Kerodon

Proof. By virtue of Remark 3.5.5.5, we may assume without loss of generality that $X$ is connected. Assume that $X$ is a $0$-groupoid; we wish to show that the projection map $X \rightarrow \Delta ^0$ is an isomorphism (the converse follows immediately from the definition). To prove this, it will suffice to show that for every pair of $m$-simplices $\sigma , \tau : \Delta ^ m \rightarrow X$, we have $\sigma = \tau $. Our proof proceeds by induction on $m$. If $m > 0$, then our inductive hypothesis guarantees that $\sigma |_{ \Lambda ^{m}_{0} } = \tau |_{ \Lambda ^{m}_{0} }$, so that $\sigma = \tau $ by virtue of our assumption that $X$ is a $0$-groupoid. It will therefore suffice to treat the case $m = 0$, so that $\sigma $ and $\tau $ can be identified with vertices $x,y \in X$. Since $X$ is a connected Kan complex, there exists an edge $e: x \rightarrow y$ with source $x$ and target $y$ (Proposition 1.2.5.10). Our assumption that $X$ is a $0$-groupoid then guarantees that $e = \operatorname{id}_{x}$, so that $x = y$ as desired. $\square$

Proof. We first show that $(1)$ implies $(2)$. For every groupoid $\mathcal{G}$, Proposition 1.3.5.2 guarantees that the simplicial set $\operatorname{N}_{\bullet }( \mathcal{G} )$ is a Kan complex. To show that $\operatorname{N}_{\bullet }( \mathcal{G} )$ is a $1$-groupoid, we must prove that if $\sigma , \tau : \Delta ^{m} \rightarrow \operatorname{N}_{\bullet }( \mathcal{G} )$ are $m$-simplices for $m \geq 2$ which have the same restriction to some horn $\Lambda ^{m}_{i} \subset \Delta ^ m$, then $\sigma = \tau $. For $m > 2$, this is immediate (since $\Lambda ^{m}_{i}$ contains the $1$-skeleton of $\Delta ^{m}$). In the case $m=2$, we can identify $m$-simplices of $\operatorname{N}_{\bullet }(\mathcal{G} )$ with commutative diagrams

in the groupoid $\operatorname{\mathcal{G}}$. The desired result then follows from the observation that any two of the morphisms $f$, $g$, and $h$ determine the third.

The implication $(3) \Rightarrow (1)$ is immediate. We will complete the proof by showing that $(2)$ implies $(3)$. Assume that $X$ is a $1$-groupoid and let $\mathcal{G} = \pi _{\leq 1}(X)$ be its fundamental groupoid. We wish to show that the tautological map $u: X \rightarrow \operatorname{N}_{\bullet }( \mathcal{G} )$ is an isomorphism: that is, it is bijective on $m$-simplices for $m \geq 0$. The proof proceeds by induction on $m$. The case $m = 0$ is immediate from the definitions. For $m \geq 2$, we have a commutative diagram

where the vertical maps are bijective (since $X$ and $\operatorname{N}_{\bullet }(\mathcal{G} )$ are $1$-groupoids) and the bottom horizontal map is bijective (by virtue of our inductive hypothesis); it follows that the upper horizontal map is bijective as well. It will therefore suffice to treat the case $m = 1$. Let $e, e': x \rightarrow y$ be edges of the simplicial set $X$ having the same source and target; we wish to show that if the homotopy classes $[e]$ and $[e']$ coincide (as morphisms in the category $\mathcal{G} = \pi _{\leq 1}(X)$), then $e = e'$. Let $\sigma $ be a $2$-simplex of $X$ which is a homotopy from $e$ to $e'$: that is, a $2$-simplex whose boundary is depicted in the diagram

(see Definition 1.4.3.1). Let $\tau $ be the right-degenerate $2$-simplex $s^{1}_{1}(e)$. Then $\sigma $ and $\tau $ have the same restriction to the horn $\Lambda ^{2}_{1} \subset \Delta ^2$. Invoking our assumption that $X$ is a $1$-groupoid, we conclude that $\sigma = \tau $. In particular, we have $e' = d^2_{1}(\sigma ) = d^{2}_{1}( \tau ) = e$. $\square$

Proof. Fix a pair of integers $0 \leq i \leq m$ with $m > n$. Let $\sigma : \Delta ^{m} \rightarrow \Delta ^{m}$ be the identity map, which we identify with its image in the normalized chain complex $\mathrm{N}_{\ast }( \Delta ^{m}; \operatorname{\mathbf{Z}})$. Then the chain complex $\mathrm{N}_{\ast }( \Delta ^{m}; \operatorname{\mathbf{Z}})$ splits as a direct sum of $\mathrm{N}_{\ast }( \Lambda ^{m}_{i}; \operatorname{\mathbf{Z}})$ with the subcomplex $C_{\ast }$ spanned by the $\sigma $ and $\partial (\sigma )$. We therefore obtain a canonical bijection

It follows that the restriction map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{m}, X ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{m}_{i}, X)$ is a bijection if and only if the abelian group $A_ m$ vanishes. The desired result follows by allowing the integers $0 \leq i \leq m$ to vary. $\square$

Proof. We first prove $(1)$. Assume that $X$ is weakly $(n-1)$-coskeletal. Suppose we are given an integer $m > n$ and a pair of $m$-simplices $\sigma , \tau : \Delta ^{m} \rightarrow X$ which satisfy $\sigma |_{ \Lambda ^{m}_{i} } = \tau |_{ \Lambda ^{m}_{i} }$ for some $0 \leq i \leq m$; we wish to show that $\sigma = \tau $. Note that $d^{m}_{i}(\sigma )$ and $d^{m}_{i}(\tau )$ are $(m-1)$-simplices of $X$ which coincide on the boundary $\operatorname{\partial \Delta }^{m-1}$. Since $X$ is weakly $(n-1)$-coskeletal, it follows that $d^{m}_{i}(\sigma ) = d^{m}_{i}(\tau )$, so that $\sigma $ and $\tau $ coincide on the boundary $\operatorname{\partial \Delta }^{m}$. Invoking our assumption that $X$ is weakly $(n-1)$-coskeletal again, we conclude that $\sigma = \tau $.

We now prove $(2)$. Suppose that $X$ is an $n$-groupoid; we wish to show that it is weakly $n$-coskeletal. By virtue of Exercise 3.5.4.11, it will suffice to show that for every integer $m > n$, the restriction map

is injective. This is clear: our assumption that $X$ is an $n$-groupoid guarantees that the composition of $\theta $ with the restriction map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\partial \Delta }^{m}, X) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{m}_{0}, X)$ is a bijection. $\square$

Proof. Combine Proposition 3.5.5.10 with Remark 3.5.4.2. $\square$

Proof. Assume first that $X$ is an $n$-groupoid and let $\sigma _0, \sigma _1: \Delta ^ n \rightarrow X$ be $n$-simplices of $X$ which are homotopic relative to $\operatorname{\partial \Delta }^ n$. Since $X$ is a Kan complex, there exists a homotopy $h: \Delta ^1 \times \Delta ^ n \rightarrow X$ from $\sigma _0$ to $\sigma _1$ which is constant along $\operatorname{\partial \Delta }^ n$ (Proposition 3.2.1.4). For $0 \leq i \leq n$, let $\alpha _{i}: [n+1] \rightarrow [1] \times [n]$ denote the nondecreasing function given by the formula

and let $\tau _{i}$ denote the $(n+1)$-simplex of $X$ given by the composition

Let $\rho _{i}, \rho '_{i}: \Delta ^{n} \rightarrow X$ be the $n$-simplices of $X$ given by $\rho _{i} = d^{n+1}_{i}( \tau _ i )$ and $\rho '_{i} = d^{n+1}_{i+1}( \tau _ i )$; by construction, we have

We will complete the proof by showing that $\rho _{i} = \rho '_{i}$ for $0 \leq i \leq n$. Using our assumption that the homotopy $h$ is constant along the boundary $\operatorname{\partial \Delta }^{n}$, we see that the degenerate $(n+1)$-simplex $s^{n}_{i}( \rho _{i} )$ coincides with $\tau _ i$ on the horn $\Lambda ^{n+1}_{i} \subset \Delta ^{n+1}$. Invoking our assumption that $X$ is an $n$-groupoid, we conclude that $\tau _ i = s^{n}_{i}( \rho _{i} )$. Applying the face operator $d^{n+1}_{i+1}$, we obtain $\rho _{i} = \rho '_{i}$.

We now prove the converse. Assume that $X$ satisfies condition $(\ast )$; we wish to show that the Kan complex $X$ is an $n$-groupoid. Fix a pair of integers $0 \leq i \leq m$ with $m > n$ and a pair of $m$-simplices $\tau _0, \tau _1: \Delta ^{m} \rightarrow X$ which coincide on the horn $\Lambda ^{m}_{i} \subset \Delta ^{m}$; we wish to show that $\tau _0 = \tau _1$. Since $X$ is weakly $n$-coskeletal, it will suffice to prove that $\tau _0$ and $\tau _1$ coincide on the boundary $\operatorname{\partial \Delta }^{m}$: that is, to show that the $(m-1)$-simplices $\sigma _0 = d^{m}_{i}(\tau _0)$ and $\sigma _1 = d^{m}_{i}(\tau _1)$ coincide. Note that $\sigma _0$ and $\sigma _1$ have the same restriction to the boundary $\operatorname{\partial \Delta }^{m-1}$. Consequently, if $m \geq n+2$, the desired result follows from our assumption that $X$ is weakly $n$-coskeletal. We may therefore assume that $m = n+1$. By virtue of $(\ast )$, it will suffice to show that the $(m-1)$-simplices $\sigma _0$ and $\sigma _1$ are homotopic relative to $\operatorname{\partial \Delta }^{m-1}$. In fact, we will prove a stronger claim: the $m$-simplices $\tau _0$ and $\tau _1$ are homotopic relative to the horn $\Lambda ^{m}_{i} \subset \Delta ^{m}$. This follows from the observation that the restriction map $\operatorname{Fun}( \Delta ^{m}, X) \rightarrow \operatorname{Fun}( \Lambda ^{m}_{i}, X)$ is a trivial Kan fibration; see Corollary 3.1.3.6. $\square$

Proof. It follows from Corollary 3.1.3.4 that $\operatorname{Fun}(K,X)$ is a Kan complex. Since $X$ is weakly $n$-coskeletal (Proposition 3.5.5.10), it follows that $\operatorname{Fun}(K,X)$ is also weakly $n$-coskeletal (Corollary 3.5.4.13). We will complete the proof by showing that $\operatorname{Fun}(K,X)$ satisfies condition $(\ast )$ of Proposition 3.5.5.12. Suppose we are given a pair of $n$-simplices $\sigma _0, \sigma _1: \Delta ^ n \rightarrow \operatorname{Fun}(K,X)$ which are homotopic relative to $\operatorname{\partial \Delta }^{n}$; we wish to show that $\sigma _0 = \sigma _1$. Let us identify $\sigma _0$ and $\sigma _1$ with morphisms $f_0, f_1: \Delta ^ n \times K \rightarrow X$. Since $X$ is weakly $n$-coskeletal, it will suffice to show that $f_0$ and $f_1$ coincide on $m$-simplices $\tau = (\tau ', \tau '')$ of $\Delta ^ n \times K$ for $m \leq n$. If $\tau '$ factors through the boundary $\operatorname{\partial \Delta }^ n$, this follows immediately from the equality $\sigma _0|_{\operatorname{\partial \Delta }^ n} = \sigma _1|_{ \operatorname{\partial \Delta }^ n}$. We may therefore assume without loss of generality that $m = n$ and that $\tau ': \Delta ^ m \rightarrow \Delta ^ n$ is the identity map. In this case, our assumption that $\sigma _0$ and $\sigma _1$ are homotopic relative to $\operatorname{\partial \Delta }^ n$ guarantees that $f_0(\tau )$ and $f_1(\tau )$ are homotopic relative to $\operatorname{\partial \Delta }^ n$, so that $f_0( \tau ) = f_1(\tau )$ by virtue of Proposition 3.5.5.12. $\square$

Proof. We first show that $(1)$ implies $(2)$. Here we may assume that $n > 0$ (otherwise, the result is a special case of Proposition 3.5.1.22). Let $\overline{\tau }$ be an $n$-simplex of $Y$, and set $\overline{\tau }_0 = \overline{\tau }|_{ \Lambda ^{n}_{0} }$. Since $f$ is bijective on $m$-simplices for $m < n$, we can lift $\overline{\tau }_0$ to a morphism $\tau _0: \Lambda ^{n}_{0} \rightarrow X$. If $f$ is a Kan fibration, then the lifting problem

admits a solution, given by an $n$-simplex $\tau $ of $X$ satisfying $f( \tau ) = \overline{\tau }$.

The implication $(2) \Rightarrow (3)$ is a special case of Corollary 3.5.2.2. We will complete the proof by showing that $(3)$ implies $(1)$. Assume that $f$ is $n$-connective and fix a pair of integers $0 \leq i \leq m$ with $m > 0$; we wish to show that every lifting problem

admits a solution. We consider three cases:

• Suppose that $m < n$. In this case, our assumption that $f$ is bijective on $m$-simplices guarantees that there is a unique $m$-simplex $\sigma $ of $X$ satisfying $f(\sigma ) = \overline{\sigma }$. By construction, we have $(f \circ \sigma )|_{ \Lambda ^{m}_{i} } = \overline{\sigma }|_{ \Lambda ^{m}_{i} } = f \circ \sigma _0$. Since $f$ is bijective on simplices of dimension $< m$, it follows that $\sigma _0 = \sigma |_{ \Lambda ^{m}_{i} }$.

• Suppose that $m > n$. In this case, our assumption that $X$ is a Kan complex guarantees that we can extend $\sigma _0$ to an $m$-simplex $\sigma $ of $X$. By construction, we have

  \[ (f \circ \sigma )|_{ \Lambda ^{m}_{i} } = f \circ \sigma _0 = \overline{\sigma }|_{ \Lambda ^{m}_{i} }. \]

  Since $Y$ is an $n$-groupoid, it follows that $f \circ \sigma = \overline{\sigma }$.

• Suppose that $m = n$. Since $f$ is bijective on $(n-1)$-simplices, the morphism $\sigma _0$ admits a unique extension $\sigma _1: \operatorname{\partial \Delta }^{n} \rightarrow X$ satisfying $f \circ \sigma _1 = \overline{\sigma }|_{ \operatorname{\partial \Delta }^{n} }$. The morphism $f$ factors as a composition

  \[ X \xrightarrow {i} X \times _{ \operatorname{Fun}( \{ 0\} , Y) } \operatorname{Fun}( \Delta ^1, Y ) \xrightarrow {q} Y, \]

  where $i$ is a homotopy equivalence and $q$ is a Kan fibration (see Example 3.1.7.10). Since $f$ is $n$-connective, the Kan fibration $q$ is also $n$-connective (Proposition 3.5.1.26). Applying Proposition 3.5.2.1, we conclude that there is an $n$-simplex $\sigma $ of $X$ satisfying $\sigma _1 = \sigma |_{ \operatorname{\partial \Delta }^{n} }$ and a homotopy from $f(\sigma )$ to $\overline{\sigma }$ which is constant when restricted to $\operatorname{\partial \Delta }^{n}$. Since $Y$ is an $n$-groupoid, Proposition 3.5.5.12 guarantees that $f(\sigma ) = \overline{\sigma }$.

Proof. It follows from Corollary 3.5.5.14 that $f$ is a Kan fibration. Applying Proposition 3.2.7.2, we deduce that $f$ is a trivial Kan fibration. In particular, $f$ admits a section $g: Y \hookrightarrow X$. To complete the proof, it will suffice to show that $g$ is an epimorphism of simplicial sets. This follows from Corollary 3.5.5.14, since $g$ is also bijective on $m$-simplices for $m < n$. $\square$

Proof. If $n = 0$, the desired result follows from Proposition 3.5.5.7. If $n = 1$, then $X$ is an $n$-groupoid if and only if it is isomorphic to the nerve $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$, where $\operatorname{\mathcal{C}}$ is a groupoid (Proposition 3.5.5.8). In this case, the assumption that $X$ has a single vertex is equivalent to the requirement that the category $\operatorname{\mathcal{C}}$ contains a single object $C$, in which case we can identify $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ with the classifying simplicial set $\mathrm{K}(G,1) =B_{\bullet }G$ for $G = \operatorname{Aut}_{\operatorname{\mathcal{C}}}(C)$. We may therefore assume that $n \geq 2$. The implication $(1) \Rightarrow (2)$ follows from Proposition 3.5.5.9. For the converse, assume that $X$ is an $n$-groupoid having a single $m$-simplex for each $m < n$. Let $x$ be the unique vertex of $X$ and set $G = \pi _{n}(X,x)$. For every $n$-simplex $\sigma $ of $X$, the restriction $\sigma |_{ \operatorname{\partial \Delta }^{n} }$ is the constant map taking the value $x$, so the homotopy class $[\sigma ]$ can be regarded as an element of the group $G$. Our assumption that $X$ is an $n$-groupoid guarantees that the assignment $\sigma \mapsto [\sigma ]$ determines a bijection from the collection of $n$-simplices of $X$ to the group $G$, which determines an isomorphism $f_0$ from the $n$-skeleton of $X$ to the $n$-skeleton of $\mathrm{K}(G,n)$. Invoking Theorem 3.2.2.10, we see that a morphism $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow X$ can be extended to an $(n+1)$-simplex of $X$ if and only if the composite map

can be extended to an $(n+1)$-simplex of $\mathrm{K}(G,n)$. Since $\mathrm{K}(G,n)$ is weakly $n$-coskeletal, it follows that $f_0$ extends uniquely to a morphism of simplicial sets $f: X \rightarrow \mathrm{K}(G,n)$ (Proposition 3.5.4.12) which is surjective on $(n+1)$-simplices. In particular, $f$ exhibits $\mathrm{K}(G,n)$ as a weak $n$-coskeleton of $X$ (Definition 3.5.4.14). Since $X$ is weakly $n$-coskeletal (Proposition 3.5.5.10), we conclude that $f$ is an isomorphism. $\square$