# Kerodon

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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

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