes

where the vertical maps are chain homotopy equivalences (Example 2.5.5.14) and the upper horizontal map is an isomorphism, so the lower horizontal map is a chain homotopy equivalence as well. $\square$

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. $\square$