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1.2.4 The Nerve of a Groupoid

According to Proposition 1.2.2.1, every category $\operatorname{\mathcal{C}}$ can be recovered, up to canonical isomorphism, from the nerve $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. In particular, any isomorphism-invariant condition on a category $\operatorname{\mathcal{C}}$ can be reformulated as a condition on the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. We now illustrate this principle with a simple example.

Definition 1.2.4.1. Let $\operatorname{\mathcal{C}}$ be a category. We say that a morphism $f: C \rightarrow D$ in $\operatorname{\mathcal{C}}$ is an isomorphism if there exists a morphism $g: D \rightarrow C$ satisfying the identities

\[ f \circ g = \operatorname{id}_{D} \quad \quad g \circ f = \operatorname{id}_{C}. \]

In this case, the morphism $g$ is uniquely determined and we write $g = f^{-1}$. We say that $\operatorname{\mathcal{C}}$ is a groupoid if every morphism in $\operatorname{\mathcal{C}}$ is invertible.

Proposition 1.2.4.2. Let $\operatorname{\mathcal{C}}$ be a category. Then $\operatorname{\mathcal{C}}$ is a groupoid (Definition 1.2.4.1) if and only if the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is a Kan complex (Definition 1.1.7.1).

Example 1.2.4.3 (The Milnor Construction). Let $G$ be a group. Then we can form a category $\operatorname{\mathcal{C}}_{G}$ having a single object $X$, where $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,X) = G$ (and the composition of morphisms in $\operatorname{\mathcal{C}}_{G}$ is given by multiplication in the group $G$). We will denote the nerve of the category $\operatorname{\mathcal{C}}_{G}$ by $\operatorname{BG}$. The geometric realization $| \operatorname{BG}|$ is a topological space called the classifying space of $G$. It can characterized (up to homotopy equivalence) by the fact that it is a CW complex with either of the following properties:

  • The space $| \operatorname{BG}|$ is connected, and its homotopy groups (with respect to any choice of base point) are given by the formula

    \[ \pi _{\ast }( |BG |) \simeq \begin{cases} G \text{ if } \ast = 1 \\ 0 & \text{ if } \ast > 1. \end{cases} \]
  • For any paracompact topological space $X$, there is a canonical bijection

    \[ \{ \text{Continuous maps $f: X \rightarrow | \operatorname{BG}|$} \} / \text{homotopy} \simeq \{ \text{$G$-torsors $P \rightarrow X$} \} / \text{isomorphism}. \]

We refer the reader to [MR0077122] for a more detailed discussion (including an extension to the setting of topological groups).

Proof of Proposition 1.2.4.2. 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 $d_0(\sigma ) = f$ and $d_1( \sigma ) = \operatorname{id}_{D}$. Setting $g = d_2(\sigma )$, 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

\[ g = \operatorname{id}_{C} \circ g = (h \circ f) \circ g = h \circ (f \circ g) = h \circ \operatorname{id}_{D} = h \]

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 $\sigma : \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$. For $0 < i < n$, this follows from Lemma 1.2.3.2 (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 $n = 0$, we have $\Lambda ^{n}_0 = \Delta ^ n$, so we can take $\sigma = \sigma _0$.

  • In the case $n = 1$, the map $\sigma _0: \Lambda ^ n_{0} \rightarrow \operatorname{\mathcal{C}}$ 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 codomain $C$ (for example, we can take $\sigma $ to be the identity map $\operatorname{id}_{C}$).

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

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

    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 $f_{i,j}: C_ i \rightarrow C_{j}$ for $i \leq j$ (as in the proof of Lemma 1.2.3.2). 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 $f_{j,k} \circ f_{i,j} = f_{i,k}$ 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_{2,3} \circ f_{1,2}) \circ f_{0,1} & = & f_{2,3} \circ (f_{1,2} \circ f_{0,1} ) \\ & = & f_{2,3} \circ f_{0,2} \\ & = & f_{0,3} \\ & = & f_{1,3} \circ f_{0,1}. \end{eqnarray*}

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

$\square$