Kerodon

Proof. Let $K$ be a simplicial set; we will show that precomposition with $f$ induces a bijection $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}(K, \operatorname{Fun}(Y,Z) ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}(K, \operatorname{Fun}(X,Z) )$. Replacing $Z$ by the $n$-groupoid $\operatorname{Fun}(K,Z)$ (Corollary 3.5.5.13), we are reduced to proving that the natural map

is a bijection.

Let $h: X \rightarrow Z$ be a morphism of Kan complexes; we wish to show that there is a unique morphism $g: Y \rightarrow Z$ satisfying $g \circ f = h$. We first claim that there is a unique morphism $g_0: \operatorname{sk}_{n}(Y) \rightarrow Z$ satisfying $g_0 \circ \operatorname{sk}_{n}(f) = h|_{ \operatorname{sk}_{n}(X) }$. Our assumption that $f$ exhibits $Y$ as a fundamental $n$-groupoid of $X$ guarantees that $\operatorname{sk}_{n}(f)$ is surjective. It will therefore suffice to show that if $\sigma $ and $\sigma '$ are $m$-simplices of $X$ for $m \leq n$ satisfying $f(\sigma ) = f(\sigma ')$, then $h(\sigma ) = h(\sigma ')$. If $m < n$, then $\sigma = \sigma '$ and the result is clear. In the case $m = n$, the assumption that $f(\sigma ) = f(\sigma ')$ guarantees that $\sigma $ and $\sigma '$ are homotopic relative to $\operatorname{\partial \Delta }^{n}$, in which case the desired result follows from Proposition 3.5.5.12.

It follows from Remark 3.5.6.2 that every $(n+1)$-simplex $\tau $ of $Y$ can be lifted to an $(n+1)$-simplex $\widetilde{\tau }$ of $X$. In particular, $g_0 \circ \tau |_{ \operatorname{\partial \Delta }^{n+1} }$ extends to an $(n+1)$-simplex of $Z$, given by $h( \widetilde{\tau } )$. Since $Z$ is weakly $n$-coskeletal (Proposition 3.5.5.10), Proposition 3.5.4.12 guarantees that $g_0$ extends uniquely to a morphism $g: Y \rightarrow Z$, which automatically satisfies the equation $g \circ f = h$. $\square$

Proof. We first prove that $\theta $ is a surjection. Fix a morphism $g: A \rightarrow \pi _{\leq n}(X)$. Using Remark 3.5.6.11 (and Proposition 1.1.4.12), we see that $g|_{ \operatorname{sk}_{n}(A) }$ can be lifted to a morphism of simplicial sets $u: \operatorname{sk}_{n}(A) \rightarrow X$. We will show that $u$ can be extended to the $(n+1)$-skeleton of $A$ (and is therefore classified by a morphism $f: A \rightarrow \operatorname{cosk}_{n}^{\circ }(X)$ satisfying $\theta (f) = g$; see Remark 3.5.4.21). By virtue of Proposition 1.1.4.12 and Variant 3.2.4.12, this is equivalent to the assertion that for every $(n+1)$-simplex $\sigma $ of $A$ having restriction $\sigma _0 = \sigma |_{ \operatorname{\partial \Delta }^{n+1} }$, the composition $(g \circ \sigma _0): \operatorname{\partial \Delta }^{n+1} \rightarrow X$ is nullhomotopic. Choose a lift of $g(\sigma )$ to an $(n+1)$-simplex of $\operatorname{cosk}_{n}^{\circ }(X)$, which we identify with a nullhomotopic map $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow X$. By construction, $g \circ \sigma _0$ and $\tau _0$ coincide after composing with the comparison map $v: X \rightarrow \pi _{\leq n}(X)$. Using Proposition 1.1.4.12 again, we see that $g \circ \sigma _0$ and $\tau _0$ are homotopic relative to the $\operatorname{sk}_{n-1}( \Delta ^{n+1} )$, so that $g \circ \sigma _0$ is also nullhomotopic.

Now suppose that we are given a pair of morphisms $f_0, f_1: A \rightarrow \operatorname{cosk}_{n}^{\circ }(X)$ satisfying $\theta (f_0) = \theta (f_1)$. We wish to show that the associated maps $u_0, u_1: \operatorname{sk}_{n}(A) \rightarrow X$ are homotopic relative to $\operatorname{sk}_{n-1}(A)$ (the converse is immediate from the definitions). Using Remark 3.5.6.11, we deduce that $u_0$ and $u_1$ coincide on $\operatorname{sk}_{n-1}(A)$. By virtue of Proposition 1.1.4.12, we are reduced to showing that for every nondegenerate $n$-simplex $\sigma $ of $A$, the compositions $u_0 \circ \sigma $ and $u_1 \circ \sigma $ are homotopic relative to $\operatorname{\partial \Delta }^{n}$. This follows from our assumption that the maps $\theta (f_0), \theta (f_1): A \rightarrow \pi _{\leq n}(X)$ coincide on the simplex $\sigma $. $\square$

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 $. $\square$

Proof. Combine Corollary 3.5.6.14 with Proposition 3.5.4.22. $\square$

Proof. By virtue of Proposition 3.5.3.23, the coskeleton $\operatorname{cosk}_{n+1}(X)$ is a Kan complex. Combining Corollary3.5.6.15 with Proposition 1.5.5.11, we conclude that $\pi _{\leq n}(X)$ is a Kan complex. To complete the proof, it will suffice to show that if $\sigma $ and $\tau $ are $m$-simplices of $\pi _{\leq n}(X)$ for some $m > n$ which satisfy $\sigma |_{ \Lambda ^{m}_{i} } = \tau |_{ \Lambda ^{m}_{i} }$ for some $0 \leq i \leq m$, then $\sigma = \tau $. Choose maps $\widetilde{\sigma }, \widetilde{\tau }: \operatorname{sk}_{n}( \Delta ^{m} ) \rightarrow X$ representing $\sigma $ and $\tau $. Using Proposition 3.5.6.12, we can choose a homotopy from $\widetilde{\sigma }|_{ \operatorname{sk}_{n}( \Lambda ^{m}_{i}) }$ to $\widetilde{\tau }|_{ \operatorname{sk}_{n}(\Lambda ^{m}_{i} ) }$ which is constant when restricted to the skeleton $\operatorname{sk}_{n-1}( \Lambda ^{m}_{i} ) = \operatorname{sk}_{n-1}( \Delta ^ m )$. If $m \geq n+2$, then $h$ is also a homotopy from $\widetilde{\sigma }|_{ \operatorname{sk}_{n}( \Delta ^ m) }$ to $\widetilde{\tau }|_{ \operatorname{sk}_{n}( \Delta ^ m ) }$, so that $\sigma = \tau $ as desired. In the case $m = n+1$, the morphisms $\widetilde{\sigma }$ and $\widetilde{\tau }$ can be extended to morphisms $\overline{\sigma }, \overline{\tau }: \Delta ^{n+1} \rightarrow X$. Using Corollary 3.1.3.6, we can extend $h$ to a homotopy $\overline{h}$ from $\overline{\sigma }$ to $\overline{\tau }$. Restricting this homotopy to the $n$-skeleton of $\Delta ^{m}$, we again conclude that $\sigma = \tau $. $\square$

Proof. It follows from Remark 3.5.6.11 that $v$ is bijective on $m$-simplices for $m < n$ and surjective on $n$-simplices. By construction, if $\sigma $ and $\sigma '$ are $n$-simplices of $X$, then $v(\sigma ) = v(\sigma ')$ if and only if $\sigma $ and $\sigma '$ are homotopic relative to $\operatorname{\partial \Delta }^{n}$. It will therefore suffice to show that $\pi _{\leq n}(X)$ is an $n$-groupoid, which follows from Corollary 3.5.6.16. $\square$