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Theorem 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.

Proof of Theorem 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 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, 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 We therefore have a commutative diagram of short exact sequences

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

where the left vertical map is a quasi-isomorphism by virtue of our inductive hypothesis. Invoking Remark, 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 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 and the upper horizontal map is an isomorphism, so the lower horizontal map is a chain homotopy equivalence as well. $\square$