# Kerodon

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

### 4.4.4 Natural Isomorphisms

Recall that, if $X$ is an arbitrary simplicial set and $\operatorname{\mathcal{C}}$ is an $\infty$-category, then the simplicial set $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$ is also an $\infty$-category (Theorem 1.4.3.7). In this section, we study isomorphisms in $\infty$-categories of the form $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$.

Definition 4.4.4.1. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $X$ be a simplicial set, and suppose we are given a pair of diagrams $f,f': X \rightarrow \operatorname{\mathcal{C}}$. A natural transformation from $f$ to $f'$ is a morphism $u: f \rightarrow f'$ in the $\infty$-category $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$. A natural isomorphism from $f$ to $f'$ is a natural transformation $u: f \rightarrow f'$ which is an isomorphism in the $\infty$-category $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$ (Definition 1.3.6.1). We say that $f$ and $f'$ are naturally isomorphic if there exists a natural isomorphism from $f$ to $f'$.

Remark 4.4.4.2. In the situation of Definition 4.4.4.1, a natural transformation from $f$ to $f'$ is simply a homotopy from $f$ to $f'$, in the sense of Definition 3.1.4.2: that is, a map of simplicial sets $h: \Delta ^1 \times X \rightarrow \operatorname{\mathcal{C}}$ satisfying $h|_{ \{ 0\} \times X} = f$ and $h|_{ \{ 1\} \times X } = f'$. However, the terminology of Definition 4.4.4.1 is intended to signal a shift in emphasis. We will generally reserve use of the term homotopy between diagrams $f,f': X \rightarrow \operatorname{\mathcal{C}}$ for the case where $\operatorname{\mathcal{C}}$ is a Kan complex, and use the term natural transformation when $\operatorname{\mathcal{C}}$ is a more general $\infty$-category.

Example 4.4.4.3. Let $\operatorname{\mathcal{C}}$ be an ordinary category, and suppose we are given a pair of diagrams $f,f': X \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. Then a natural transformation from $f$ to $f'$ can be identified with a collection of morphisms $\{ u_{x}: f(x) \rightarrow f'(x) \} _{x \in X}$ with the following property: for every edge $e: x \rightarrow y$ of the simplicial set $X$, the diagram

$\xymatrix@R =50pt@C=50pt{ f(x) \ar [r]^-{ u_ x } \ar [d]^{ f(e) } & f'(x) \ar [d]^{ f'(e) } \\ f(y) \ar [r]^-{u_ y} & f'(y) }$

commutes (in the category $\operatorname{\mathcal{C}}$).

In particular, if $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are ordinary categories and we are given a pair of functors $f,f': \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$, then giving a natural transformation from $f$ to $f'$ (in the sense of classical category theory) is equivalent to giving a natural transformation from $\operatorname{N}_{\bullet }(f): \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ to $\operatorname{N}_{\bullet }(f'): \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$.

Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $X$ be an arbitrary simplicial set. For every vertex $x \in X$, evaluation at $x$ determines a functor

$\operatorname{ev}_{x}: \operatorname{Fun}(X, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \{ x\} , \operatorname{\mathcal{C}}) \simeq \operatorname{\mathcal{C}}.$

In particular, if $u: f \rightarrow f'$ is an isomorphism in the $\infty$-category $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$, then $\operatorname{ev}_{x}(u): f(x) \rightarrow f'(x)$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}$. Our goal in this section is to prove the converse:

Theorem 4.4.4.4. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $f,f': X \rightarrow \operatorname{\mathcal{C}}$ be diagrams in $\operatorname{\mathcal{C}}$ indexed by a simplicial set $X$, and let $u: f \rightarrow f'$ be a natural transformation. Then $u$ is a natural isomorphism if and only if, for every vertex $x \in X$, the induced map $\operatorname{ev}_{x}(u): f(x) \rightarrow f'(x)$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}$.

Remark 4.4.4.5. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be $\infty$-categories and suppose we are given a pair of functors $F,G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$, which restrict to functors between their cores $F^{\simeq }, G^{\simeq }: \operatorname{\mathcal{C}}^{\simeq } \rightarrow \operatorname{\mathcal{D}}^{\simeq }$ (see Remark 4.4.3.4). Let $u$ be a natural transformation from $F$ to $G$, which we identify with a map of simplicial sets $u: \Delta ^1 \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$. If $u$ is a natural isomorphism, then it restricts to a map of simplicial sets $u_0: \Delta ^1 \times \operatorname{\mathcal{C}}^{\simeq } \rightarrow \operatorname{\mathcal{D}}^{\simeq }$, which we can regard as a homotopy from $F^{\simeq }$ to $G^{\simeq }$. In particular, if the functors $F$ and $G$ are naturally isomorphic, then the morphisms $F^{\simeq }$ and $G^{\simeq }$ are homotopic.

Corollary 4.4.4.6. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. Then the functor

$(\operatorname{Set_{\Delta }})^{\operatorname{op}} \rightarrow \operatorname{Set_{\Delta }}\quad \quad X \mapsto \operatorname{Fun}(X, \operatorname{\mathcal{C}})^{\simeq }$

preserves limits (that is, it carries colimits in the category of simplicial sets to limits of Kan complexes).

The proof of Theorem 4.4.4.4 will use the following combinatorial assertion:

Lemma 4.4.4.7. Let $m \geq 0$ and $n \geq 2$ be integers. Then there exists a sequence of simplicial subsets

$X(0) \subset X(1) \subset X(2) \subset \cdots \subset X(t) = \Delta ^{m} \times \Delta ^{n}$

with the following properties:

$(1)$

The simplicial subset $X(0) \subseteq \Delta ^{m} \times \Delta ^{n}$ is the union of $\Delta ^{m} \times \Lambda ^{n}_{0}$ and $\operatorname{\partial \Delta }^{m} \times \Delta ^{n}$.

$(2)$

For each $0 < s \leq t$, there exist integers $q \geq 2$ and $p < q$ and a pushout diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{q}_{p} \ar [r] \ar [d] & X(s-1) \ar [d] \\ \Delta ^{q} \ar [r]^-{\sigma } & X(s). }$

Moreover, if $p=0$, then the map $\sigma : \Delta ^{q} \rightarrow X(s) \subseteq \Delta ^{m} \times \Delta ^{n}$ satisfies $\sigma (0) = (0,0)$ and $\sigma (1) = (0,1)$.

Proof. Let $\sigma$ be a nondegenerate $q$-simplex of the product $\Delta ^{m} \times \Delta ^{n}$, given by a chain

$(i_0, j_0) < (i_1, j_1) < \cdots < (i_ q, j_ q).$

We will say that $\sigma$ is free if the composite maps

$\Delta ^{q} \xrightarrow {\sigma } \Delta ^{m} \times \Delta ^{n} \twoheadrightarrow \Delta ^{m} \quad \quad \Delta ^{q} \xrightarrow {\sigma } \Delta ^{m} \times \Delta ^{n} \twoheadrightarrow \Delta ^{n}$

are surjective and there exists an integer $0 \leq p < q$ such that $(i_ p, j_ p) = (p,0)$ and $(i_{p+1}, j_{p+1} ) = (p, 1)$. If this condition is satisfied, then the integer $p$ is uniquely determined; we will refer to $p$ as the index of $\sigma$ and denote it by $p(\sigma )$. We also denote the dimension $q$ of $\sigma$ by $q(\sigma )$.

Let $\{ \sigma _1, \sigma _{2}, \cdots , \sigma _{t} \}$ be an enumeration of the collection of all free simplices of the product $\Delta ^{m} \times \Delta ^{n}$. Without loss of generality, we may assume that that this enumeration satisfies the following pair of conditions:

• For $1 \leq s \leq s' \leq t$, we have $q( \sigma _ s ) \leq q( \sigma _{s'} )$.

• If $1 \leq s \leq s' \leq t$ are integers satisfying $q( \sigma _ s ) \leq q( \sigma _{s'} )$, then $p( \sigma _ s ) \geq p( \sigma _{s'} )$.

Let $X(0)$ denote the union $(\Delta ^{m} \times \Lambda ^{n}_{0}) \cup (\operatorname{\partial \Delta }^{m} \times \Delta ^{n}) \subseteq \Delta ^{m} \times \Delta ^{n}$. For $0 < s \leq t$, we let $X(s)$ denote the smallest simplicial subset of $\Delta ^{m} \times \Delta ^{n}$ which contains $X(0)$ together with the simplices $\{ \sigma _1, \sigma _2, \ldots , \sigma _{s} \}$. We will show that the sequence

$X(0) \subset X(1) \subset \cdots \subset X(t)$

satisfies the requirements of Lemma 4.4.4.7.

We first claim that $X(t) = \Delta ^{m} \times \Delta ^{n}$. Let $\sigma$ be an arbitrary nondegenerate $q$-simplex of $\Delta ^{m} \times \Delta ^{n}$, which we will identify with a sequence

$(i_0, j_0) < (i_1, j_1) < \cdots < (i_ q, j_ q)$

of elements of the partially ordered set $[m] \times [n]$. We wish to show that $\sigma$ is contained in $X(t)$. Without loss of generality, we may assume that the sequence $(i_0, i_1, \ldots , i_ q)$ contains every element of the set $[m] = \{ 0 < 1 < \cdots < m \}$. (otherwise, $\sigma$ is contained in the simplicial subset $\operatorname{\partial \Delta }^ m \times \Delta ^ n \subseteq X(0) \subseteq X(t)$). Similarly, we may assume that that the sequence $(j_0, j_1, \ldots , j_ q)$ contains every element of the set $\{ 1 < 2 < \cdots < n \}$ (otherwise, $\sigma$ is is contained in the simplicial subset $\Delta ^ m \times \Lambda ^{n}_{0} \subseteq X(0) \subseteq X(t)$). In particular, the sequence $\sigma$ contains $(p,1)$, for some integer $0 \leq p \leq n$. Let us assume that $p$ is chosen as small as possible. In this case, there are two possibilities:

• The sequence $\sigma$ also contains the pair $(p,0)$. In this case, $\sigma$ is a free simplex of $\Delta ^{m} \times \Delta ^{n}$, and therefore belongs to $X(t)$.

• The sequence $\sigma$ does not contain $(p,0)$, and therefore has the form

$(0,0) < (1,0) < \cdots < (p-1, 0) < (p,1) < (i_{p+1}, j_{p+1} ) < \cdots < (i_ q, j_ q).$

We can then identify $\sigma$ with the $p$th face of the $(q+1)$-simplex $\sigma '$ given by the sequence

$(0,0) < (1,0) < \cdots < (p-1, 0) < (p,0) < (p,1) < (i_{p+1}, j_{p+1} ) < \cdots < (i_ q, j_ q).$

The simplex $\sigma '$ is free and therefore belongs to $X(t)$, so that $\sigma$ belongs to $X(t)$ as well.

We now complete the proof by verifying requirement $(2)$ of Lemma 4.4.4.7. Fix an integer $0 < s \leq t$ and let $\sigma = \sigma _{s}$ be the corresponding free simplex of $\Delta ^{m} \times \Delta ^{n}$. Let $q = q(\sigma )$ be the dimension of $\sigma$ and let $p = p(\sigma )$ be the index of $\sigma$, so that $0 \leq p < q$ and $\sigma$ has the form

$(0,0) < (1,0) < \cdots < (p,0) < (p,1) < (i_{p+2}, j_{p+2} ) < \cdots < (i_ q, j_ q).$

By construction, the simplicial subset $X(s) \subseteq \Delta ^{m} \times \Delta ^{n}$ is the union of $X(s-1)$ with the image of $\sigma$. We therefore have a pushout diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ K \ar [r] \ar [d] & X(s-1) \ar [d] \\ \Delta ^{q} \ar [r]^-{\sigma } & X(s), }$

where $K \subseteq \Delta ^{q}$ denotes the inverse image of $X(s-1)$. We will complete the proof by showing that $K$ is equal to the horn $\Lambda ^{q}_{p} \subseteq \Delta ^{q}$.

We first show that the horn $\Lambda ^{q}_{p}$ is contained in $K$. For this, it will suffice to show that for every integer $0 \leq p' \leq q$ satisfying $p' \neq p$, the face $\tau = d_{p'}( \sigma )$ is contained in $X(s-1)$. We consider three cases:

• For $p' < p$, the simplex $\tau$ is given by the sequence

$(0,0) < \cdots < (p'-1,0) < (p'+1,0) < \cdots < (p,0) < (p,1) < \cdots < (i_ q, j_ q),$

which is contained in the simplicial subset $\operatorname{\partial \Delta }^{m} \times \Delta ^{n} \subseteq X(0) \subseteq X(s-1)$.

• For $p' = p+1$, the simplex $\tau$ is given by the sequence

$(0,0) < (1,0) < \cdots < (p,0) < (i_{p+2}, j_{p+2} ) < \cdots < (i_ q, j_ q).$

If $j_{p+2} \geq 2$, then $\tau$ belongs to the simplicial subset $\Delta ^ m \times \Lambda ^{n}_{0} \subseteq X(0) \subseteq X(s-1)$. Otherwise, we must have $(i_{p+2}, j_{p+2} ) = (p+1, 1)$, so that $\tau$ occurs as a face of the free simplex $\sigma '$ given by the sequence

$(0,0) < (1,0) < \cdots < (p,0) < (p+1,0) < (p+1, 1) < \cdots < (i_ q, j_ q),$

which has dimension $q$ and index $p+1$. By construction, $\sigma '$ belongs to the set $\{ \sigma _1, \sigma _2, \ldots , \sigma _{s-1} \}$, and is therefore contained in the simplicial subset $X(s-1) \subseteq \Delta ^{m} \times \Delta ^ n$.

• For $p' > p+1$, the simplex $\tau$ is given by the sequence

$(0,0) < \cdots < (p,0) < (p,1) < \cdots < (i_{p'-1}, j_{p'-1} ) < (i_{p'+1}, j_{p'+1} ) < \cdots < (i_ q, j_ q).$

It follows that $\tau$ is either contained in the simplicial subset $X(0) = (\Delta ^{m} \times \Lambda ^{n}_{0}) \cup (\operatorname{\partial \Delta }^{m} \times \Delta ^{n})$ or that it is a free simplex of $\Delta ^{m} \times \Delta ^{n}$ having dimension $q-1$. In the latter case, $\tau$ must belong to the set $\{ \sigma _1, \ldots , \sigma _{s-1} \}$, and is therefore contained in the simplicial subset $X(s-1) \subseteq \Delta ^{m} \times \Delta ^{n}$.

To show that the inclusion $\Lambda ^{q}_{p} \subseteq K$ is an equality, it will suffice to show that $K$ does not contain the $p$th face of $\Delta ^{q}$. Let $\tau = d_{p}(\sigma )$ be the $p$th face of $\sigma$, given by the sequence

$(0,0) < (1,0) < \cdots < (p-1, 0) < (p,1) < (i_{p+1}, j_{p+1} ) < \cdots < (i_ q, j_ q).$

We wish to show that $\tau$ is not contained in $X(s-1)$. Assume otherwise. Since $\tau$ is not contained in $X(0)$, we conclude that $\tau$ is contained in some free simplex $\sigma ' \in \{ \sigma _1, \sigma _2, \ldots , \sigma _{s-1} \}$. Note that $\tau \neq \sigma '$ (since $\tau$ is not free), so we have inequalities

$q-1 = q(\tau ) < q(\sigma ') \leq q(\sigma ) = q.$

It follows that $\sigma '$ is a free $q$-simplex of $\Delta ^{m} \times \Delta ^{n}$ which contains $\tau$ and is not equal to $\sigma$, and is therefore necessarily given by the sequence

$(0,0) < (1,0) < \cdots < (p-1, 0) < (p-1,1) < (p,1) < (i_{p+1}, j_{p+1} ) < \cdots < (i_ q, j_ q).$

We therefore have $p(\sigma ') = p-1 < p = p(\sigma )$, which contradicts our assumption regarding the choice of enumeration $\{ \sigma _1, \sigma _2, \ldots , \sigma _ t \}$. $\square$

Lemma 4.4.4.8. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $A \subseteq B$ be simplicial sets, let $n \geq 2$ be an integer, and suppose we are given a morphism of simplicial sets

$F_0: (A \times \Delta ^ n) \coprod _{ (A \times \Lambda ^{n}_{0} ) } (B \times \Lambda ^{n}_0) \rightarrow \operatorname{\mathcal{C}}.$

Assume that, for every vertex $x \in B$, the composite map

$\Delta ^1 \simeq \{ x\} \times \operatorname{N}_{\bullet }( \{ 0,1\} ) \hookrightarrow B \times \Lambda ^{n_0} \xrightarrow {F_0} \operatorname{\mathcal{C}}$

corresponds to an isomorphism in $\operatorname{\mathcal{C}}$. Then $F_0$ can be extended to a morphism of simplicial sets $F: B \times \Delta ^{n} \rightarrow \operatorname{\mathcal{C}}$.

Proof. Let $P$ denote the collection of all pairs $(K, F_ K)$, where $K \subseteq B$ is a simplicial subset containing $A$ and $F_ K: K \times \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$ is a morphism of simplicial sets satisfying

$F_ K |_{A \times \Delta ^ n} = F_0 |_{ A \times \Delta ^ n } \quad \quad F_{K} |_{ K \times \Lambda ^ n_0} = F_0 |_{ K \times \Lambda ^ n_0}.$

We regard $P$ as partially ordered set, where $(K, F_ K) \leq (K', F_{K'} )$ if $K \subseteq K'$ and $F_ K = F_{K'} |_{K' \times \Delta ^ n}$. The partially ordered set $P$ satisfies the hypotheses of Zorn's lemma, and therefore has a maximal element $(K_{\mathrm{max}}, F_{K_{\mathrm{max}}})$. We will complete the proof by showing that $K_{\mathrm{max}} = B$. Assume otherwise. Then there exists some nondegenerate $m$-simplex $\tau : \Delta ^{m} \rightarrow B$ whose image is not contained in $K_{ \mathrm{max} }$. Choosing $m$ as small as possible, we can assume that $\tau$ carries the boundary $\operatorname{\partial \Delta }^{m}$ into $B$. Let $K' \subseteq B$ be the union of $K_{\mathrm{max}}$ with the image of $\tau$, so that we have a pushout diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [d] \ar [r] & K_{\mathrm{max}} \ar [d] \\ \Delta ^{m} \ar [r] & K'. }$

We will complete the proof by showing that the lifting problem

$\xymatrix@R =50pt@C=50pt{ (K_{\mathrm{max}} \times \Delta ^ n) \coprod _{ (K_{\mathrm{max}} \times \Lambda ^ n_0 )} (K' \times \Lambda ^ n_0) \ar [r]^-{(F_ K, F_0|_{K' \times \Lambda ^{n_0})}} \ar [d] & \operatorname{\mathcal{C}}\ar [d] \\ K' \times \Delta ^ n \ar [r] \ar@ {-->}[ur] & \Delta ^0 }$

admit a solution (contradicting the maximality of the pair $(K_{\mathrm{max}}, F_{K_{\mathrm{max}}} )$). To prove this, we can replace the inclusion $K_{\mathrm{max}} \hookrightarrow K'$ by $\operatorname{\partial \Delta }^ m \hookrightarrow \Delta ^ m$. We are therefore reduced to proving Lemma 4.4.4.8 in the special case where $B = \Delta ^ m$ is a simplex and $A = \operatorname{\partial \Delta }^ m$ is its boundary.

Choose a sequence of simplicial subsets

$X(0) \subset X(1) \subset X(2) \subset \cdots \subset X(t) = \Delta ^{m} \times \Delta ^{n}$

satisfying the requirements of Lemma 4.4.4.7, so that $F_0$ can be identified with a morphism $X(0) \rightarrow \operatorname{\mathcal{C}}$. We will show that, for $0 \leq s \leq t$, there exists a morphism of simplicial sets $F_{s}: X(s) \rightarrow \operatorname{\mathcal{C}}$ extending $F_0$ (taking $s = t$, this will complete the proof of Lemma 4.4.4.8). We proceed by induction on $s$, the case $s=0$ being vacuous. Assume that $s > 0$ and that we have already constructed a morphism $F_{s-1}: X(s-1) \rightarrow \operatorname{\mathcal{C}}$ satisfying $F_{s-1}|_{X(0)} = F_0$. By construction, there exists integers $q \geq 2$, $p > 0$, and a pushout diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{q}_{p} \ar [r]^-{\sigma _0} \ar [d] & X(s-1) \ar [d] \\ \Delta ^{q} \ar [r]^-{\sigma } & X(s). }$

Moreover, in the special case $p=0$, we can assume that $\sigma (0) = (0,0)$ and $\sigma (1) = (0,1)$, so that the composite map

$\Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ 0 < 1 \} ) \hookrightarrow \Lambda ^{q}_{p} \xrightarrow {\sigma _0} X(s-1) \xrightarrow { F_{s-1} } \operatorname{\mathcal{C}}$

corresponds to an isomorphism in $\operatorname{\mathcal{C}}$. To construct the desired extension $F_{s}: X(s) \rightarrow \operatorname{\mathcal{C}}$, it will suffice to show that the composite map

$\Lambda ^{q}_{p} \xrightarrow {\sigma _0} X(s-1) \xrightarrow { F_{s-1} } \operatorname{\mathcal{C}}$

can be extended to a $q$-simplex of $\operatorname{\mathcal{C}}$. In the case $0 < p < q$, the existence of the desired extension follows from our assumption that $\operatorname{\mathcal{C}}$ is an $\infty$-category. In the special case $p=0$, it follows instead from the characterization of isomorphisms given by Theorem 4.4.2.5. $\square$

Proof of Theorem 4.4.4.4. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $X$ be a simplicial set, and let $u: f \rightarrow f'$ be a natural transformation between diagrams $f,f': X \rightarrow \operatorname{\mathcal{C}}$ having the property that, for every vertex $x \in X$, the induced map $u_ x: f(x) \rightarrow f'(x)$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}$. We wish to show that $u$ is an isomorphism in the functor $\infty$-category $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$. We will prove this by verifying the criterion of Theorem 4.4.2.5. Let $n \geq 2$ be an integer, and suppose we are given a morphism of simplicial sets $\sigma _0: \Lambda ^{n}_{0} \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{C}})$ for which the composite map

$\Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ 0,1\} ) \hookrightarrow \Lambda ^ n_0 \xrightarrow {\sigma _0} \operatorname{Fun}(X, \operatorname{\mathcal{C}})$

is equal to $u$. We wish to show that $\sigma _0$ can be extended to an $n$-simplex $\sigma$ of the diagram $\infty$-category $\operatorname{Fun}(X, \operatorname{\mathcal{C}})$. Let us identify $\sigma _0$ with a morphism of simplicial sets $\rho _0: X \times \Lambda ^{n}_{0} \rightarrow \operatorname{\mathcal{C}}$; we wish to show that $\rho _0$ can be extended to a morphism $\rho : X \times \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$. This follows by applying Lemma 4.4.4.8 in the special case $B = X$ and $A = \emptyset$. $\square$