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.