# Kerodon

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Lemma 2.5.9.13. Let $n$ be a nonnegative integer and let $I$ denote the set $\{ 1, 2, \cdots , n \}$, endowed with its usual ordering. Then we have an equality

$\partial [ \operatorname{\raise {0.1ex}{\square }}^{I} ] = \sum _{i=1}^{n} (-1)^{i} ([ \{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{ I \setminus \{ i\} } ] - [ \{ 1\} \times \operatorname{\raise {0.1ex}{\square }}^{ I \setminus \{ i\} } ] )$

in the abelian group $\mathrm{N}_{n-1}( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}})$.

Proof of Lemma 2.5.9.13. Using the description of $[ \operatorname{\raise {0.1ex}{\square }}^{I} ]$ as a shuffle product (Remark 2.5.9.7) and the fact that the shuffle product satisfies the Leibniz rule (Proposition 2.5.7.10), we compute

\begin{eqnarray*} \partial [ \operatorname{\raise {0.1ex}{\square }}^{I} ] & = & \partial ( [ \Delta ^1 ] \triangledown \cdots \triangledown [ \Delta ^1] ) \\ & = & \sum _{i=1}^{n} (-1)^{i-1} [ \operatorname{\raise {0.1ex}{\square }}^{ i-1} ] \triangledown \partial ( [ \Delta ^1] ) \triangledown [\operatorname{\raise {0.1ex}{\square }}^{n-i}] \\ & = & \sum _{i=1}^{n} (-1)^{i} [ \operatorname{\raise {0.1ex}{\square }}^{ i-1} ] \triangledown (d_1[ \Delta ^1] - d_0[ \Delta ^1]) \triangledown [\operatorname{\raise {0.1ex}{\square }}^{n-i}] \\ & = & \sum _{i=1}^{n} (-1)^{i} ([ \{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{ I \setminus \{ i\} } ] - [ \{ 1\} \times \operatorname{\raise {0.1ex}{\square }}^{ I \setminus \{ i\} } ] ). \end{eqnarray*}
$\square$