# Kerodon

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Proposition 1.2.2.1. The nerve functor $\operatorname{N}_{\bullet }: \operatorname{Cat}\rightarrow \operatorname{Set_{\Delta }}$ is fully faithful.

Proof of Proposition 1.2.2.1. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{C}}'$ be categories. We wish to show that the nerve functor $\operatorname{N}_{\bullet }$ induces a bijection

$\theta : \operatorname{Hom}_{\operatorname{Cat}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{C}}' ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}), \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}') ).$

Here the source of $\theta$ is the set of all functors from $\operatorname{\mathcal{C}}$ to $\operatorname{\mathcal{C}}'$. We first note that $\theta$ is injective: a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$ is determined by its behavior on the objects and morphisms of $\operatorname{\mathcal{C}}$, and therefore by the behavior of $\theta (F)$ on the vertices and edges of the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ (see Example 1.2.1.4). Let us prove the surjectivity of $\theta$. Let $f: \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}')$ be a morphism of simplicial sets; we wish to show that there exists a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$ such that $f = \theta (F)$. For each $n \geq 0$, the morphism $f$ determines a map of sets $\operatorname{N}_{n}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{n}(\operatorname{\mathcal{C}}')$, which we will also denote by $f$. In the case $n = 0$, this map carries each object $C \in \operatorname{\mathcal{C}}$ to an object of $\operatorname{\mathcal{C}}'$, which we will denote by $F(C)$. For every pair of objects $C, D \in \operatorname{\mathcal{C}}$, the map $f$ carries each morphism $u: C \rightarrow D$ to a morphism $f(u)$ in the category $\operatorname{\mathcal{C}}'$. Since $f$ commutes with face maps, the morphism $f(u)$ has source $F(C)$ and target $F(D)$ (see Example 1.2.1.4), and can therefore be regarded as an element of $\operatorname{Hom}_{\operatorname{\mathcal{C}}'}( F(C), F(D) )$; we denote this element by $F(u)$. We will complete the proof by verifying the following:

$(a)$

The preceding construction determines a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$.

$(b)$

We have an equality $f = \theta (F)$ of maps from $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ to $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}')$.

To prove $(a)$, we first note that the compatibility of $f$ with degeneracy maps implies that we have $F(\operatorname{id}_{C}) = \operatorname{id}_{F(C)}$ for each $C \in \operatorname{\mathcal{C}}$ (see Example 1.2.1.4). It will therefore suffice to show that for every pair of composable morphisms $u: C \rightarrow D$ and $v: D \rightarrow E$ in the category $\operatorname{\mathcal{C}}$, we have $F(v) \circ F(u) = F( v \circ u )$ as elements of the set $\operatorname{Hom}_{\operatorname{\mathcal{C}}'}( F(C), F(E) )$. For this, we observe that the diagram $C \xrightarrow {u} D \xrightarrow {v} E$ can be identified with a $2$-simplex $\sigma$ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. Using the equality $d_ i( f(\sigma ) ) = f( d_ i(\sigma ) )$ for $i = 0,2$, we see that $f(\sigma )$ corresponds to the diagram $F(C) \xrightarrow {F(u)} F(D) \xrightarrow {F(v)} F(E)$ in $\operatorname{\mathcal{C}}'$. We now compute

$F(v) \circ F(u) = d_1( f(\sigma ) ) = f( d_1(\sigma ) ) = F( v \circ u ).$

This completes the proof of $(a)$. To prove $(b)$, we must show that $f(\tau ) = \theta (F)(\tau )$ for each $n$-simplex $\tau$ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. This follows by construction in the case $n \leq 1$, and follows in general since an $n$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}')$ is determined by its $1$-dimensional faces (see Remark 1.2.1.3). $\square$