Kerodon

Proof. Combine Theorem 2.3.4.1 with Corollary 2.3.2.8. $\square$

Proof. Let $\operatorname{\mathcal{D}}$ be a category, which we regard as a $2$-category having only identity morphisms. We wish to show that every morphism of simplicial sets $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ factors uniquely through the morphism $\operatorname{N}_{\bullet }^{\operatorname{D}}(F)$. By virtue of Theorem 2.3.4.1, this is equivalent to the assertion that every strictly unitary lax functor $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ factors uniquely through $F$, which follows from Proposition 2.2.8.3. $\square$

Proof of Theorem 2.3.4.1. By virtue of Proposition 2.2.7.7, we may assume without loss of generality that the $2$-categories $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are strictly unitary (this assumption will simplify some of the notation in what follows). Let $U: \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$ be a map of simplicial sets. Then:

• Each object $X$ of $\operatorname{\mathcal{C}}$ can be identified with a vertex of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ (Example 2.3.1.13), whose image under $U$ is a vertex of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$. This vertex can be identified with an object of $\operatorname{\mathcal{D}}$, which we denote by $U_0(X)$.

• Each $1$-morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ can be identified with an edge of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ (Example 2.3.1.14), whose image under $U$ is an edge of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$. This edge can be identified with a $1$-morphism of $\operatorname{\mathcal{D}}$, which we will denote by $U_1(f): U_0(X) \rightarrow U_0(Y)$.

• Let $f: X \rightarrow Y$, $g: Y \rightarrow Z$, and $h: X \rightarrow Z$ be $1$-morphisms of $\operatorname{\mathcal{C}}$, and let $\gamma : g \circ f \Rightarrow h$ be a $2$-morphism of $\operatorname{\mathcal{C}}$. The $2$-morphism $\gamma $ determines a $2$-simplex of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ (Example 2.3.1.15). The image of this $2$-simplex under $U$ is a $2$-simplex of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$, which we can identify with a $2$-morphism $U_2(\gamma ): U_1(g) \circ U_1(f) \Rightarrow U_1( h )$ in $\operatorname{\mathcal{D}}$. Beware that this notation is slightly abusive: the $2$-morphism $U_2(\gamma )$ is a priori dependent not only on $\gamma $, but also on the factorization of the source of $\gamma $ as a composition $g \circ f$.

Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a strictly unitary lax functor. Unwinding the definitions, we see that the induced map of simplicial sets $\operatorname{N}_{\bullet }^{\operatorname{D}}(F): \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$ coincides with $U$ if and only if the following conditions are satisfied:

$(0)$

For every object $X \in \operatorname{\mathcal{C}}$, we have $F(X) = U_0(X)$ (as objects of $\operatorname{\mathcal{D}}$).

$(1)$

For every $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we have $F(f) = U_1(f)$ (as $1$-morphisms from $F(X) = U_0(X)$ to $F(Y) = U_0(Y)$ in $\operatorname{\mathcal{D}}$).

$(2)$

For every triple of $1$-morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$, and $h: X \rightarrow Z$ in $\operatorname{\mathcal{C}}$ and every $2$-morphism $\gamma : g \circ f \Rightarrow h$, the $2$-morphism $U_2(\gamma ): U_1(g) \circ U_1(f) \Rightarrow U_1(h)$ of $\operatorname{\mathcal{D}}$ is given by the (vertical) composition

\[ U_1(g) \circ U_1(f) = F(g) \circ F(f) \xRightarrow { \mu _{g,f} } F( g \circ f) \xRightarrow { F(\gamma )} F( h ) = U_1(h), \]

Let us note two special cases of condition $(2)$. Taking $h = g \circ f$ and $\gamma : g \circ f \Rightarrow h$ to be the identity $2$-morphism, we obtain the following:

$(2_0)$

For every pair of composable $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$ of $\operatorname{\mathcal{C}}$, the composition constraint $\mu _{g,f}: F(g) \circ F(f) \Rightarrow F(g \circ f)$ coincides with the $2$-morphism $U_2( \operatorname{id}_{g \circ f} )$.

Taking $g$ to be the identity morphism $\operatorname{id}_{Y}: Y \rightarrow Y$ and invoking our assumption that $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are strictly unitary, we also obtain:

$(2_1)$

For every pair of $1$-morphisms $f,h: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$ and every $2$-morphism $\gamma : f \Rightarrow h$, we have

\[ U_2( \gamma ) = F( \gamma ) \mu _{ \operatorname{id}_{Y}, f} = F(\gamma ) \]

(here the second identity follows from Remark 2.2.7.5, since the $2$-categories $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are strictly unitary).

We wish to show that there is a unique strictly unitary lax functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ satisfying conditions $(0)$, $(1)$, and $(2)$. The uniqueness is clear: by virtue of the analysis above, the functor $F$ must be given on objects, $1$-morphisms, and $2$-morphisms of $\operatorname{\mathcal{C}}$ by the formulae

(where, in the third formula, we identify the domain of each $2$-morphism $\gamma : f \Rightarrow h$ in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y)$ with the composition $\operatorname{id}_{Y} \circ f$), and the composition constraint $\mu _{g,f}: F(g) \circ F(f) \Rightarrow F(g \circ f)$ must be given by $\mu _{g,f} = U_2( \operatorname{id}_{g \circ f})$. To complete the proof, it will suffice to show that these formulae supply a well-defined lax functor $F: \operatorname{\mathcal{C}}\Rightarrow \operatorname{\mathcal{D}}$, and that $F$ satisfies condition $(2)$ above (note that $F$ satisfies conditions $(0)$ and $(1)$ by construction).

We first show that $F$ satisfies condition $(2)$. Suppose we are given a triple of $1$-morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$, and $h: X \rightarrow Z$, together with a $2$-morphism $\gamma : g \circ f \Rightarrow h$ in the $2$-category $\operatorname{\mathcal{C}}$. Consider the map $\partial \Delta ^3 \rightarrow \operatorname{N}^{\operatorname{D}}_{\bullet }(\operatorname{\mathcal{C}})$ represented by the pair of diagrams

(see Example 2.3.1.16). Using the identity $\alpha _{ \operatorname{id}_ Z, g, f} = \operatorname{id}_{ g \circ f}$ (Remark 2.2.7.3), we see that these diagrams satisfy the compatibility condition of Example 2.3.1.16, and can therefore be regarded as a $3$-simplex of $\operatorname{N}^{\operatorname{D}}_{\bullet }(\operatorname{\mathcal{C}})$. Applying the map of simplicial sets $U$, we deduce that the diagrams

determine a $3$-simplex of $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$: that is, we have a commutative diagram

By virtue of Remark 2.2.7.3, we see that this is equivalent to the identity $U_2(\gamma ) = F(\gamma ) \mu _{g,f}$ asserted by $(2)$.

Note that from condition $(2)$, we can deduce that $F$ satisfies the dual of condition $(2_1)$: that is, for every $2$-morphism $\gamma : g \Rightarrow h$ in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y)$, we have $F(\gamma ) = U_2(\gamma )$, where the right hand side is computed by regarding $\gamma $ as a $2$-morphism with domain $g \circ \operatorname{id}_{X}$. It follows that the construction of $F$ from $U$ is invariant under the operation of replacing $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ by the opposite $2$-categories $\operatorname{\mathcal{C}}^{\operatorname{op}}$ and $\operatorname{\mathcal{D}}^{\operatorname{op}}$ (this will be useful in what follows, since it reduces the number of identities that we need to check).

We now show that, for every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the construction of $F$ on $1$-morphisms and $2$-morphisms determines a functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( X, Y) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$. For this, we must establish the following:

• For each $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we have $F( \operatorname{id}_{f} ) = \operatorname{id}_{ F(f)}$ (as $2$-morphisms from $F(f)$ to itself in $\operatorname{\mathcal{D}}$). By definition, this is equivalent to the identity $U_2( \operatorname{id}_{f} ) = \operatorname{id}_{ F(f)}$, which follows from the compatibility of the map $U: \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$ with the degeneracy operators

  \[ s^{1}_1: \operatorname{N}_{1}^{\operatorname{D}}(\operatorname{\mathcal{C}}) \Rightarrow \operatorname{N}_{2}^{\operatorname{D}}(\operatorname{\mathcal{C}}) \quad \quad s^{1}_1: \operatorname{N}_{1}^{\operatorname{D}}(\operatorname{\mathcal{D}}) \Rightarrow \operatorname{N}_{2}^{\operatorname{D}}(\operatorname{\mathcal{D}}). \]

• For every triple of $1$-morphisms $f,g,h: X \Rightarrow Y$ in $\operatorname{\mathcal{C}}$ and every pair of $2$-morphisms $\gamma : f \Rightarrow g$, $\delta : g \Rightarrow h$, we have $F(\delta \gamma ) = F(\delta ) F(\gamma )$. To prove this, consider the map $\partial \Delta ^3 \rightarrow \operatorname{N}^{\operatorname{D}}_{\bullet }(\operatorname{\mathcal{C}})$ represented by the pair of diagrams

  \[ \xymatrix@C =100pt@R=50pt{ & Y \ar [r]^-{ \operatorname{id}_ Y } \ar@ {=>}[]+<5pt,-10pt>;+<15pt,-25pt>^-{\gamma } & Y \ar [dr]^{ \operatorname{id}_ Y } \ar@ {=>}[]+<-20pt,-20pt>;+<-45pt,-45pt>^{\delta } & \\ X \ar [ur]^{ f} \ar [urr]_{ g } \ar [rrr]^{h} & & & Y } \]
  \[ \xymatrix@C =100pt@R=50pt{ & Y \ar [r]^-{ \operatorname{id}_ Y} \ar [drr]_{ \operatorname{id}_ Y } \ar@ {=>}[]+<20pt,-20pt>;+<45pt,-45pt>_{\delta \gamma } & Y \ar [dr]^{ \operatorname{id}_ Y } \ar@ {=>}[]+<-5pt,-10pt>;+<-15pt,-25pt>^{ \operatorname{id}_{ \operatorname{id}_ Y} } & \\ X \ar [ur]^{f} \ar [rrr]^{h} & & & Y } \]

  (see Example 2.3.1.16). It follows from Remark 2.2.7.3 that the associativity constraint $\alpha _{ \operatorname{id}_ Y, \operatorname{id}_{Y}, f }$ is the identity, so that the diagrams above satisfy the compatibility condition of Example 2.3.1.16 and therefore determine a $3$-simplex of $\operatorname{N}^{\operatorname{D}}_{\bullet }(\operatorname{\mathcal{C}})$. Applying the map of simplicial sets $U$, we deduce that there exists a $3$-simplex of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{\mathcal{D}}}$ whose boundary is given by the diagrams

  \[ \xymatrix@C =70pt@R=50pt{ & F(Y) \ar [r]^-{ \operatorname{id}_{F(Y)} } \ar@ {=>}[]+<5pt,-10pt>;+<15pt,-25pt>^-{F(\gamma ) } & F(Y) \ar [dr]^{ \operatorname{id}_{F(Y)} } \ar@ {=>}[]+<-20pt,-20pt>;+<-45pt,-45pt>^{F(\delta ) } & \\ F(X) \ar [ur]^{ F(f)} \ar [urr]_{ F(G) } \ar [rrr]^{F(h)} & & & F(Y) } \]
  \[ \xymatrix@C =70pt@R=50pt{ & F(Y) \ar [r]^-{ \operatorname{id}_{F(Y)}} \ar [drr]_{ \operatorname{id}_{F(Y)} } \ar@ {=>}[]+<20pt,-20pt>;+<45pt,-45pt>_{F(\delta \gamma ) } & F(Y) \ar [dr]^{ \operatorname{id}_{F(Y)} } \ar@ {=>}[]+<-5pt,-10pt>;+<-15pt,-25pt>^{ \operatorname{id}_{ \operatorname{id}_{F(Y)}} } & \\ F(X) \ar [ur]^{F(f)} \ar [rrr]^{F(h)} & & & F(Y). } \]

  Using the criterion of Example 2.3.1.16, we see that this is equivalent to the identity $F(\delta \gamma ) = F(\delta ) F(\gamma )$.

We now show that, for every triple of objects $X,Y,Z \in \operatorname{\mathcal{C}}$, the composition constraints $\mu _{g,f}: F(g) \circ F(f) \Rightarrow F(g \circ f)$ depends functorially on $f \in \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y)$ and $g \in \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(Y,Z)$. We will argue that for fixed $f$, the construction $g \mapsto \mu _{g,f}$ is functorial; functoriality in $g$ will then follow by symmetry. Suppose we are given a $2$-morphism $\gamma : g \Rightarrow h$ in $\operatorname{\mathcal{C}}$; we wish to show that the diagram $\tau :$

commutes in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(X), F(Z) )$. To prove this, we consider the map $\operatorname{\partial }\Delta ^3 \rightarrow \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ represented by the pair of diagrams

Using the identity $\alpha _{ \operatorname{id}_{Z}, g, f} = \operatorname{id}_{ g \circ f}$ supplied by Remark 2.2.7.3, we see that this diagram defines a $3$-simplex of $\operatorname{N}^{\operatorname{D}}_{\bullet }(\operatorname{\mathcal{C}})$. Applying the map of simplicial sets $U$, we deduce that there is a $3$-simplex of $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$ whose boundary is represented by the pair of diagrams

This translates to the commutativity of the diagram

which (again by virtue of Remark 2.2.7.3) is equivalent to the commutativity of the diagram $\tau $.

To complete the proof, it will suffice to show that $F$ and $\mu $ satisfy conditions $(a)$, $(b)$, and $(c)$ of Definition 2.2.4.5. Condition $(a)$ is immediate from the construction, and $(b)$ follows by symmetry. To verify $(c)$, suppose we are given a triple of composable $1$-morphisms $W \xrightarrow {f} X \xrightarrow {g} Y \xrightarrow {h} Z$ in the $2$-category $\operatorname{\mathcal{C}}$. Consider the $3$-simplex of $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ represented by the pair of diagrams

Applying $U$, we obtain a $3$-simplex of $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$ represented by the pair of diagrams

which is equivalent to the commutativity of the pentagon appearing in the diagram

in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(W), F(Z) )$. Since the triangle on the lower left commutes by virtue of $(2)$, it follows that the outer cycle of the diagram commutes, as desired. $\square$