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1.3.4 Characterization of Nerves

We now describe the essential image of the functor $\operatorname{N}_{\bullet }: \operatorname{Cat}\rightarrow \operatorname{Set_{\Delta }}$.

Proposition Let $S$ be a simplicial set. Then $S$ is isomorphic to the nerve of a category if and only if it satisfies the following condition:

$(\ast ')$

For every pair of integers $0 < i < n$ and every map of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow S$, there exists a unique map $\sigma : \Delta ^{n} \rightarrow S$ such that $\sigma _0 = \sigma |_{ \Lambda ^{n}_{i} }$.

The proof of Proposition will require some preliminaries. We begin by establishing the necessity of condition $(\ast ')$.

Lemma Let $\operatorname{\mathcal{C}}$ be a category. Then the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ satisfies condition $(\ast ')$ of Proposition

Proof. Choose integers $0 < i < n$ together with a map of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$; we wish to show that $\sigma _0$ can be extended uniquely to a $n$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. For $0 \leq j \leq n$, let $C_{j} \in \operatorname{\mathcal{C}}$ denote the image under $\sigma _0$ of the $j$th vertex of $\Delta ^ n$ (which belongs to the horn $\Lambda ^{n}_{i}$). We first consider the case where $n \geq 3$. In this case, $\Lambda ^{n}_{i}$ contains every edge of $\Delta ^ n$. For $0 \leq j \leq k \leq n$, let $f_{k,j}: C_ j \rightarrow C_ k$ denote the $1$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ obtained by evaluating $\sigma _0$ on the edge of $\Delta ^ n$ corresponding to the pair $(j,k)$. We claim that the construction

\[ j \mapsto C_ j \quad \quad (j \leq k) \mapsto f_{k,j} \]

determines a functor $[n] \rightarrow \operatorname{\mathcal{C}}$, which we can then identify with an $n$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ having the desired properties. It is easy to see that $f_{j,j} = \operatorname{id}_{ C_ j}$ for each $0 \leq j \leq n$, so it will suffice to show that $f_{\ell ,k} \circ f_{k,j} = f_{\ell ,j}$ for every triple $0 \leq j \leq k \leq \ell \leq n$. The triple $(j,k, \ell )$ determines a $2$-simplex $\tau $ of $\Delta ^ n$. If $\tau $ is contained in $\Lambda ^{n}_{i}$, then $\tau ' = \sigma _0( \tau )$ is a $2$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ satisfying $d^{2}_0( \tau ') = f_{\ell ,k}$, $d^{2}_1( \tau ') = f_{\ell ,j}$, and $d^{2}_2( \tau ') = f_{k,j}$, so that $\tau '$ “witnesses” the identity $f_{\ell ,k} \circ f_{k,j} = f_{\ell ,j}$. It will therefore suffice to treat the case where the simplex $\tau $ does not belong to the $\Lambda ^{n}_{i}$. In this case, our assumption that $n \geq 3$ guarantees that we must have $\{ j, k, \ell \} = [n] \setminus \{ i \} $. It follows that $n = 3$, so that either $i = 1$ or $i = 2$. We will treat the case $i = 1$ (the case $i =2$ follows by a similar argument). Note that $\Lambda ^3_1$ contains all of the nondegenerate $2$-simplices of $\Delta ^3$ other than $\tau $; applying the map $\sigma _0$, we obtain $2$-simplices of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ which witness the identities

\[ f_{3,0} = f_{3,1} \circ f_{1,0} \quad \quad f_{3,1} = f_{3,2} \circ f_{2,1} \quad \quad f_{2,0} = f_{2,1} \circ f_{1,0}. \]

We now compute

\[ f_{3,0} = f_{3,1} \circ f_{1,0} = ( f_{3,2} \circ f_{2,1} ) \circ f_{1,0} = f_{3,2} \circ ( f_{2,1} \circ f_{1,0} ) = f_{3,2} \circ f_{2,0} \]

so that $f_{\ell ,j} = f_{\ell ,k} \circ f_{k,j}$, as desired.

It remains to treat the case $n =2$. In this case, the inequality $0 < i < n$ guarantees that $i = 1$. The morphism $\sigma _0: \Lambda ^ n_ i \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ can then be identified with a pair of composable morphisms $f_{1,0}: C_0 \rightarrow C_1$ and $f_{2,1}: C_1 \rightarrow C_{2}$ in the category $\operatorname{\mathcal{C}}$. This data extends uniquely to a $2$-simplex $\sigma $ of $\operatorname{\mathcal{C}}$ satisfying $d^{2}_1(\sigma ) = f_{2,1} \circ f_{1,0}$ (see Remark $\square$

Lemma Let $f: S \rightarrow T$ be a morphism of simplicial sets which is bijective on both vertices and edges. If both $S$ and $T$ satisfy condition $(\ast ')$ of Proposition, then $f$ is an isomorphism.

Proof. We claim that, for every simplicial set $K$, composition with $f$ induces a bijection

\[ \theta _{K}: \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( K, S) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( K, T). \]

Writing $K$ as a union of its skeleta $\operatorname{sk}_ n(K)$, we can reduce to the case where $K$ has dimension $\leq n$, for some integer $n \geq -1$ (see Definition We now proceed by induction on $n$. The case $n = -1$ is trivial (since a simplicial set of dimension $\leq -1$ is empty). Let us therefore assume that $n \geq 0$, so that Proposition supplies a pushout diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ \coprod \operatorname{\partial \Delta }^{n} \ar [r] \ar [d] & \coprod \Delta ^{n} \ar [d] \\ \operatorname{sk}_{n-1}(K) \ar [r] & K. } \]

It follows from our inductive hypothesis that the maps $\theta _{\operatorname{\partial \Delta }^{n}}$ and $\theta _{ \operatorname{sk}_{n-1}(K)}$ are bijective. Consequently, to show that $\theta _{ K }$ is bijective, it will suffice to show that $\theta _{ \Delta ^{n} }$ is bijective: that is, that $f$ induces a bijection on $n$-simplices. For $n \leq 1$, this follows from our hypothesis. To handle the case $n \geq 2$, we observe that there is a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{n}, S) \ar [d]^{ \theta _{ \Delta ^ n} } \ar [r] & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{n}_{1}, S ) \ar [d]^{ \theta _{ \Lambda ^{n}_{1} }} \\ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{n}, T) \ar [r] & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{n}_{1}, T). } \]

Here the right vertical map is bijective by virtue of our inductive hypothesis, and the horizontal maps are bijective by virtue of our assumption that both $S$ and $T$ satisfy condition $(\ast ')$. It follows that the left vertical map is also bijective, as desired. $\square$

Proof of Proposition Let $S$ be a simplicial set satisfying condition $(\ast ')$ of Proposition; we will show that there is a category $\operatorname{\mathcal{C}}$ and an isomorphism of simplicial sets $u: S \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ (the converse follows from Lemma It follows from Proposition that the category $\operatorname{\mathcal{C}}$ is uniquely determined (up to isomorphism), and from the proof of Proposition we can extract an explicit construction of $\operatorname{\mathcal{C}}$:

  • The objects of $\operatorname{\mathcal{C}}$ are the vertices of $S$.

  • Given a pair of objects $C, D \in \operatorname{\mathcal{C}}$, we let $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,D)$ denote the collection of edges $e$ of $S$ having source $C$ and target $D$.

  • For each object $C \in \operatorname{\mathcal{C}}$, we define the identity morphism $\operatorname{id}_{C} \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,C)$ to be the degenerate edge $s^{0}_0(C)$.

  • Given a triple of objects $C,D,E \in \operatorname{\mathcal{C}}$ and a pair of morphisms $f \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,D)$ and $g \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}( D, E)$, we can apply hypothesis $(\ast ')$ (in the special case $n = 2$ and $i = 1$) to conclude that there is a unique $2$-simplex $\sigma $ of $S_{\bullet }$ satisfying $d^{2}_2(\sigma ) = f$ and $d^{2}_0(\sigma ) = g$. We define the composition $g \circ f \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,E)$ to be the edge $d^{2}_1(\sigma )$.

We claim that $\operatorname{\mathcal{C}}$ is a category. For this, we must check the following:

  • The composition law on $\operatorname{\mathcal{C}}$ is unital: for every morphism $f: C \rightarrow D$ in $\operatorname{\mathcal{C}}$, we have equalities

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

    Let us verify the identity on the left; the proof in the other case is similar. For this, we must construct a $2$-simplex $\sigma $ of $S$ such that $d^{2}_0(\sigma ) = \operatorname{id}_{D}$ and $d^{2}_1(\sigma ) = d^{2}_2(\sigma ) = f$. The degenerate $2$-simplex $s^{1}_1(f)$ has these properties.

  • The composition law on $\operatorname{\mathcal{C}}$ is associative. That is, for every triple of composable morphisms

    \[ f: W \rightarrow X \quad \quad g: X \rightarrow Y \quad \quad h: Y \rightarrow Z \]

    in $\operatorname{\mathcal{C}}$, we have an identity $h \circ (g \circ f) = (h \circ g) \circ f$ in $\operatorname{\mathcal{C}}$. Applying condition $(\ast ')$ repeatedly, we deduce the following:

    • There is a unique $2$-simplex $\sigma _0$ of $S$ satisfying $d^{2}_0(\sigma _0) = h$ and $d^{2}_2(\sigma _0) = g$ (it follows that $d^{2}_1(\sigma _0) = h \circ g$).

    • There is a unique $2$-simplex $\sigma _3$ of $S$ satisfying $d^{2}_0(\sigma _3) = g$ and $d^{2}_2(\sigma _3) = f$ (it follows that $d^{2}_1(\sigma _3) = g \circ f$).

    • There is a unique $2$-simplex $\sigma _2$ of $S$ satisfying $d^{2}_0( \sigma _2) = h \circ g$ and $d^{2}_2( \sigma _2) = f$ (it follows that $d^{2}_1( \sigma _2) = (h \circ g) \circ f$).

    • There is a unique $3$-simplex $\tau $ of $S$ satisfying $d^{3}_0( \tau ) = \sigma _0$, $d^{3}_2(\tau ) = \sigma _2$, and $d^{3}_3(\tau ) = \sigma _3$ (this follows by applying $(\ast ')$ to the horn inclusion $\Lambda ^{3}_{1} \hookrightarrow \Delta ^3$).

    The $3$-simplex $\tau $ can be depicted in the following diagram

    \[ \xymatrix@C =70pt@R=70pt{ & X \ar [r]^-{g} \ar [drr]_{ h \circ g} & Y \ar [dr]^{ h} & \\ W \ar [ur]^{f} \ar [urr]_{g \circ f} \ar [rrr]^{ (h \circ g) \circ f } & & & Z. } \]

    Set $\sigma _1 = d^{3}_1( \tau )$. Then $\sigma _1$ is a $2$-simplex of $S$ satisfying $d^{2}_0( \sigma _1) = h$, $d^{2}_1( \sigma _1) = (h \circ g) \circ f$, and $d^{2}_2(\sigma _1) = g \circ f$. It follows that $\sigma _1$ “witnesses” the identity $h \circ (g \circ f) = (h \circ g) \circ f$.

Note that every $n$-simplex $\sigma : \Delta ^{n} \rightarrow S$ determines a functor $[n] \rightarrow \operatorname{\mathcal{C}}$, given on objects by the values of $\sigma $ on the vertices of $\Delta ^ n$ and on morphisms by the values of $\sigma $ on the edges of $\Delta ^ n$. This construction determines a map of simplicial sets $u: S \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ which is bijective on simplices of dimension $\leq 1$. Since the simplicial sets $S$ and $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ both satisfy condition $(\ast ')$ (Lemma, it follows from Lemma that $u$ is an isomorphism. $\square$

Remark The characterization of Proposition has many variants. For example, one can replace condition $(\ast ')$ by the following a priori weaker condition:

$(\ast '_{0})$

For every $n \geq 2$ and every morphism of simplicial sets $\sigma _0: \Lambda ^{n}_{1} \rightarrow S$, there is a unique $n$-simplex $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$ satisfying $\sigma _0 = \sigma |_{ \Lambda ^{n}_{1} }$.

See Corollary for a closely related result.