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2.5.7 The Shuffle Product

Let $\operatorname{ Ab }_{\Delta } = \operatorname{Fun}( \operatorname{{\bf \Delta }}^{\operatorname{op}}, \operatorname{ Ab })$ denote the category of simplicial abelian groups. We will regard $\operatorname{ Ab }_{\Delta }$ as a monoidal category with respect to the “levelwise” tensor product of (Example 2.1.2.16): if $A_{\bullet }$ and $B_{\bullet }$ are simplicial abelian groups, then their tensor product $A_{\bullet } \otimes B_{\bullet }$ is the simplicial abelian group given by the construction $([n] \in \operatorname{{\bf \Delta }}^{\operatorname{op}} ) \mapsto A_{n} \otimes B_{n}$. The category of chain complexes $\operatorname{Ch}(\operatorname{\mathbf{Z}})$ is also equipped with a monoidal structure (Construction 2.5.1.17); we denote the tensor product of chain complexes $X_{\ast }$ and $Y_{\ast }$ by $X_{\ast } \boxtimes Y_{\ast }$ or $(X \boxtimes Y)_{\ast }$; given chains $x \in X_{p}$ and $y \in Y_{q}$, we will write $x \boxtimes y$ for the image of $(x,y)$ in the abelian group $(X \boxtimes Y)_{p+q}$. According to Theorem 2.5.6.1, the normalized Moore complex functor $A_{\bullet } \mapsto \mathrm{N}_{\ast }(A)$ determines a fully faithful embedding $\mathrm{N}_{\ast }: \operatorname{ Ab }_{\Delta } \hookrightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}})$. Beware that this functor does not commute with the formation of tensor products. Nevertheless, we have the following result:

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

    \[ \xymatrix { \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \ar [r]^-{ \triangledown } \ar [d]^{ \mathrm{N}_{\ast }(f) \times \mathrm{N}_{\ast }(g)} & \mathrm{N}_{p+q}( A \otimes B) \ar [d]^{ \mathrm{N}_{\ast }(f \otimes g) } \\ \mathrm{N}_{\ast }(A') \times \mathrm{N}_{\ast }(B') \ar [r]^-{ \triangledown } & \mathrm{N}_{\ast }( A' \otimes B') } \]

    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).

For simplicial abelian groups $A_{\bullet }$ and $B_{\bullet }$ and integer $p,q \in \operatorname{\mathbf{Z}}$, we will refer to the map

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

of Proposition 2.5.7.1 as the shuffle product. We begin by giving an explicit construction of this map, following Eilenberg and MacLane (see [MR0056295]).

Notation 2.5.7.2 ($(p,q)$-Shuffles). Let $p$ and $q$ be nonnegative integers. A $(p,q)$-shuffle is a strictly increasing map of partially ordered sets $\sigma : [p+q] \rightarrow [p] \times [q]$, which we will often identify with a nondegenerate $(p+q)$-simplex of the Cartesian product $\Delta ^{p} \times \Delta ^{q}$.

If $\sigma $ is a $(p,q)$-shuffle, we let $\sigma _{-}: [p+q] \rightarrow [p]$ and $\sigma _{+}: [p+q] \rightarrow [q]$ denote the nondecreasing maps given by the components of $\sigma $ (so that $\sigma (i) = ( \sigma _{-}(i), \sigma _{+}(i) )$ for $0 \leq i \leq p+q$). Let $I_{-}$ denote the set of integers $1 \leq i \leq p+q$ satisfying $\sigma _{-}(i-1) < \sigma _{-}(i)$ (or equivalently $\sigma _{+}(i-1) = \sigma _{+}(i)$), and let $I_{+}$ the set of integers $1 \leq i \leq p+q$ satisfying $\sigma _{+}(i-1) < \sigma _{+}(i)$ (or equivalently $\sigma _{-}(i-1)= \sigma _{-}(i)$). We let $(-1)^{\sigma }$ denote the product

\[ \prod _{ (i,j) \in I_{-} \times I_{+} } \begin{cases} 1 & \text{ if } i < j \\ -1 & \text{ if } i > j. \end{cases} \]

We will refer to $(-1)^{\sigma }$ as the sign of the $(p,q)$-shuffle $\sigma $.

Construction 2.5.7.3 (The Unnormalized Shuffle Product). Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups, and suppose we are given elements $a \in A_{p}$ and $b \in B_{q}$. We let $a \bar{\triangledown } b$ denote the sum

\[ \sum _{\sigma } (-1)^{\sigma } \sigma _{-}^{\ast }(a) \otimes \sigma _{+}^{\ast }(b) \in (A \otimes B)_{p+q} \]

Here the sum is taken over all $(p,q)$-shuffles $\sigma = (\sigma _{-}, \sigma _{+})$ (Notation 2.5.7.2), and we write $\sigma _{-}^{\ast }: A_{p} \rightarrow A_{p+q}$ and $\sigma _{+}^{\ast }: B_{q} \rightarrow B_{p+q}$ for the structure morphisms of the simplicial abelian groups $A_{\bullet }$ and $B_{\bullet }$, respectively. We will refer to $a \bar{\triangledown } b$ as the unnormalized shuffle product of $a$ and $b$.

We now summarize some essential properties of Construction 2.5.7.3.

Remark 2.5.7.4 (Unitality of the Shuffle Product). Let $\operatorname{\mathbf{Z}}[ \Delta ^{0} ]$ be the constant simplicial abelian group taking the value $\operatorname{\mathbf{Z}}$, and let us identify the integer $1$ with the corresponding $0$-simplex of $\operatorname{\mathbf{Z}}[ \Delta ^{0} ]$. Then, for any simplicial abelian group $A_{\bullet }$, the canonical isomorphisms $A_{\bullet } \simeq (A \otimes \operatorname{\mathbf{Z}}[\Delta ^{0}])_{\bullet }$ and $A_{\bullet } \simeq (\operatorname{\mathbf{Z}}[ \Delta ^{0} ] \otimes A_{\bullet } )$ are given by $a \mapsto a \bar{\triangledown } 1$ and $a \mapsto 1 \bar{\triangledown } a$, respectively.

Remark 2.5.7.5 (Commutativity of the Shuffle Product). Let $\sigma : [p+q] \rightarrow [p] \times [q]$ be a $(p,q)$-shuffle, and let $\sigma ': [p+q] \rightarrow [q] \times [p]$ denote the composition of $\sigma $ with the isomorphism $[p] \times [q] \simeq [q] \times [p]$ given by permuting the factors. Then $\sigma '$ is a $(q,p)$-shuffle, whose sign is given by $(-1)^{\sigma '} = (-1)^{pq} \cdot (-1)^{\sigma }$. Consequently, if $A_{\bullet }$ and $B_{\bullet }$ are simplicial abelian groups containing simplices $a \in A_{p}$ and $b \in B_{q}$, then the canonical isomorphism $(A \otimes B)_{p+q} \simeq (B \otimes A)_{p+q}$ carries $a \bar{\triangledown } b$ to $(-1)^{pq} ( b \bar{\triangledown } a)$.

Remark 2.5.7.6 (Associativity of the Shuffle Product). Let $A_{\bullet }$, $B_{\bullet }$, and $C_{\bullet }$ be simplicial abelian groups containing simplices $a \in A_{p}$, $b \in B_{q}$, and $c \in C_{r}$. Then the canonical isomorphism $( A \otimes (B \otimes C))_{p+q+r} \simeq ( (A \otimes B) \otimes C)_{p+q+r}$ carries $a \bar{\triangledown } (b \bar{\triangledown } c)$ to $( a \bar{\triangledown } b) \bar{\triangledown } c$. Both of these iterated shuffle products can be described concretely as the sum

\[ \sum _{ \sigma } (-1)^{\sigma } \sigma _{-}^{\ast }(a) \otimes \sigma _{0}^{\ast }(b) \otimes \sigma _{+}^{\ast }(c), \]

where the sum is taken over all strictly increasing maps $\sigma = (\sigma _{-}, \sigma _{0}, \sigma _{+}): [p+q+r] \rightarrow [p] \times [q] \times [r]$, and $(-1)^{\sigma }$ denotes the product

\[ \prod _{1 \leq i < j \leq p+q+r} \begin{cases} -1 & \text{ if } \sigma _{-}(j-1) < \sigma _{-}(j) \text{ and } \sigma _{-}(i-1) = \sigma _{-}(i) \\ -1 & \text{ if } \sigma _+(j-1) = \sigma _+(j) \text{ and } \sigma _+(i-1) < \sigma _{+}(i) \\ 1 & \text{ otherwise.} \end{cases} \]

Proposition 2.5.7.7. Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. Then the unnormalized shuffle product $\bar{\triangledown }: A_{p} \times B_{q} \rightarrow (A \otimes B)_{p+q}$ satisfies the Leibniz rule

\[ \partial ( a \bar{\triangledown } b) = (\partial a) \bar{\triangledown } b + (-1)^{p} a \bar{\triangledown } (\partial b). \]

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 $\tau = d_ j(\sigma )$. 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

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

    here $\delta ^{a}: [p-1] \hookrightarrow [p]$ denotes the unique monomorphism whose image does not contain $a$ (Notation 1.1.1.8). 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

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

    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_ 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_ 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.

$\square$

We now adapt the shuffle product to the setting of normalized Moore complexes. For every simplicial abelian group $A_{\bullet }$, let $\mathrm{D}_{\ast }(A) \subseteq \mathrm{C}_{\ast }(A)$ be the subcomplex generated by the degenerate simplices of $A_{\bullet }$ (see Proposition 2.5.5.6).

Proposition 2.5.7.8. Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. Then the unnormalized shuffle product

\[ \bar{\triangledown }: \mathrm{C}_ p(A) \times \mathrm{C}_ q(B) \rightarrow \mathrm{C}_{p+q}(A \otimes B) \]

carries the subsets $\mathrm{D}_{p}(A) \times \mathrm{C}_{q}(B)$ and $ \mathrm{C}_{p}(A) \boxtimes \mathrm{D}_{q}(B)$ into the subgroup $\mathrm{D}_{p+q}(A \otimes B) \subseteq \mathrm{C}_{p+q}( A \otimes B)$.

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 $a = s_ i(a')$ 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

\[ A_{p+q} \xrightarrow {d_ j} A_{p+q-1} \xrightarrow {s_ j} A_{p+q} \quad \quad B_{p+q} \xrightarrow {d_ j} B_{p+q-1} \xrightarrow {s_ j} B_{p+q}, \]

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

\[ \sum _{\sigma } (-1)^{\sigma } \sigma _{-}^{\ast }(a) \otimes \sigma _{+}^{\ast }(b) \]

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

Construction 2.5.7.9 (The Shuffle Product). Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. It follows from Proposition 2.5.7.8 that for every pair of integers $p,q \in \operatorname{\mathbf{Z}}$, there is a unique bilinear map $\triangledown : \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B)$ for which the diagram

\[ \xymatrix { \mathrm{C}_{p}(A) \boxtimes \mathrm{C}_{q}(B) \ar [r]^-{ \bar{\triangledown } } \ar@ {->>}[d] & \mathrm{C}_{p+q}(A \otimes B) \ar@ {->>}[d] \\ \mathrm{N}_{p}(A) \boxtimes \mathrm{N}_{q}(B) \ar [r]^-{ \triangledown } & \mathrm{N}_{p+q}(A \otimes B) } \]

commutes. We will refer to $\triangledown : \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B)$ as the shuffle product map. Given elements $a \in \mathrm{N}_{p}(A)$ and $b \in \mathrm{N}_{q}(B)$, we will write $a \triangledown b$ for the image of the pair $(a,b)$ under the shuffle product map, which we refer to as the shuffle product of $a$ and $b$.

We now summarize some properties of the properties of Construction 2.5.7.9, which follow immediately from the corresponding results for the unnormalized shuffle product (Remarks 2.5.7.4, 2.5.7.5, 2.5.7.6, and Proposition 2.5.7.7).

Proposition 2.5.7.10. Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. Then:

$(1)$

The canonical isomorphisms $\mathrm{N}_{\ast }(A) \simeq \mathrm{N}_{\ast }(A \otimes \operatorname{\mathbf{Z}}[ \Delta ^0] )$ and $\mathrm{N}_{\ast }(A) \simeq \mathrm{N}_{\ast }(\operatorname{\mathbf{Z}}[ \Delta ^0] \otimes A )$ are given by $a \mapsto a \triangledown 1$ and $a \mapsto 1 \triangledown a$, respectively; here we identify the integer $1$ with its image in $\mathrm{N}_{\ast }( \Delta ^0; \operatorname{\mathbf{Z}}) \simeq \operatorname{\mathbf{Z}}$.

$(2)$

For $a \in \mathrm{N}_{p}(A)$ and $b \in \mathrm{N}_{q}(B)$, we have $a \triangledown b = (-1)^{pq} (b \triangledown a)$; here we abuse notation by identifying $a \triangledown b$ with its image under the canonical isomorphism $\mathrm{N}_{p+q}(A \otimes B) \simeq \mathrm{N}_{p+q}( B \otimes A)$.

$(3)$

Let $C_{\bullet }$ be another simplicial abelian group, and suppose we are given elements $a \in \mathrm{N}_{p}(A)$, $b \in \mathrm{N}_{q}(B)$, and $c \in \mathrm{N}_{r}(C)$. Then $a \triangledown (b \triangledown c) = (a \triangledown b) \triangledown c$; here we abuse notation by identifying $a \triangledown (b \triangledown c)$ with its image under the canonical isomorphism $\mathrm{N}_{p+q+r}( A \otimes (B \otimes C) ) \simeq \mathrm{N}_{p+q+r}( (A \otimes B) \otimes C)$.

$(4)$

The shuffle product $\triangledown : \mathrm{N}_{p}(A) \times \mathrm{N}_{q}(B) \rightarrow \mathrm{N}_{p+q}(A \otimes B)$ satisfies the Leibniz rule

\[ \partial ( a \triangledown b) = (\partial a) \triangledown b + (-1)^{p} a \triangledown (\partial b). \]

Notation 2.5.7.11 (The Eilenberg-Zilber Homomorphism). Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. It follows from assertion $(4)$ of Proposition 2.5.7.10 that there is a unique chain map

\[ \mathrm{EZ}: \mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B) \rightarrow \mathrm{N}_{\ast }(A \otimes B) \]

satisfying $\mathrm{EZ}( a \boxtimes b) = a \triangledown b$ (see Exercise 2.5.1.15). We will refer to $\mathrm{EZ}$ as the Eilenberg-Zilber homomorphism (see Remark 2.5.7.16). It follows from assertions $(1)$ and $(3)$ of Proposition 2.5.7.10 that the collection of chain maps

\[ \{ \mathrm{EZ}: \mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B) \rightarrow \mathrm{N}_{\ast }(A \otimes B) \} _{A_{\bullet }, B_{\bullet } \in \operatorname{ Ab }_{\Delta }} \]

determine a lax monoidal structure (Definition 2.1.5.8) on the normalized Moore complex functor $\mathrm{N}_{\ast }: \operatorname{ Ab }_{\Delta } \rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}})$, with unit given by the canonical isomorphism of chain complexes $\operatorname{\mathbf{Z}}[0] \simeq \mathrm{N}_{\ast }( \operatorname{\mathbf{Z}}[\Delta ^0] )$ (in fact, it is even a lax symmetric monoidal structure in the sense of Definition : this follows from assertion $(2)$ of Proposition 2.5.7.10).

Example 2.5.7.12. Let $S_{\bullet }$ and $T_{\bullet }$ be simplicial sets, and let $\operatorname{\mathbf{Z}}[ S_{\bullet } ]$ and $\operatorname{\mathbf{Z}}[ T_{\bullet } ]$ denote the free simplicial abelian groups generated by $S_{\bullet }$ and $T_{\bullet }$, respectively. Then the tensor product $\operatorname{\mathbf{Z}}[ S_{\bullet }] \otimes \operatorname{\mathbf{Z}}[ T_{\bullet } ]$ can be identified with the free simplicial abelian group $\operatorname{\mathbf{Z}}[ S_{\bullet } \times T_{\bullet } ]$ generated by the Cartesian product $S_{\bullet } \times T_{\bullet }$. Invoking Construction 2.5.7.9, we obtain shuffle product maps

\[ \triangledown : \mathrm{N}_{p}(S; \operatorname{\mathbf{Z}}) \times \mathrm{N}_{q}(T; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{p+q}(S \times T; \operatorname{\mathbf{Z}}) \]

which induce a map of chain complexes $\mathrm{EZ}: \mathrm{N}_{\ast }(S; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{\ast }(S \times T; \operatorname{\mathbf{Z}})$. Allowing $S_{\bullet }$ and $T_{\bullet }$ to vary, these chain maps furnish a lax (symmetric) monoidal structure on the functor

\[ \mathrm{N}_{\ast }( -; \operatorname{\mathbf{Z}}): \operatorname{Set_{\Delta }}\rightarrow \operatorname{Ch}(\operatorname{\mathbf{Z}}) \quad \quad S_{\bullet } \mapsto \mathrm{N}_{\ast }(S; \operatorname{\mathbf{Z}}). \]

Remark 2.5.7.13. The Eilenberg-Zilber homomorphism of Example 2.5.7.12 admits a topological interpretation. Recall that, for every simplicial set $S_{\bullet }$, the topological space $| S_{\bullet } |$ is a CW complex (Remark 1.1.8.14). More precisely, $| S_{\bullet } |$ admits a CW decomposition with one cell $e_{\sigma }$ for each nondegenerate simplex $\sigma : \Delta ^ n \rightarrow S_{\bullet }$, where $e_{\sigma }$ is defined as the image of the composite map

\[ | \Delta ^ n |^{\circ } \hookrightarrow | \Delta ^ n | \xrightarrow { | \sigma |} | S_{\bullet } |; \]

here $| \Delta ^ n |^{\circ } = \{ (t_0, \ldots , t_ n) \in \operatorname{\mathbf{R}}_{>0}: t_0 + \cdots + t_ n = 1 \} $ denotes the interior of the topological $n$-simplex. The chain complex $\mathrm{N}_{\ast }( S; \operatorname{\mathbf{Z}})$ of Construction 2.5.5.9 can then be identified with the cellular chain complex associated to this cell decomposition of $| S_{\bullet } |$.

When $S_{\bullet } = S'_{\bullet } \times S''_{\bullet }$ factors a product of two other simplicial sets $S'_{\bullet }$ and $S''_{\bullet }$, the topological space $| S_{\bullet } |$ admits a different CW structure, whose cells are given by $\varphi ^{-1}( e_{\sigma '} \times e_{\sigma ''} )$; here $\varphi $ denotes the canonical map $| S_{\bullet } | \rightarrow | S'_{\bullet } | \times | S''_{\bullet } |$, and $\sigma '$ and $\sigma ''$ range over the collection of nondegenerate simplices of $S'_{\bullet }$ and $S''_{\bullet }$, respectively. The cellular chain complex associated to this cell decomposition can be identified with the tensor product complex $\mathrm{N}_{\ast }( S'_{\bullet }; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }( S''_{\bullet }; \operatorname{\mathbf{Z}})$.

It is not difficult to see that if $\sigma ' \in S'_{p}$ and $\sigma '' \in S''_{q}$ are nondegenerate simplices of $S'_{\bullet }$ and $S''_{\bullet }$, respectively, then the subset $\varphi ^{-1}( e_{\sigma '} \times e_{\sigma ''} ) \subseteq | S_{\bullet } |$ can be written as a finite union of cells of the form $e_{\sigma }$ (where $\sigma $ is a nondegenerate simplex of $S_{\bullet }$). Writing $[\sigma ']$ and $[\sigma '']$ for the corresponding generators of $\mathrm{N}_{p}( S'; \operatorname{\mathbf{Z}})$ and $\mathrm{N}_{q}(S''; \operatorname{\mathbf{Z}})$, the shuffle product is given by

\[ [\sigma '] \triangledown [\sigma ''] = \sum _{\sigma } \pm [ \sigma ] \in \mathrm{N}_{p+q}(S), \]

where the sum is taken over all nondegenerate $(p+q)$-simplices $\sigma $ of $S_{\bullet }$ satisfying $e_{\sigma } \subseteq \varphi ^{-1}( e_{\sigma '} \times e_{\sigma ''})$; note that every such simplex $\sigma $ can be written uniquely as a composition

\[ \Delta ^{p+q} \xrightarrow {\tau } \Delta ^{p} \times \Delta ^{q} \xrightarrow {\sigma ' \times \sigma ''} S'_{\bullet } \times S''_{\bullet } = S_{\bullet } \]

where $\tau $ is a $(p,q)$-shuffle in the sense of Notation 2.5.7.2. Moreover, the sign $(-1)^{\tau }$ also admits a topological interpretation: it is the degree of the open embedding $\varphi |_{ e_{\sigma }}: e_{\sigma } \hookrightarrow e_{\sigma '} \times e_{\sigma ''}$ (with respect to certain standard orientations of the cells $e_{\sigma }$, $e_{\sigma '}$, and $e_{\sigma ''}$).

Theorem 2.5.7.14. Let $A_{\bullet }$ and $B_{\bullet }$ be simplicial abelian groups. Then the Eilenberg-Zilber homomorphism

\[ \mathrm{EZ}: \mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }(B) \rightarrow \mathrm{N}_{\ast }(A \otimes B) \]

is a quasi-isomorphism: that is, it induces an isomorphism on homology.

Corollary 2.5.7.15. Let $S_{\bullet }$ and $T_{\bullet }$ be simplicial sets. Then the Eilenberg-Zilber homomorphism

\[ \mathrm{EZ}: \mathrm{N}_{\ast }(S; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{\ast }(S \times T; \operatorname{\mathbf{Z}}) \]

is a quasi-isomorphism.

Remark 2.5.7.16. Corollary 2.5.7.15 is essentially due to Eilenberg and Zilber. More precisely, in [MR52767], Eilenberg and Zilber proved that there exists a collection of quasi-isomorphisms $G_{S,T}: \mathrm{N}_{\ast }(S; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{\ast }(S \times T; \operatorname{\mathbf{Z}})$ depending functorially on the the simplicial sets $S_{\bullet }$ and $T_{\bullet }$. The proof given in [MR52767] uses the method of acyclic models and does not provide a concrete description of the maps $G_{S,T}$. However, it is not difficult to see that such a collection of chain maps $\{ G_{S,T} \} $ must coincide up to sign with the Eilenberg-Zilber homomorphisms of Example 2.5.7.12 (see Exercise 2.5.7.18 below).

Variant 2.5.7.17. Let $S_{\bullet }$ and $T_{\bullet }$ be simplicial sets containing simplicial subsets $S'_{\bullet }$ and $T'_{\bullet }$, respectively. Applying Theorem 2.5.7.14 to the simplicial abelian groups $\operatorname{\mathbf{Z}}[ S_{\bullet } ] / \operatorname{\mathbf{Z}}[ S'_{\bullet } ]$ and $\operatorname{\mathbf{Z}}[ T_{\bullet } ] / \operatorname{\mathbf{Z}}[ T'_{\bullet } ]$, we obtain a quasi-isomorphism

\[ \mathrm{EZ}: \mathrm{N}_{\ast }(S, S'; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T, T'; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{\ast }(S \times T, (S' \times T) \cup (S \times T'); \operatorname{\mathbf{Z}}), \]

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 2.5.6.7). We will say that $M_{\ast }$ is good if the Eilenberg-Zilber map

\[ \mathrm{N}_{\ast }(A) \boxtimes M_{\ast } \simeq \mathrm{N}_{\ast }(A) \boxtimes \mathrm{N}_{\ast }( \mathrm{K}(M_{\ast })) \xrightarrow { \mathrm{EZ} } \mathrm{N}_{\ast }(A \otimes \mathrm{K}(M_{\ast }) ) \]

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

\[ 0 \rightarrow M'_{\ast } \rightarrow M_{\ast } \rightarrow T[n] \rightarrow 0, \]

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

\[ 0 \rightarrow \mathrm{K}( M'_{\ast } ) \rightarrow \mathrm{K}( M_{\ast } ) \rightarrow \mathrm{K}( T[n] ) \rightarrow 0 \]

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

\[ \xymatrix { 0 \ar [r] & \mathrm{N}_{\ast }(A) \boxtimes M'_{\ast } \ar [r] \ar [d] & \mathrm{N}_{\ast }(A) \boxtimes M_{\ast } \ar [r] \ar [d] & \mathrm{N}_{\ast }(A) \boxtimes T[n] \ar [r] & 0 \\ 0 \ar [r] & \mathrm{N}_{\ast }(A \otimes \mathrm{K}(M'_{\ast }) ) \ar [r] & \mathrm{N}_{\ast }(A \otimes \mathrm{K}(M_{\ast }) ) \ar [r] & \mathrm{N}_{\ast }(A \otimes \mathrm{K}(T[n]) ) & 0,} \]

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

\[ \mathrm{EZ}: \mathrm{N}_{\ast }(\Delta ^ m; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(\Delta ^ n; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}_{\ast }(\Delta ^ m \otimes \Delta ^ n; \operatorname{\mathbf{Z}}) \]

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

\[ \xymatrix { \operatorname{\mathbf{Z}}[0] \boxtimes \operatorname{\mathbf{Z}}[0] \ar [r]^-{\sim } \ar [d]^{ [u] \boxtimes [v] } & \operatorname{\mathbf{Z}}[0] \ar [d]^{ [w] } \\ \mathrm{N}_{\ast }(\Delta ^ m; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(\Delta ^ n; \operatorname{\mathbf{Z}}) \ar [r] & \mathrm{N}_{\ast }(\Delta ^ m \otimes \Delta ^ n; \operatorname{\mathbf{Z}}) } \]

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

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

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

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

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. $\square$

Exercise 2.5.7.18. For every pair of simplicial sets $S_{\bullet }$ and $T_{\bullet }$, let

\[ G_{S,T}: \mathrm{N}_{\ast }( S; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T; \operatorname{\mathbf{Z}}) \rightarrow \mathrm{N}( S \times T; \operatorname{\mathbf{Z}}) \]

be a chain map. Assume that the maps $G_{S,T}$ depend functorially on $S_{\bullet }$ and $T_{\bullet }$: that is, for all maps of simplicial sets $f: S_{\bullet } \rightarrow S'_{\bullet }$ and $g: T_{\bullet } \rightarrow T'_{\bullet }$, the diagram of chain complexes

\[ \xymatrix { \mathrm{N}_{\ast }( S; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T; \operatorname{\mathbf{Z}}) \ar [r]^-{ G_{S,T}} \ar [d]^{ \mathrm{N}_{\ast }(f; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(g; \operatorname{\mathbf{Z}}) } & \mathrm{N}_{\ast }(S \times T; \operatorname{\mathbf{Z}}) \ar [d]^{ \mathrm{N}_{\ast }(f \times g; \operatorname{\mathbf{Z}}) } \\ \mathrm{N}_{\ast }( S'; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }(T'; \operatorname{\mathbf{Z}}) \ar [r]^-{ G_{S',T'}} & \mathrm{N}_{\ast }(S' \times T'; \operatorname{\mathbf{Z}}) } \]

is commutative. Adapt the proof Proposition 2.5.7.1 to show that there exists an integer $n$ (not depending on $S_{\bullet }$ and $T_{\bullet }$) such that $G_{S,T} = n \mathrm{EZ}$, where $\mathrm{EZ}$ is the Eilenberg-Zilber homomorphism of Example 2.5.7.12.