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2.3.5 Twisted Arrows and the Nerve of $\operatorname{Corr}(\operatorname{\mathcal{C}})^{\operatorname{c}}$

Let $\operatorname{\mathcal{E}}$ be a category which admits fiber products and let $\operatorname{Corr}(\operatorname{\mathcal{E}})$ denote the $2$-category of correspondences in $\operatorname{\mathcal{E}}$ (Example 2.2.2.1). Our goal in this section is to give an explicit description of the Duskin nerve of the conjugate $2$-category $\operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}}$ (Corollary 2.3.5.8 and Remark 2.3.5.9). We will obtain this description by formulating a universal property of $\operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}}$ as an object of the category $\operatorname{2Cat}_{\operatorname{ULax}}$. First, we need a brief digression.

Construction 2.3.5.1 (The Twisted Arrow Category). Let $\operatorname{\mathcal{C}}$ be a category. We define a new category $\operatorname{Tw}(\operatorname{\mathcal{C}})$ as follows:

• An object of $\operatorname{Tw}(\operatorname{\mathcal{C}})$ is a morphism $f: C \rightarrow D$ in $\operatorname{\mathcal{C}}$.

• Let $f: C \rightarrow D$ and $f': C' \rightarrow D'$ be objects of $\operatorname{Tw}(\operatorname{\mathcal{C}})$. A morphism from $f$ to $f'$ in $\operatorname{Tw}(\operatorname{\mathcal{C}})$ is a pair of morphisms $u: C \rightarrow C'$, $v: D' \rightarrow D$ in $\operatorname{\mathcal{C}}$ satisfying $f = v \circ f' \circ u$, so that we have a commutative diagram

$\xymatrix { C \ar [r]^-{f} \ar [d]^{u} & D \\ C' \ar [r]^-{f'} & D'. \ar [u]_{v} }$
• Let $f: C \rightarrow D$, $f': C' \rightarrow D'$, and $f'': C'' \rightarrow D''$ be objects of $\operatorname{Tw}(\operatorname{\mathcal{C}})$. If $(u,v)$ is a morphism from $f$ to $f'$ in $\operatorname{Tw}(\operatorname{\mathcal{C}})$ and $(u',v')$ is a morphism from $f'$ to $f''$ in $\operatorname{\mathcal{C}}$, then the composition $(u',v') \circ (u, v)$ in $\operatorname{Tw}(\operatorname{\mathcal{C}})$ is the pair $(u' \circ u, v \circ v')$.

We will refer to $\operatorname{Tw}(\operatorname{\mathcal{C}})$ as the twisted arrow category of $\operatorname{\mathcal{C}}$.

Remark 2.3.5.2. Let $[1] = \{ 0 < 1 \}$ denote a linearly ordered set with two elements. For any category $\operatorname{\mathcal{C}}$, we can identify morphisms of $\operatorname{\mathcal{C}}$ with functors $F: [1] \rightarrow \operatorname{\mathcal{C}}$. The collection of such functors can be organized into a category $\operatorname{Fun}( [1], \operatorname{\mathcal{C}})$, which we refer to as the arrow category of $\operatorname{\mathcal{C}}$. The arrow category $\operatorname{Fun}( [1], \operatorname{\mathcal{C}})$ has the same objects as the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$. However, the morphisms are different: if $f: C \rightarrow D$ and $f': C' \rightarrow D'$ are morphisms of $\operatorname{\mathcal{C}}$, then morphisms from $f$ to $f'$ in $\operatorname{Fun}( [1], \operatorname{\mathcal{C}})$ can be identified with commutative diagrams

$\xymatrix { C \ar [r]^-{f} \ar [d] & D \ar [d] \\ C' \ar [r]^-{f'} & D'. }$

Remark 2.3.5.3. Let $\operatorname{\mathcal{C}}$ be a category and let $\operatorname{Tw}(\operatorname{\mathcal{C}})$ denote its twisted arrow category. There is an evident forgetful functor $\operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{C}}\times \operatorname{\mathcal{C}}^{\operatorname{op}}$, given on objects by the construction $(f: C \rightarrow D) \mapsto (C,D)$.

Example 2.3.5.4. Let $Q$ be a partially ordered set, which we regard as a category. Then the twisted arrow category $\operatorname{Tw}(Q)$ can be identified (via the forgetful functor of Remark 2.3.5.3) with the partially ordered set

$\{ (p,q) \in Q \times Q^{\operatorname{op}}: p \leq q \} \subseteq Q \times Q^{\operatorname{op}}.$

Remark 2.3.5.5 (Functoriality). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of categories. Then $F$ induces a functor of twisted arrow categories $\operatorname{Tw}(F): \operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{Tw}(\operatorname{\mathcal{D}})$. This construction is compatible with composition, and can therefore be regarded as a functor $\operatorname{Tw}: \operatorname{Cat}\rightarrow \operatorname{Cat}$ from the category of (small) categories to itself.

Construction 2.3.5.6. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{E}}$ be categories, where $\operatorname{\mathcal{E}}$ admits fiber products, and let $F: \operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{E}}$ be a functor. We define a strictly unitary lax functor $F^{+}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}}$ as follows:

• For each $C \in \operatorname{\mathcal{C}}$, we define $F^{+}(C) = F( \operatorname{id}_{C} )$; here we regard the identity morphism $\operatorname{id}_ C: C \rightarrow C$ as an object of the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$.

• For each morphism $f: C \rightarrow D$ in $\operatorname{\mathcal{C}}$, we define $F^{+}(f)$ to be the $1$-morphism of $\operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}}$ given by the correspondence

$F( \operatorname{id}_ C ) \xleftarrow { F(\operatorname{id}_ C, f) } F(f) \xrightarrow { F( f, \operatorname{id}_ D) } F( \operatorname{id}_ D );$

this determines the values of $F^{+}$ on $2$-morphisms, since every $2$-morphism in $\operatorname{\mathcal{C}}$ is an identity $2$-morphism.

• For every pair of composable morphisms $C \xrightarrow {f} D \xrightarrow {g} E$, the composition constraint $\mu _{g,f}: F^{+}(g) \circ F^{+}(f) \Rightarrow F^{+}(g \circ f)$ is the $2$-morphism of $\operatorname{Corr}(\operatorname{\mathcal{C}})^{\operatorname{c}}$ corresponding to the map $F(g \circ f) \rightarrow F(f) \times _{ F(\operatorname{id}_ D) } F(g)$ classified by the commutative diagram

$\xymatrix { & F(g \circ f) \ar [dl]_{ F(\operatorname{id}_ C, g)} \ar [dr]^{ F( f, \operatorname{id}_ E) } & \\ F(f) \ar [dr]_{ F(f, \operatorname{id}_ D) } & & F(g) \ar [dl]^{ F( \operatorname{id}_ D, g) } \\ & F( \operatorname{id}_ D) & }$

in the category $\operatorname{\mathcal{E}}$.

Theorem 2.3.5.7. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{E}}$ be categories where $\operatorname{\mathcal{E}}$ admits fiber products. Then Construction 2.3.5.6 induces a bijection of sets

$\{ \textnormal{Functors F: \operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{E}}} \} \xrightarrow {\sim } \{ \textnormal{Strictly Unitary Lax Functors F^{+}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}}} \} .$

Corollary 2.3.5.8. Let $\operatorname{\mathcal{E}}$ be a category which admits fiber products and let $\operatorname{Corr}(\operatorname{\mathcal{E}})$ denote the $2$-category of correspondences in $\operatorname{\mathcal{E}}$ (Example 2.2.2.1). Then the simplicial set $\operatorname{N}_{\bullet }^{\operatorname{D}}( \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}} )$ is given concretely by the formula

$( [n] \in \operatorname{{\bf \Delta }}) \mapsto \{ \textnormal{Functors \operatorname{Tw}([n]) \rightarrow \operatorname{\mathcal{E}}} \} .$

Remark 2.3.5.9. Let $\operatorname{\mathcal{E}}$ be a category which admits fiber products. Combining Corollary 2.3.5.8 with Example 2.3.5.4, we see that $n$-simplices of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}( \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}} )$ can be identified with diagrams

$\{ (i,j) \in [n] \times [n]^{\operatorname{op}}: i \leq j \} \rightarrow \operatorname{\mathcal{E}},$

which we can represent graphically as

$\xymatrix { & & & X_{0,n} \ar [dl] \ar [dr] & & & \\ & & \cdots \ar [dl] \ar [dr] & & \cdots \ar [dl] \ar [dr] & & \\ & X_{0,1} \ar [dl] \ar [dr] & & \cdots \ar [dl]\ar [dr] & & X_{n-1,n} \ar [dl] \ar [dr] & \\ X_{0,0} & & X_{1,1} & \cdots & X_{n-1,n-1} & & X_{n,n}. }$

In §, we will use this description to extend the definition of $\operatorname{Corr}(\operatorname{\mathcal{E}})$ to the case where $\operatorname{\mathcal{E}}$ is an $\infty$-category.

Example 2.3.5.10. Let $\operatorname{\mathcal{E}}$ be a category which admits fiber products. Then $2$-simplices $\sigma$ of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}( \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}} )$ can be identified with commutative diagrams

$\xymatrix { & & X_{0,2} \ar [dl] \ar [dr] & & \\ & X_{0,1} \ar [dl] \ar [dr] & & X_{1,2} \ar [dl] \ar [dr] & \\ X_{0,0} & & X_{1,1} & & X_{2,2} }$

in the category $\operatorname{\mathcal{E}}$. Such a diagram corresponds to a thin $2$-simplex of $\operatorname{N}_{\bullet }^{\operatorname{D}}( \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}} )$ (in the sense of Definition 2.3.2.3) if and only if the square appearing in the diagram is a pullback: that is, it induces an isomorphism $X_{0,2} \rightarrow X_{0,1} \times _{ X_{1,1} } X_{1,2}$.

Proof of Theorem 2.3.5.7. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{E}}$ be categories, where $\operatorname{\mathcal{E}}$ admits fiber products, and let $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{Corr}(\operatorname{\mathcal{E}})^{\operatorname{c}}$ be a strictly unitary lax functor of $2$-categories. For every morphism $f: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, we can identify $G(f)$ with a correspondence from $G(C)$ to $G(D)$ in the category $\operatorname{\mathcal{E}}$, given by a diagram we will denote by $G(C) \xleftarrow { u(f) } M(f) \xrightarrow { v(f) } G(D)$. Our assumption that $G$ is strictly unitary guarantees the following:

$(\ast )$

For each object $C \in \operatorname{\mathcal{C}}$, the object $M( \operatorname{id}_ C )$ is equal to $G(C)$, and the maps $u(\operatorname{id}_ C): M( \operatorname{id}_ C) \rightarrow G(C)$ and $v( \operatorname{id}_ C): M( \operatorname{id}_ C) \rightarrow G(C)$ are the identity morphisms from $G(C)$ to itself in the category $\operatorname{\mathcal{E}}$.

For every pair of composable $1$-morphisms $C \xrightarrow {f} D \xrightarrow {g} E$, the composition constraint $\mu _{g,f}$ for the lax functor $G$ can be identified with a morphism from $M(g \circ f)$ to the fiber product $M(f) \times _{ G(D)} M(g)$ in the category $\operatorname{\mathcal{E}}$, or equivalently with a pair of morphisms

$p(g,f): M(g \circ f) \rightarrow M(f) \quad \quad q(g,f): M(g \circ f) \rightarrow M(g)$

satisfying $v(f) \circ p(g,f) = u(g) \circ q(g,f)$. The axioms for a lax functor (Definition 2.2.4.5) then translate to the following additional conditions:

$(a)$

For every $1$-morphism $f: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, $p( \operatorname{id}_ D, f)$ is the identity morphism from $M(f)$ to itself.

$(b)$

For every $1$-morphism $f: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, $q( f, \operatorname{id}_ C )$ is the identity morphism from $M(f)$ to itself.

$(c)$

For every composable triple of $1$-morphisms $W \xrightarrow {f} X \xrightarrow {g} Y \xrightarrow {h} Z$ in the category $\operatorname{\mathcal{C}}$, we have

$p(h \circ g, f) = p(g,f) \circ p( h, g \circ f) \quad \quad q(h, g \circ f ) = q(h,g) \circ q(h \circ g, f)$
$q(g,f) \circ p( h, g \circ f) = p( h, g) \circ q(h \circ g, f).$

We wish to show that there exists a unique functor of ordinary categories $F: \operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{E}}$ such that $G = F^{+}$, where $F^{+}$ is the lax functor associated to $F$ by Construction 2.3.5.6. For this condition to be satisfied, the functor $F$ must have the following properties:

$(0)$

For each object $C \in \operatorname{\mathcal{C}}$, we have $F( \operatorname{id}_ C ) = G(C)$ (this guarantees that $G$ and $F^{+}$ coincide on objects).

$(1)$

For each morphism $f: C \rightarrow D$ in $\operatorname{\mathcal{C}}$ (regarded as an object of $\operatorname{Tw}(\operatorname{\mathcal{C}})$), we have $F(f) = M(f)$, and the morphisms $u(f)$ and $v(f)$ are given by $F( \operatorname{id}_ C, f)$ and $F(f, \operatorname{id}_ D)$, respectively (this guarantees that $G$ and $F^{+}$ coincide on $1$-morphisms, and therefore also on $2$-morphisms).

$(2)$

For every pair of composable $1$-morphisms $C \xrightarrow {f} D \xrightarrow {g} E$, the morphisms $p(g,f): M(g \circ f) \rightarrow M(f)$ and $q(g,f): M(g \circ f) \rightarrow M(g)$ are given by $F(\operatorname{id}_ C,g): F(g \circ f) \rightarrow F(f)$ and $F(f, \operatorname{id}_ E): F(g \circ f) \rightarrow F(g)$, respectively (this guarantees that the composition constraints on $G$ and $F^{+}$ coincide).

Note that the value of $F$ on each object of $\operatorname{Tw}(\operatorname{\mathcal{C}})$ is determined by condition $(1)$. Moreover, if $(u,v)$ is a morphism from $f: C \rightarrow D$ to $f': C' \rightarrow D'$ in the category $\operatorname{Tw}(\operatorname{\mathcal{C}})$, then condition $(2)$ guarantees that $F(u,v)$ must be equal to the composition

$F(f) = M(v \circ f' \circ u) \xrightarrow { p(v, f' \circ u)} M(f' \circ u) \xrightarrow {q(f',u)} M(f') = F(f').$

This proves the uniqueness of the functor $F$.

To prove existence, we define $F$ on objects $f$ of $\operatorname{Tw}(\operatorname{\mathcal{C}})$ by the formula $F(f) = M(f)$, and on morphisms $(u,v): f \rightarrow f'$ by the formula $F(u,v) = q(f',u) \circ p(v, f' \circ u)$. For any morphism $f: C \rightarrow D$ in $\operatorname{\mathcal{C}}$, we can use $(a)$ and $(b)$ to compute

$F(\operatorname{id}_ C, \operatorname{id}_ D) = q( f, \operatorname{id}_ C) \circ p( \operatorname{id}_ D, f \circ \operatorname{id}_ C) = \operatorname{id}_{M(f)} \circ \operatorname{id}_{ M(f)} = \operatorname{id}_{ F(f) },$

so that $F$ carries identity morphisms in $\operatorname{Tw}(\operatorname{\mathcal{C}})$ to identity morphisms in $\operatorname{\mathcal{E}}$. To complete the proof that $F$ is a functor, we note that for every pair of composable morphisms

$(f: C \rightarrow D) \xrightarrow { (u,v)} (f': C' \rightarrow D') \xrightarrow { (u',v')} (f'': C'' \rightarrow D'')$

in $\operatorname{Tw}(\operatorname{\mathcal{C}})$, the identities given in $(c)$ allow us to compute

\begin{eqnarray*} F(u', v') \circ F(u,v) & = & q( f'', u') \circ p(v', f'' u') \circ q(v' f'' u', u) \circ p( v, v' f'' u' u) \\ & = & q(f'', u') \circ q(f'' u', u) \circ p( v', f'' u' u ) \circ p(v, v' f'' u' u) \\ & = & q( f'', u' u ) \circ p( v' v, f'' u' u) \\ & = & F(u' \circ u, v \circ v' ). \end{eqnarray*}

We now complete the proof by showing that the functor $F$ satisfies conditions $(0)$, $(1)$, and $(2)$. Condition $(0)$ is an immediate consequence of $(\ast )$. To prove $(2)$, we note that for any pair of composable morphisms $C \xrightarrow {f} D \xrightarrow {g} E$, identities $(a)$ and $(b)$ yield equalities

$F( \operatorname{id}_ C, g) = q(f,\operatorname{id}_ C) \circ p( g, f \circ \operatorname{id}_ C) = p(g,f) \quad \quad F(f, \operatorname{id}_ E) = q(g,f) \circ p( \operatorname{id}_ E, g \circ f) = q(g,f).$

For $(1)$, we note that if $f: C \rightarrow D$ is a morphism in $\operatorname{\mathcal{C}}$, then we have

\begin{eqnarray*} F(\operatorname{id}_ C, f) & = & q( \operatorname{id}_ C, \operatorname{id}_ C) \circ p( f, \operatorname{id}_ C \circ \operatorname{id}_ C) \\ & = & \operatorname{id}_{G(C)} \circ p(f, \operatorname{id}_ C ) \\ & = & v( \operatorname{id}_ C ) \circ p( f, \operatorname{id}_ C) \\ & = & u(f) \circ q(f, \operatorname{id}_ C) \\ & = & u(f) \end{eqnarray*}

and a similar calculation yields $F(f, \operatorname{id}_ D) = v(f)$. $\square$

We close this section by describing a variant of Theorem 2.3.5.7 which will be useful in §4.6.1 (see Example 4.6.1.12).

Construction 2.3.5.11. Let $\operatorname{\mathcal{C}}$ be a category, let $\operatorname{\mathcal{D}}$ be a $2$-category containing objects $X$ and $Y$, and let $F: \operatorname{Tw}(\operatorname{\mathcal{C}})^{\operatorname{op}} \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)$ be a functor. We define a strictly unitary lax functor $U_ F: [1] \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ as follows:

$(1)$

The lax functor $U_{F}$ is given on objects by $U_ F(0,A) = X$ and $U_ F(1,A) = Y$ for each $A \in \operatorname{\mathcal{C}}$.

$(2)$

Let $f: A \rightarrow B$ be a morphism in the category $\operatorname{\mathcal{C}}$, which we also regard as an object of the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$. For $0 \leq i \leq j \leq 1$, we let $f_{ji}$ denote the corresponding morphism from $(i,A)$ to $(j,B)$ in the product category $[1] \times \operatorname{\mathcal{C}}$. Then the lax functor $U_{F}$ is given on $1$-morphisms by the formula

$U_{F}( f_{ji} ) = \begin{cases} \operatorname{id}_{X} & \text{ if } i = j = 0 \\ \operatorname{id}_{Y} & \text{ if } i = j = 1 \\ F(f) & \text{ if } 0 = i < j = 1. \end{cases}$
$(3)$

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be composable morphisms in the category $\operatorname{\mathcal{C}}$, and let $0 \leq i \leq j \leq k \leq 1$. Then the composition constraint $\mu _{g_{kj}, f_{ji} }$ for the lax functor $U_ F$ is given as follows:

• If $i=j=k=0$, then $\mu _{g_{kj},f_{ji} }$ is the unit constraint $\upsilon _{X}: \operatorname{id}_{X} \circ \operatorname{id}_{X} \xRightarrow {\sim } \operatorname{id}_{X}$ of the $2$-category $\operatorname{\mathcal{D}}$.

• If $i=0$ and $j=k=1$, then $\mu _{g_{kj},f_{ji} }$ is given by the composition

$\operatorname{id}_{Y} \circ F(f) \xRightarrow { \lambda _{F(f)} } F(f) \xRightarrow { F(\operatorname{id}_ A, g) } F( g \circ f),$

where $\lambda _{F(f)}$ is the left unit constraint of Construction 2.2.1.11 and we regard the pair $(\operatorname{id}_ A,g)$ as an element of $\operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}( g \circ f,f)$.

• If $i=j=0$ and $k=1$, then $\mu _{g_{kj},f_{ji} }$ is given by the composition

$F(g) \circ \operatorname{id}_{X} \xRightarrow { \rho _{F(g)} } F(g) \xRightarrow { F(f, \operatorname{id}_ C) } F( g \circ f),$

where $\rho _{F(g)}$ is the right unit constraint of Construction 2.2.1.11 and we regard the pair $(f, \operatorname{id}_ C)$ as an element of $\operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}( g \circ f,g)$.

• If $i=j=k=1$, then $\mu _{g_{kj},f_{ji} }$ is equal to the unit constraint $\upsilon _{Y}: \operatorname{id}_{Y} \circ \operatorname{id}_{Y} \xRightarrow {\sim } \operatorname{id}_{Y}$ of the $2$-category $\operatorname{\mathcal{D}}$.

Exercise 2.3.5.12. Show that Construction 2.3.5.11 is well-defined. That is, given a functor $F: \operatorname{Tw}(\operatorname{\mathcal{C}})^{\operatorname{op}} \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)$ as in Construction 2.3.5.11, show that there is a unique strictly unitary lax functor $U_{F}$ satisfying properties $(1)$, $(2)$, and $(3)$ of Construction 2.3.5.11.

Theorem 2.3.5.13. Let $\operatorname{\mathcal{C}}$ be a category and let $\operatorname{\mathcal{D}}$ be a $2$-category containing objects $X$ and $Y$. Then the assignment $F \mapsto U_{F}$ of Construction 2.3.5.11 induces a monomorphism of sets

$\xymatrix { \{ \textnormal{Functors F: \operatorname{Tw}(\operatorname{\mathcal{C}})^{\operatorname{op}} \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)} \} \ar [d] \\ \{ \textnormal{Strictly unitary lax functors U: [1] \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}} \} . }$

The image of this monomorphism consists of those strictly unitary lax functors $U: [1] \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ having the property that $U|_{ \{ 0\} \times \operatorname{\mathcal{C}}}$ and $U|_{ \{ 1\} \times \operatorname{\mathcal{C}}}$ are the constant functors taking the values $X$ and $Y$, respectively.

Corollary 2.3.5.14. Let $\operatorname{\mathcal{C}}$ be a category, let $\operatorname{\mathcal{D}}$ be a $2$-category containing objects $X$ and $Y$, let $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$ denote the Duskin nerve of $\operatorname{\mathcal{D}}$, and let $\operatorname{Hom}_{ \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})}( X,Y)$ denote the iterated fiber product

$\{ X \} \times _{ \operatorname{Fun}( \{ 0\} , \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}}) ) } \operatorname{Fun}( \Delta ^1, \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}}) ) \times _{ \operatorname{Fun}( \{ 1\} , \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}}) ) } \{ Y \}$

(see Construction 4.6.1.1). Then Construction 2.3.5.11 determines a bijection

$\xymatrix { \{ \textnormal{Functors F: \operatorname{Tw}(\operatorname{\mathcal{C}})^{\operatorname{op}} \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)} \} \ar [d] \\ \{ \textnormal{Morphisms \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{Hom}_{ \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})}( X,Y)} \} .}$

Proof. By definition, a morphism from $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ to $\operatorname{Hom}_{ \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})}(X,Y)$ can be identified with a morphism of simplicial sets

$U: \operatorname{N}_{\bullet }([1] \times \operatorname{\mathcal{C}}) \simeq \Delta ^{1} \times \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{D}})$

such that $U|_{ \operatorname{N}_{\bullet }( \{ 0\} \times \operatorname{\mathcal{C}})}$ and $U|_{ \operatorname{N}_{\bullet }( \{ 1\} \times \operatorname{\mathcal{D}}) }$ are the constant maps taking the values $X$ and $Y$, respectively. The desired result now follows by combining Theorems 2.3.5.13 and 2.3.4.1. $\square$

Proof of Theorem 2.3.5.13. Let $\operatorname{\mathcal{C}}$ be an ordinary category, let $\operatorname{\mathcal{D}}$ be a $2$-category containing objects $X$ and $Y$, and let $U: [1] \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a strictly unitary lax functor having the property that $U|_{ \{ 0\} \times \operatorname{\mathcal{C}}}$ and $U|_{ \{ 1\} \times \operatorname{\mathcal{C}}}$ are the constant functors taking the values $X$ and $Y$, respectively. We wish to show that there exists a unique functor of ordinary categories $F: \operatorname{Tw}(\operatorname{\mathcal{C}})^{\operatorname{op}} \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)$ such that $U$ is equal to the strictly unitary lax functor $U_{F}$ given by Construction 2.3.5.11. To prove this, we may assume without loss of generality that the $2$-category $\operatorname{\mathcal{D}}$ is strictly unitary (Proposition 2.2.7.7). Given a morphism $f: A \rightarrow B$ in the category $\operatorname{\mathcal{C}}$ and a pair of integers $0 \leq i \leq j \leq 1$, we write $f_{ji}: (i,A) \rightarrow (j,B)$ for the corresponding morphism in the product category $[1] \times \operatorname{\mathcal{C}}$. Unwinding the definitions, we see that the identity $U = U_{F}$ imposes the following requirements on the functor $F$:

$(2)$

Let $f: A \rightarrow B$ be a morphism in the category $\operatorname{\mathcal{C}}$, which we identify with both with an object in the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$. Then $F(f)$ is equal to $U( f_{10} ) \in \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)$.

$(3)$

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be composable morphisms in the category $\operatorname{\mathcal{C}}$, and regard the pairs $(\operatorname{id}_ A, g)$ and $(f, \operatorname{id}_ C)$ element of $\operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}( g \circ f, f)$ and $\operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}( g \circ f, g)$, respectively. Then $F( \operatorname{id}_ A, g)$ and $F(f, \operatorname{id}_ C)$ are equal to the composition contraints $\mu _{ g_{11}, f_{10} }$ and $\mu _{ g_{10}, f_{00} }$ for the lax functor $U$, respectively.

We now establish the uniqueness of the functor $F$. The value of $F$ on objects is determined by condition $(2)$. If $f: A \rightarrow B$ and $f': A' \rightarrow B'$ are objects of the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$, then an element of $\operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}(f', f)$ can be identified with a pair $(u,v)$ where $u \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(A',A)$ and $v \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(B,B')$ satisfy $f' = v \circ f \circ u$. In this case, the morphism $(u,v)$ factors as a composition $(\operatorname{id}_{A'}, v) \circ (u, \operatorname{id}_{B} )$, so condition $(3)$ guarantees the identity

$F(u,v) = F(u, \operatorname{id}_{B'} ) \circ F( \operatorname{id}_{A}, v) = \mu _{ (v \circ f)_{01}, u_{00} } \circ \mu _{ v_{11}, f_{10} }.$

This proves the uniqueness of $F$ on morphisms.

To prove existence, we define the functor $F$ on objects $f \in \operatorname{Tw}(\operatorname{\mathcal{C}})$ by setting $F(f) = U( f_{10} )$, and on morphisms $(u,v) \in \operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}(f',f)$ by the formula

$F(u,v) = \mu _{ f_{10} \circ v_{00}, u_{00} } \circ \mu _{ v_{11}, f_{10} }.$

Note that this prescription automatically satisfies condition $(2)$. Since $U$ is a strictly unitary functor between strictly unitary $2$-categories, its composition constraints $\mu _{g,h}$ are the identity whenever either $g$ or $h$ is an identity morphism (Remark 2.2.7.5), which shows that $F$ satisfies condition $(3)$ and carries identity morphisms to identity morphisms. We will complete the proof by showing that $F$ is compatible with composition. Let $f: A \rightarrow B$, $f': A' \rightarrow B'$, and $f'': A'' \rightarrow B''$ be morphisms in the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$, and suppose we are given morphisms $(u,v) \in \operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}( f', f)$ and $(u',v') \in \operatorname{Hom}_{\operatorname{Tw}(\operatorname{\mathcal{C}})}( f'', f')$. We wish to prove an equality $F( u \circ u', v' \circ v) = F(u',v') \circ F(u,v)$ of morphisms from $F(f)$ to $F(f'')$ in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)$. Unwinding the definitions, this is equivalent to the commutativity of the outer cycle of the diagram

$\xymatrix@R =50pt@C=50pt{ & & F(vfu) \ar@ {=>}[dr]^{ \mu _{v'_{11}, (vfu)_{10} } } & & \\ & F(vf) \ar@ {=>}[ur]^{ \mu _{ (vf)_{01}, u_{00}}} \ar@ {=>}[dr]^{ \mu _{v'_{11}, (vf)_{00} } } & & F(v'vfu) \ar@ {=>}[dr]^{ \mu _{ (v'vfu)_{10}, u_{11} } } & \\ F(f) \ar@ {=>}[ur]^{ \mu _{v_{11}, f_{10} } }\ar@ {=>}[rr]^{ \mu _{ (vv')_{11}, f_{10} }} & & F(v'vf) \ar@ {=>}[ur]^{ \mu _{ (v'vf)_{10}, u_{00} }} \ar@ {=>}[rr]^{ \mu _{ (v'vf)_{10}, (uu')_{00} }} & & F( v'vfuu') }$

in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(X,Y)$. In fact, the entire diagram commutes. The commutativity of the upper square follows by applying property $(c)$ of Definition 2.2.4.5 to the composable triple of morphisms

$(0,A') \xrightarrow {u_{00} } (0,A) \xrightarrow { (vf)_{01} } (1,B') \xrightarrow { v'_{11} } (1,B'')$

in the product category $[1] \times \operatorname{\mathcal{C}}$. The commutativity of the lower left triangle follows by applying property $(c)$ to the composable triple of morphisms

$(0,A) \xrightarrow {f_{10}} (1,B) \xrightarrow { v_{11} } (1,B') \xrightarrow { v'_{11} } (1,B'')$

and noting that the composition constraint $\mu _{ v'_{11}, v_{11} }$ is equal to the identity (by virtue of our assumption that the lax functor $U|_{ \{ 1\} \times \operatorname{\mathcal{C}}}$ is constant). Similarly, the commutativity of the lower right triangle follows by applying $(c)$ to the compostable triple of morphisms

$(0,A'') \xrightarrow { u'_{00} } (0,A') \xrightarrow {u_{00} } (0,A) \xrightarrow { (v'vf)_{01} } (1,B'')$

and noting that the composition constraint $\mu _{ u_{00}, u'_{00} }$ is equal to the identity (by virtue of our assumption that the lax functor $U|_{ \{ 0\} \times \operatorname{\mathcal{C}}}$ is constant). $\square$