Kerodon

Proof. Fix an integer $m \geq 0$; we wish to show that every lifting problem

admits a solution.

Let $\sigma $ be any $m$-simplex of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ satisfying $q( \sigma ) = \overline{\sigma }$. By virtue of Remark 4.8.4.10, the commutativity of the diagram (4.81) guarantees that $\sigma _0$ and $\sigma $ coincide on the $(n-1)$-skeleton of $\operatorname{\partial \Delta }^{m}$. Consequently, if $m \leq n$, then $\sigma $ is a solution to the the lifting problem (4.81). We will therefore assume that $m > n$. In this case, the boundary $\operatorname{\partial \Delta }^{m}$ contains the $n$-skeleton of $\Delta ^{m}$. It will therefore suffice to show that $\sigma _0$ can be extended to an $m$-simplex $\sigma '$ of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$: the commutativity of the diagram (4.81) guarantees that any such extension satisfies the identity $q( \sigma ' ) = \overline{\sigma }$ (Proposition 4.8.4.11). If $m \geq n+2$, then the existence of $\sigma '$ is automatic (since $\operatorname{cosk}^{\circ }_{n}(\operatorname{\mathcal{C}})$ is $(n+1)$-coskeletal). It will therefore suffice to treat the case $m = n+1$. In this case, we can identify $\sigma _0$ with a diagram $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow \operatorname{\mathcal{C}}$, and we wish to show that $\tau _0$ can be extended to an $(n+1)$-simplex of $\operatorname{\mathcal{C}}$. Note that $\sigma |_{ \operatorname{\partial \Delta }^{n+1} }$ determines another diagram $\tau _1: \operatorname{\partial \Delta }^{n+1} \rightarrow \operatorname{\mathcal{C}}$. Moreover, the commutativity of the diagram (4.81) guarantees that $\tau _0$ and $\tau _1$ are isomorphic relative to $\operatorname{sk}_{n-1}( \Delta ^{n+1} )$ (Proposition 4.8.4.11). Using Corollary 4.4.5.3, we are reduced to showing that $\tau _1$ can be extended to an $(n+1)$-simplex of $\operatorname{\mathcal{C}}$, which follows from the existence of $\sigma $. $\square$