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Remark Let $n$ be a nonnegative integer. It follows from Lemma that the boundary $\partial [ \operatorname{\raise {0.1ex}{\square }}^{n} ]$ belongs to the subcomplex $\mathrm{N}_{\ast }( \operatorname{\partial \raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}}) \subset \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}})$. In other words, the image of the fundamental chain $[ \operatorname{\raise {0.1ex}{\square }}^{n} ]$ in the relative chain complex

\[ \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n}, \operatorname{\partial \raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}}) = \mathrm{N}_{\ast }(\operatorname{\raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}}) / \mathrm{N}_{\ast }( \operatorname{\partial \raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}}) \]

is a cycle. In fact, one can be more precise: the construction $1 \mapsto [ \operatorname{\raise {0.1ex}{\square }}^{n} ]$ determines a quasi-isomorphism of chain complexes $u_ n: \operatorname{\mathbf{Z}}[n] \rightarrow \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n}, \operatorname{\partial \raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}})$. To prove this, we proceed by induction on $n$: the case $n=0$ is trivial, and the inductive step follows by identifying $u$ with the composition

\begin{eqnarray*} \operatorname{\mathbf{Z}}[n] & \simeq & \operatorname{\mathbf{Z}}[1] \boxtimes \operatorname{\mathbf{Z}}[n-1] \\ & \xrightarrow {\operatorname{id}\boxtimes u_{n-1}} & \operatorname{\mathbf{Z}}[1] \boxtimes \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n-1}, \operatorname{\partial \raise {0.1ex}{\square }}^{n-1}; \operatorname{\mathbf{Z}}) \\ & \simeq & \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{1}, \operatorname{\partial \raise {0.1ex}{\square }}^{1}; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n-1}, \operatorname{\partial \raise {0.1ex}{\square }}^{n-1}; \operatorname{\mathbf{Z}}) \\ & \xrightarrow { \mathrm{EZ} }& \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n}, \operatorname{\partial \raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}}) \end{eqnarray*}

where $\mathrm{EZ}$ denotes the Eilenberg-Zilber map of Variant (which is a quasi-isomorphism, by virtue of Theorem Note that this property characterizes the fundamental chain $[ \operatorname{\raise {0.1ex}{\square }}^{n}]$ up to sign (since the quotient map $\mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}}) \twoheadrightarrow \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n}, \operatorname{\partial \raise {0.1ex}{\square }}^{n}; \operatorname{\mathbf{Z}})$ is an isomorphism in degree $n$).