Kerodon

Proof. Fix an integer $n \geq 0$ and a diagram of simplicial sets

we wish to show that the map $\sigma _0$ admits an extension $\sigma : \Delta ^{n+1} \rightarrow \operatorname{N}^{\operatorname{hc}}_{\bullet }( \operatorname{\mathcal{C}}^{\Delta } )$ as indicated, rendering the diagram commutative. Let us abuse notation by identifying $\sigma _0$ with a simplicial functor from $\operatorname{Path}[ \partial \Delta ^{n+1} ]_{\bullet }$ to $\operatorname{\mathcal{C}}^{\Delta }$. Set $X = \sigma _0(0)$, $Y = \sigma _0(n+1)$, and $I = \{ 1, 2, \cdots , n \} $, so that $\sigma _0$ determines a morphism of simplicial sets

(see Proposition 2.4.6.12), which we will identify with a chain map $f_0: \mathrm{N}_{\ast }( \operatorname{\partial \raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}}) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\ast }$. By virtue of Corollary 2.4.6.13, choosing an extension of $\sigma _0$ to a map $\sigma : \Delta ^{n+1} \rightarrow \operatorname{N}^{\operatorname{hc}}_{\bullet }( \operatorname{\mathcal{C}}^{\Delta } )$ is equivalent to choosing an extension of $u_0$ to a map of simplicial sets $u: \operatorname{\raise {0.1ex}{\square }}^{I} \rightarrow \mathrm{K}( \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) )$, or an extension of $f_0$ to a chain map $f: \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}}) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\ast }$.

Endow $I = \{ 1, \cdots , n \} $ with the opposite of its usual ordering and let $[ \operatorname{\raise {0.1ex}{\square }}^{I} ]$ denote the fundamental chain of Construction 2.5.9.6. Note that the boundary $\partial [ \operatorname{\raise {0.1ex}{\square }}^{n} ]$ belongs to the subcomplex $\mathrm{N}_{\ast }( \operatorname{\partial \raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}}) \subset \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}})$ (see Lemma 2.5.9.13). Unwinding the definitions, we see that $\tau $ supplies a chain $z \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{n}$ satisfying $\partial (z) = f_0( \partial [ \operatorname{\raise {0.1ex}{\square }}^{I}] ) \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{n-1}$. Let $M_{\ast }$ denote the subcomplex of $\mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}})$ generated by $\mathrm{N}_{\ast }( \operatorname{\partial \raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}})$ together with the fundamental chain $[ \operatorname{\raise {0.1ex}{\square }}^{I} ]$, so that $f_0$ extends uniquely to a chain map $f_1: M_{\ast } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\ast }$ satisfying $f_1( [ \operatorname{\raise {0.1ex}{\square }}^{I} ] ) = z$. Unwinding the definitions, we see that if $f: \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}}) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\ast }$ is a map of chain complexes extending $f_0$, then the corresponding extension $\sigma : \Delta ^{n+1} \rightarrow \operatorname{N}^{\operatorname{hc}}_{\bullet }( \operatorname{\mathcal{C}}^{\Delta } )$ of $\sigma _0$ satisfies $\mathfrak {Z} \circ \sigma = \tau $ if and only if $f|_{ M_{\ast } } = f_1$. We will complete the proof by showing that $M_{\ast }$ is a direct summand of $\mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}})$ (so that any map $f_1: M_{\ast } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\ast }$ can be extended to $\mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}})$). To prove this, note that we have an exact sequence of chain complexes

where the first map is a quasi-isomorphism (Variant 2.5.7.17). It follows that the chain complex $\mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{I}; \operatorname{\mathbf{Z}}) / M_{\ast }$ is acyclic and free in each degree, so that the exact sequence

splits by virtue of Proposition 2.5.1.10. $\square$