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 $\operatorname{cosk}_{n}^{\circ }(X)$ satisfying $q( \sigma ) = \overline{\sigma }$. By virtue of Remark 3.5.6.11, the commutativity of the diagram (3.77) 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 (3.77). 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 $\operatorname{cosk}_{n}^{\circ }(X)$: the commutativity of the diagram (3.77) guarantees that any such extension satisfies the identity $q( \sigma ' ) = \overline{\sigma }$ (Proposition 3.5.6.12). If $m \geq n+2$, then the existence of $\sigma '$ is automatic (since $\operatorname{cosk}^{\circ }_{n}(X)$ 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 morphism $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow X$, and we wish to show that $\tau _0$ is nullhomotopic. Note that $\sigma |_{ \operatorname{\partial \Delta }^{m} }$ determines a morphism $\tau _1: \operatorname{\partial \Delta }^{n+1} \rightarrow X$. Moreover, the commutativity of the diagram (3.77) guarantees that $\tau _0$ and $\tau _1$ are homotopic relative to $\operatorname{sk}_{n-1}( \Delta ^{n+1} )$ (Proposition 3.5.6.12). It will therefore suffice to show that $\tau _1$ is nullhomotopic, which follows from the existence of $\sigma $. $\square$