Kerodon

Proof of Proposition 1.3.5.2. Suppose first that $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is a Kan complex; we wish to show that $\operatorname{\mathcal{C}}$ is a groupoid. Let $f: C \rightarrow D$ be a morphism in $\operatorname{\mathcal{C}}$. Using the surjectivity of the map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^2, \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{2}_{2}, \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) )$, we see that there exists a $2$-simplex $\sigma $ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ satisfying $d^{2}_0(\sigma ) = f$ and $d^{2}_1( \sigma ) = \operatorname{id}_{D}$. Setting $g = d^{2}_2(\sigma )$, we conclude that $f \circ g = \operatorname{id}_{D}$: that is, $g$ is a left inverse to $f$. Similarly, the surjectivity of the map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^2, \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{2}_{0}, \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) )$ allows us to construct a map $h: D \rightarrow C$ satisfying $h \circ f = \operatorname{id}_{C}$. The calculation

then shows that $g = h$ is an inverse of $f$, so that $f$ is invertible as desired.

Now suppose that $\operatorname{\mathcal{C}}$ is a groupoid. We wish to show that, for $0 \leq i \leq n$, every map $\sigma _0: \Lambda ^{n}_{i} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. For $0 < i < n$, this follows from Lemma 1.3.4.2 (and does not require the assumption that $\operatorname{\mathcal{C}}$ is a groupoid). We will treat the case where $i = 0$; the case $i = n$ follows by similar reasoning. We consider several cases:

• In the case $n = 1$, the map $\sigma _0: \Lambda ^ n_{0} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ can be identified with an object $C \in \operatorname{\mathcal{C}}$. In this case, we can take $\sigma $ to be an edge of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ corresponding to any morphism with target $C$ (for example, we can take $\sigma $ to be the identity morphism $\operatorname{id}_{C}$).

• In the case $n = 2$, we can identify $\sigma _0$ with a pair of morphisms in $\operatorname{\mathcal{C}}$ having the same source, which we can depict as a diagram

  \[ \xymatrix@R =50pt@C=50pt{ & D \ar@ {-->}[dr] & \\ C \ar [ur]^{f} \ar [rr]^{g} & & E. } \]

  Our assumption that $\operatorname{\mathcal{C}}$ is a groupoid guarantees that we can extend this diagram to a $2$-simplex of $\operatorname{\mathcal{C}}$, whose $0$th face is given by the morphism $g \circ f^{-1}: D \rightarrow E$.

• In the case $n \geq 3$, the map $\sigma _0$ determines a collection of objects $\{ C_ i \} _{0 \leq i \leq n }$ and morphisms $f_{j,i}: C_ i \rightarrow C_{j}$ for $i \leq j$ (as in the proof of Lemma 1.3.4.2). We wish to show that these morphisms determine a functor $[n] \rightarrow \operatorname{\mathcal{C}}$ (which we can then identify with an $n$-simplex $\sigma $ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ satisfying $\sigma |_{ \Lambda ^{n}_0} = \sigma _0$). For this, we must verify the identity $f_{k,j} \circ f_{j,i} = f_{k,i}$ for $0 \leq i \leq j \leq k \leq n$. Note that this identity is satisfied whenever the triple $(i \leq j \leq k)$ determines a $2$-simplex of $\Delta ^ n$ belonging to the horn $\Lambda ^{n}_0$. This is automatic unless $n = 3$ and $(i,j,k) = (1,2,3)$. To handle this exceptional case, we compute

  \begin{eqnarray*} (f_{3,2} \circ f_{2,1}) \circ f_{1,0} & = & f_{3,2} \circ (f_{2,1} \circ f_{1,0} ) \\ & = & f_{3,2} \circ f_{2,0} \\ & = & f_{3,0} \\ & = & f_{3,1} \circ f_{1,0}. \end{eqnarray*}

  Since $\operatorname{\mathcal{C}}$ is a groupoid, composing with $f_{1,0}^{-1}$ on the right yields the desired identity $f_{3,2} \circ f_{2,1} = f_{3,1}$.