Proposition 2.3.1.9 (00A2)

Proposition 2.3.1.9. Let $\operatorname{\mathcal{C}}$ be a $2$-category and let $n$ be a nonnegative integer. Suppose we are given the following data:

$(0)$: A collection of objects $\{ X_ i \} _{ 0 \leq i \leq n}$ of the $2$-category $\operatorname{\mathcal{C}}$.
$(1')$: A collection of $1$-morphisms $\{ f_{j,i}: X_ i \rightarrow X_ j \} _{0 \leq i < j \leq n }$ in the $2$-category $\operatorname{\mathcal{C}}$
$(2')$: A collection of $2$-morphisms $\{ \mu _{k,j,i}: f_{k,j} \circ f_{j,i} \Rightarrow f_{k,i} \} _{0 \leq i < j < k \leq n}$ in the $2$-category $\operatorname{\mathcal{C}}$.

This data can be extended uniquely to an $n$-simplex of the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ (as described in Remark 2.3.1.8) if and only if the following condition is satisfied:

$(c')$

For $0 \leq i < j < k < \ell \leq n$, we have a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ f_{\ell , k} \circ (f_{k,j} \circ f_{j,i} ) \ar@ {=>}[rr]^-{\alpha _{f_{\ell ,k}, f_{k,j}, f_{j,i} } } \ar@ {=>}[d]_{ \operatorname{id}_{ f_{\ell ,k}} \circ \mu _{k,j,i} } & & ( f_{\ell ,k} \circ f_{k,j} ) \circ f_{j,i} \ar@ {=>}[d]^{ \mu _{\ell ,k,j} \circ \operatorname{id}_{ f_{j,i} }} \\ f_{\ell , k} \circ f_{k,i} \ar@ {=>}[dr]_{ \mu _{\ell ,k,i} } & & f_{\ell , j} \circ f_{j,i} \ar@ {=>}[dl]^{ \mu _{\ell , j, i} } \\ & f_{\ell , i} & } \]

in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( X_ i, X_{\ell } )$.

Proof. We wish to show that there is a unique way to choose $1$-morphisms $f_{j,i}: X_ i \rightarrow X_ j$ for $i = j$ and $2$-morphisms $\mu _{k,j,i}: f_{k,j} \circ f_{j,i} \Rightarrow f_{k,i}$ for $i = j \leq k$ and $i \leq j = k$ so that conditions $(a)$, $(b)$, and $(c)$ of Remark 2.3.1.8 are satisfied. The uniqueness is clear: to satisfy condition $(a)$, we must have $f_{i,i} = \operatorname{id}_{X_ i}$ for $0 \leq i \leq n$, and to satisfy condition $(b)$ we must have $\mu _{k,j,i} = \rho _{ f_{j,i} }$ when $i = j$ and $\mu _{k,j,i} = \lambda _{ f_{k,j} }$ when $j = k$. To complete the proof, it will suffice to verify the following:

$(I)$

The prescription above is consistent. That is, when $i = j = k$, we have $\rho _{f_{j,i}} = \lambda _{ f_{k,j} }$ (as morphisms of the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( X_ i, X_ k)$).

$(II)$

The prescription above satisfies condition $(c)$ of Remark 2.3.1.8. That is, the diagram

\[ \xymatrix@R =50pt@C=50pt{ f_{\ell , k} \circ (f_{k,j} \circ f_{j,i} ) \ar@ {=>}[rr]^-{\alpha _{f_{\ell ,k}, f_{k,j}, f_{j,i} }} \ar@ {=>}[d]_{ \operatorname{id}_{ f_{\ell ,k} } \circ \mu _{k,j,i} } & & ( f_{\ell ,k} \circ f_{k,j} ) \circ f_{j,i} \ar@ {=>}[d]^{ \mu _{\ell ,k,j} \circ \operatorname{id}_{ f_{j,i} } } \\ f_{\ell , k} \circ f_{k,i} \ar@ {=>}[dr]_{ \mu _{\ell ,k,i} } & & f_{\ell , j} \circ f_{j,i} \ar@ {=>}[dl]^{ \mu _{\ell , j, i} } \\ & f_{\ell , i} & } \]

commutes in the special cases $0 \leq i = j \leq k \leq \ell \leq n$, $0 \leq i \leq j = k \leq \ell \leq n$, and $0 \leq i \leq j \leq k = \ell \leq n$.

Assertion $(I)$ follows from Corollary 2.2.1.15. Assertion $(II)$ follows from the triangle identity in $\operatorname{\mathcal{C}}$ in the case $j = k$, and from Proposition 2.2.1.16 in the cases $i = j$ and $k = \ell $. $\square$