# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Remark 8.1.3.5. In the situation of Example 8.1.3.4, let us identify $\operatorname{Tw}( \Delta ^{n-1} )$ with the nerve of the partially ordered subset $Q' \subset Q$ consisting of pairs $(i,j)$ satisfying $i \leq j < n$. Suppose we are given an object of the $\infty$-category $\operatorname{Fun}_{ / \Delta ^ n}( \operatorname{Tw}(\Delta ^ n), \operatorname{\mathcal{E}})$, corresponding to a functor $F: \operatorname{N}_{\bullet }(Q) \rightarrow \operatorname{\mathcal{E}}$ having the property that $F' = F|_{ \operatorname{N}_{\bullet }(Q') }$ satisfies conditions $(1)$ and $(2)$ of Theorem 8.1.3.1, when regarded as an object of the $\infty$-category $\operatorname{Fun}_{ / \Delta ^{n-1} }( \operatorname{Tw}( \Delta ^{n-1} ), \Delta ^{n-1} \times _{ \Delta ^{n} } \operatorname{\mathcal{E}})$. Then the analogous conditions for $F$ can be restated as follows:

$(1)$

The image $F(n,n)$ is an initial object of the $\infty$-category $\operatorname{\mathcal{E}}_{n}$. Equivalently, $F(n,n)$ is a $U$-initial object of the $\infty$-category $\operatorname{\mathcal{E}}$. Since the set $\{ q \in Q': q < (n,n) \}$ is empty, this is equivalent to the requirement that $F$ is $U$-left Kan extended from $\operatorname{N}_{\bullet }(Q')$ at the element $(n,n) \in Q$.

$(2)$

For $0 \leq i \leq n-1$, the functor $F$ determines a $U$-cocartesian morphism $F(i,n-1) \rightarrow F(i,n)$ in the $\infty$-category $\operatorname{\mathcal{E}}$. Since the partially ordered set $\{ q \in Q': q < (i,n) \}$ has a largest element $(i,n-1)$, this is equivalent to the requirement that $F$ is $U$-left Kan extended from $\operatorname{N}_{\bullet }(Q')$ at the element $(i,n) \in Q$ (Corollary 7.2.2.5).

In particular, $F$ satisfies both of these conditions if and only if it is $U$-left Kan extended from $\operatorname{N}_{\bullet }(Q')$.