# Kerodon

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### 4.4.2 Isomorphisms and Lifting Properties

Recall that a morphism of simplicial sets $X \rightarrow S$ is a Kan fibration if and only if it is both a left fibration and a right fibration (Example 4.2.1.5). In the special case $S = \Delta ^0$, either one of these conditions is individually sufficient.

Proposition 4.4.2.1 (Joyal [MR1935979]). Let $X$ be a simplicial set. The following conditions are equivalent:

$(a)$

The projection map $X \rightarrow \Delta ^0$ is a Kan fibration.

$(b)$

The simplicial set $X$ is a Kan complex.

$(c)$

The simplicial set $X$ is an $\infty$-category and the homotopy category $\mathrm{h} \mathit{X}$ is a groupoid.

$(d)$

The simplicial set $X$ is an $\infty$-category and every morphism in $X$ is an isomorphism.

$(e)$

The projection map $X \rightarrow \Delta ^0$ is a left fibration.

$(f)$

The projection map $X \rightarrow \Delta ^0$ is a right fibration.

Corollary 4.4.2.2 (Duskin [MR1897816]). Let $\operatorname{\mathcal{C}}$ be a $2$-category. Then $\operatorname{\mathcal{C}}$ is a $2$-groupoid (in the sense of Definition 2.2.8.24) if and only if the Duskin nerve $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ is a Kan complex.

Proof. The $2$-category $\operatorname{\mathcal{C}}$ is a $2$-groupoid if and only if it is a $(2,1)$-category and the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ is a groupoid (Remark 2.2.8.25). The first condition is equivalent to the requirement that $\operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}})$ is an $\infty$-category (Theorem 2.3.2.1). If this condition is satisfied, then Corollary 2.3.4.6 supplies an isomorphism $\mathrm{h} \mathit{\operatorname{\mathcal{C}}} \simeq \mathrm{h} \mathit{ \operatorname{N}_{\bullet }^{\operatorname{D}}(\operatorname{\mathcal{C}}) }$. The desired equivalence now follows from Proposition 4.4.2.1. $\square$

Corollary 4.4.2.3. Let $q: X \rightarrow S$ be morphism of simplicial sets which is either a left or a right fibration. Then, for every vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a Kan complex.

Corollary 4.4.2.4. Suppose we are given a commutative diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ A \ar [r]^-{f} \ar [d]^{i} & X \ar [d]^{q} \\ B \ar [r]^-{g} \ar@ {-->}[ur]^{ \overline{f} } & S, }$

where $i$ is a monomorphism. Then:

• If $q$ is either a left or right fibration, then the simplicial set $\operatorname{Fun}_{A/ \, /S}(B, X)$ of Construction 3.1.3.7 is a Kan complex.

• If $q$ is a left fibration and $i$ is left anodyne, then the Kan complex $\operatorname{Fun}_{A/ \, /S}(B, X)$ is contractible.

• If $q$ is a right fibration and $i$ is right anodyne, then the Kan complex $\operatorname{Fun}_{A/ \, /S}(B, X)$ is contractible.

Proof. Without loss of generality, we may assume that $q$ is a left fibration. By virtue of Remark 3.1.3.11, the simplicial set $\operatorname{Fun}_{A/ \, /S}( B, X)$ can be identified with a fiber of the restriction map

$\theta : \operatorname{Fun}(B,X) \rightarrow \operatorname{Fun}(A,X) \times _{ \operatorname{Fun}(A,S) } \operatorname{Fun}(B,S).$

Proposition 4.2.5.1 asserts that $\theta$ is a left fibration of simplicial sets, so its fibers are Kan complexes (Corollary 4.4.2.3). If $i$ is left anodyne, then $\theta$ is a trivial Kan fibration (Proposition 4.2.5.4), so its fibers are contractible Kan complexes. $\square$

Corollary 4.4.2.5. Let $q: X \rightarrow S$ and $g: B \rightarrow S$ be morphisms of simplicial sets. If $q$ is either a left fibration or a right fibration, then the simplicial set $\operatorname{Fun}_{/S}(B,X)$ is a Kan complex.

Proof. Apply Corollary 4.4.2.4 in the special case $A = \emptyset$. $\square$

Our proof of Proposition 4.4.2.1 is based on the following characterization of isomorphisms in an $\infty$-category $\operatorname{\mathcal{C}}$:

Theorem 4.4.2.6 (Joyal). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $u: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. The following conditions are equivalent:

$(1)$

The morphism $u$ is an isomorphism.

$(2)$

Let $n \geq 2$ and let $\sigma _0: \Lambda ^{n}_{0} \rightarrow \operatorname{\mathcal{C}}$ be a morphism of simplicial sets for which the initial edge

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

is equal to $u$. Then $\sigma _0$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$.

$(3)$

Let $n \geq 2$ and let $\sigma _0: \Lambda ^{n}_{n} \rightarrow \operatorname{\mathcal{C}}$ be a morphism of simplicial sets for which the final edge

$\Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ n-1 < n\} ) \hookrightarrow \Lambda ^ n_ n \xrightarrow {\sigma _0} \operatorname{\mathcal{C}}$

is equal to $u$. Then $\sigma _0$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$.

Proof of Proposition 4.4.2.1 from Theorem 4.4.2.6. Let $X$ be a simplicial set. By definition, the projection map $X \rightarrow \Delta ^{0}$ is a left fibration if and only if, for every pair of integers $0 \leq i < n$, every morphism of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow X$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$. This condition is automatically satisfied when $n=1$ (we can identify $\sigma _0$ with a vertex $x \in X$, and take $\sigma$ to be the degenerate edge $\operatorname{id}_{x}$), and is satisfied for $0 < i < n$ if and only if $X$ is an $\infty$-category. Assuming that $X$ is an $\infty$-category, it is satisfied for $i = 0$ if and only if every morphism in $X$ is an isomorphism (by virtue of Theorem 4.4.2.6). This proves the equivalence $(d) \Leftrightarrow (e)$, and the equivalence $(d) \Leftrightarrow (f)$ follows by applying the same reasoning to the opposite simplicial set $X^{\operatorname{op}}$. In particular, $(e)$ and $(f)$ are equivalent to one another, and therefore equivalent to $(a)$ (see Example 4.2.1.5). The equivalences $(a) \Leftrightarrow (b)$ and $(c) \Leftrightarrow (d)$ are immediate from the definitions. $\square$

The proof of Theorem 4.4.2.6 will require some preliminaries.

Definition 4.4.2.7. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be $\infty$-categories. We will say that a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is conservative if it satisfies the following condition:

• Let $u: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$. If $F(u): F(X) \rightarrow F(Y)$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{D}}$, then $u$ is an isomorphism.

Example 4.4.2.8. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. Then the canonical map $\operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ is conservative.

Example 4.4.2.9. Let $\operatorname{\mathcal{D}}$ be an $\infty$-category, and let $\operatorname{\mathcal{C}}\subseteq \operatorname{\mathcal{D}}$ be a replete subcategory (Example 4.4.1.11). Then the inclusion map $\operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{D}}$ is conservative. That is, if $u: X \rightarrow Y$ is a morphism of $\operatorname{\mathcal{C}}$ which is an isomorphism in $\operatorname{\mathcal{D}}$, then $u$ is an isomorphism in $\operatorname{\mathcal{C}}$. To prove this, we observe that if $v: Y \rightarrow X$ is a homotopy inverse of $u$ in the $\infty$-category $\operatorname{\mathcal{D}}$, then the morphism $v$ also belongs to $\operatorname{\mathcal{C}}$ (by virtue of our assumption that $\operatorname{\mathcal{C}}$ is a replete subcategory of $\operatorname{\mathcal{D}}$) and is also a homotopy inverse to $u$ in $\operatorname{\mathcal{C}}$.

Remark 4.4.2.10. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ and $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be functors between $\infty$-categories, where $G$ is conservative. Then $F$ is conservative if and only if the composition $(G \circ F): \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ is conservative.

Proposition 4.4.2.11. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between $\infty$-categories. If $F$ is a left or a right fibration, then $F$ is conservative.

Proof. Without loss of generality, we may assume that $F$ is a left fibration. Let $u: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$, and suppose that $F(u)$ is an isomorphism in $\operatorname{\mathcal{D}}$. Let $\overline{v}: F(Y) \rightarrow F(X)$ is a homotopy inverse to $F(u)$, so that there exists a $2$-simplex $\overline{\sigma }$ of $\operatorname{\mathcal{D}}$ as depicted in the following diagram:

$\xymatrix@R =50pt@C=50pt{ & F(Y) \ar [dr]^{\overline{v}} & \\ F(X) \ar [ur]^{F(u)} \ar [rr]^{F( \operatorname{id}_ X) } & & F(X). }$

Invoking our assumption that $F$ is a left fibration, we can lift $\overline{\sigma }$ to a diagram

$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{v} & \\ X \ar [ur]^{u} \ar [rr]^{\operatorname{id}_{X}} & & X }$

in the $\infty$-category $\operatorname{\mathcal{C}}$. This lift supplies a morphism $v: Y \rightarrow X$ and witnesses $\operatorname{id}_{X}$ as a composition of $v$ with $u$, so that $v$ is a left homotopy inverse to $u$. Moreover, the image $F(v) = \overline{v}$ is an isomorphism in $\operatorname{\mathcal{D}}$. Repeating the preceding argument (with $u: X \rightarrow Y$ replaced by $v: Y \rightarrow X$), we deduce that there exists a morphism $w: X \rightarrow Y$ which is left homotopy inverse to $v$. It follows that $u$ and $w$ are homotopic, so that $v$ is a homotopy inverse to $u$ (Remark 1.3.6.6). In particular, $u$ is an isomorphism. $\square$

Corollary 4.4.2.12. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a conservative functor of $\infty$-categories and let $q: K \rightarrow \operatorname{\mathcal{C}}$ be a diagram in $\operatorname{\mathcal{C}}$. Then the induced functors

$F_{q/}: \operatorname{\mathcal{C}}_{q/} \rightarrow \operatorname{\mathcal{D}}_{(F \circ q)/} \quad \quad F_{/q}: \operatorname{\mathcal{C}}_{/q} \rightarrow \operatorname{\mathcal{D}}_{/(F \circ q)}$

are also conservative.

Proof. We will show that the functor $F_{/q}$ is conservative; the conservativity of $F_{q/}$ follows by a similar argument. Let $\pi : \operatorname{\mathcal{C}}_{/q} \rightarrow \operatorname{\mathcal{C}}$ and $\pi ': \operatorname{\mathcal{D}}_{/(F \circ q)} \rightarrow \operatorname{\mathcal{D}}$ denote the projection maps. Then $\pi$ and $\pi '$ are right fibrations of $\infty$-categories (Proposition 4.3.6.1), and therefore conservative (Proposition 4.4.2.11). Since $F$ is conservative, from Remark 4.4.2.10 that the functor $F \circ \pi = \pi ' \circ F_{/q}$ is also conservative. Applying Remark 4.4.2.10 again, we conclude that $F_{/q}$ is conservative. $\square$

Proposition 4.4.2.13. Let $q: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories, let $u$ be an isomorphism in $\operatorname{\mathcal{C}}$, let $n \geq 2$ be an integer, and suppose we are given a lifting problem

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{n} \ar [r]^-{ \sigma _0} \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{q} \\ \Delta ^{n} \ar@ {-->}[ur]^{\sigma } \ar [r]^-{ \overline{\sigma } } & \operatorname{\mathcal{D}}. }$

If the composite map

$\Delta ^{1} \simeq \operatorname{N}_{\bullet }( \{ n-1 < n \} ) \hookrightarrow \Lambda ^{n}_ n \xrightarrow {\sigma _0} \operatorname{\mathcal{C}}$

is equal to $u$, then there exists an $n$-simplex $\sigma : \Delta ^{n} \rightarrow \operatorname{\mathcal{C}}$ rendering the diagram commutative.

Proof. Using Lemma 4.3.6.14, we can identify the horn $\Lambda ^{n}_{n}$ with the pushout

$(\Delta ^{n-2} \star \{ 1\} ) \coprod _{ ( \operatorname{\partial \Delta }^{n-2} \star \{ 1\} ) } ( \operatorname{\partial \Delta }^{n-2} \star \Delta ^1) \subseteq \Delta ^{n-2} \star \Delta ^1 \simeq \Delta ^{n}.$

Set $f= \sigma _0|_{ \Delta ^{n-2} }$ and $f_0 = \sigma _0|_{ \operatorname{\partial \Delta }^{n-2} }$, and let $\operatorname{\mathcal{E}}$ denote the fiber product $\operatorname{\mathcal{C}}_{f_0/} \times _{\operatorname{\mathcal{D}}_{ (q \circ f_0)/}} \operatorname{\mathcal{D}}_{( q \circ f)/}$. Note that there is an evident projection map $\theta : \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$, given by the composition

$\operatorname{\mathcal{E}}\xrightarrow {\theta '} \operatorname{\mathcal{C}}_{f_0/} \xrightarrow {\theta ''} \operatorname{\mathcal{C}}.$

The morphism $\theta ''$ is a left fibration (Proposition 4.3.6.1), and the morphism $\theta '$ is a pullback of the restriction map $\operatorname{\mathcal{D}}_{(q \circ f)/} \rightarrow \operatorname{\mathcal{D}}_{(q \circ f_0)/}$ and is therefore also a left fibration (Corollary 4.3.6.13). It follows that $\theta : \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a left fibration (Remark 4.2.1.11), and in particular $\operatorname{\mathcal{E}}$ is an $\infty$-category (Remark 4.1.1.9).

Note that the restriction of $\sigma _0$ to $\Delta ^{n-2} \star \{ 1\}$ can be identified with an object $Y$ of the coslice $\infty$-category $\operatorname{\mathcal{C}}_{f/}$. Let

$\rho : \operatorname{\mathcal{C}}_{f/} \rightarrow \operatorname{\mathcal{C}}_{f_0/} \times _{ \operatorname{\mathcal{D}}_{ ( q \circ f_0)/ }} \operatorname{\mathcal{D}}_{(q \circ f)/} = \operatorname{\mathcal{E}}$

be the left fibration of Proposition 4.3.6.8, and set $\overline{Y} = \rho (Y) \in \operatorname{\mathcal{E}}$. Then the restriction $\sigma _0|_{ \operatorname{\partial \Delta }^{n-2} \star \Delta ^1}$ and $\overline{\sigma }$ determine a morphism $\overline{v}: \overline{X} \rightarrow \overline{Y}$ in the $\infty$-category $\operatorname{\mathcal{E}}$. Unwinding the definitions, we see that choosing an $n$-simplex $\sigma : \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$ satisfying the requirements of Proposition 4.4.2.13 is equivalent to choosing a morphism $v: X \rightarrow Y$ in $\operatorname{\mathcal{C}}_{f/}$ satisfying $\rho (v) = \overline{v}$. Since $\rho$ is a left fibration, it is an isofibration (Example 4.4.1.10). Consequently, to prove the existence of $v$, it will suffice to show that $\overline{v}$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{E}}$. Since $\theta$ is a left fibration, this follows from our assumption that $u = \theta ( \overline{v} )$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}$ (Proposition 4.4.2.11). $\square$

Proof of Theorem 4.4.2.6. The implication $(1) \Rightarrow (3)$ is a special case of Proposition 4.4.2.13. We will complete the proof by showing that $(3) \Rightarrow (1)$ (a similar argument shows that $(1)$ and $(2)$ are equivalent). Let $u: X \rightarrow Y$ be a morphism in an $\infty$-category $\operatorname{\mathcal{C}}$, and consider the map $\sigma _0: \Lambda ^{2}_{2} \rightarrow \operatorname{\mathcal{C}}$ depicted in the diagram

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{u} & \\ Y \ar@ {-->}[ur]^{v} \ar [rr]^{ \operatorname{id}_ Y} & & Y. }$

If $u$ satisfies condition $(3)$, then we can complete $\sigma _0$ to a $2$-simplex of $\operatorname{\mathcal{C}}$, which witnesses the morphism $v = d_2(\sigma )$ as a right homotopy inverse of $u$. The tuple $(\sigma , s_0(u), s_1(u), \bullet )$ then determines a morphism of simplicial sets $\tau _0: \Lambda ^3_3 \rightarrow \operatorname{\mathcal{C}}$ (see Exercise 1.1.2.14). Invoking assumption $(3)$ again, we can extend $\tau _0$ to a $3$-simplex $\tau$ of $\operatorname{\mathcal{C}}$. The face $d_3(\tau )$ then witnesses that $v$ is also a left homotopy inverse to $u$, so that $u$ is an isomorphism as desired. $\square$

We close this section by recording another useful consequence of Proposition 4.4.2.13:

Proposition 4.4.2.14. Let $q: X \rightarrow S$ be a morphism of simplicial sets. The following conditions are equivalent:

$(1)$

The morphism $q$ is a trivial Kan fibration.

$(2)$

The morphism $q$ is a left fibration and, for every vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a contractible Kan complex.

$(3)$

The morphism $q$ is a right fibration and, for every vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a contractible Kan complex.

We will deduce Proposition 4.4.2.14 from the following more precise assertion:

Lemma 4.4.2.15. Let $q: X \rightarrow S$ be a left fibration of simplicial sets, let $s \in S$ be a vertex having the property that the Kan complex $X_{s} = \{ s\} \times _{S} X$ is contractible, and let $\overline{\sigma }: \Delta ^ n \rightarrow S$ be an $n$-simplex of $S$ satisfying $\overline{\sigma }(n) = s$. Then every lifting problem

$\xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{n} \ar [r]^-{ \sigma _0 } \ar [d] & X \ar [d]^{q} \\ \Delta ^{n} \ar [r]^-{ \overline{\sigma } } \ar@ {-->}[ur]^-{\sigma } & S }$

Proof. When $n=0$, the desired result follows from the fact that the fiber $X_{s}$ is nonempty. We may therefore assume without loss of generality that $n > 0$. Replacing $q$ by the projection map $\Delta ^{n} \times _{S} X \rightarrow \Delta ^ n$, we may further reduce to the special case where $S = \Delta ^ n$ and $\overline{\sigma }$ is the identity map. In this case, our assumption that $q$ is a left fibration guarantees that $X$ is an $\infty$-category (Remark 4.1.1.9).

Let $\overline{h}: \Delta ^{1} \times \Delta ^{n} \rightarrow \Delta ^ n$ be the morphism given on vertices by $h(i,j) = \begin{cases} j & \textnormal{ if } i=0 \\ n & \textnormal{ if } i=1. \end{cases}$ Since the inclusion $\{ 0\} \times \operatorname{\partial \Delta }^ n \hookrightarrow \Delta ^1 \times \operatorname{\partial \Delta }^{n}$ is left anodyne (Proposition 4.2.5.3), our assumption that $q$ is a left fibration guarantees the existence of a morphism $h': \Delta ^1 \times \operatorname{\partial \Delta }^ n \rightarrow X$ satisfying $h'|_{ \{ 0\} \times \operatorname{\partial \Delta }^ n} = \sigma _0$ and $q \circ h' = \overline{h}|_{ \Delta ^1 \times \operatorname{\partial \Delta }^ n}$. We will complete the proof by showing that $h'$ can be extended to a map $h: \Delta ^1 \times \Delta ^ n \rightarrow X$ satisfying $q \circ h = \overline{h}$ (in this case, our original lifting problem admits the solution $\sigma = h|_{ \{ 0\} \times \Delta ^ n}$).

Let $Y(0) \subset Y(1) \subset Y(2) \subset \cdots \subset Y(n+1) = \Delta ^1 \times \Delta ^ n$ denote the filtration constructed in the proof of Lemma 3.1.2.11. Then $Y(0)$ can be described as the pushout

$(\Delta ^1 \times \operatorname{\partial \Delta }^ n) \coprod _{( \{ 1\} \times \operatorname{\partial \Delta }^ n) } ( \{ 1\} \times \Delta ^ n ).$

Using our assumption that the fiber $X_{s}$ is a contractible Kan complex, we see that $h'$ can be extended to a morphism of simplicial sets $h_0: Y(0) \rightarrow X$ satisfying $q \circ h_0 = \overline{h}|_{Y(0)}$. We claim that $h_0$ can be extended to a compatible sequence of maps $h_{i}: Y(i) \rightarrow X$ satisfying $q \circ h_ i = \overline{h}|_{Y(i)}$. To prove this, we recall that each $Y(i+1)$ can be realized as a pushout of the horn inclusion $\Lambda ^{n+1}_{i+1} \hookrightarrow \Delta ^{n+1}$, so that the construction of $h_{i+1}$ from $h_{i}$ can be rephrased as a lifting problem

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{n+1}_{i+1} \ar [r]^-{f_ i} \ar [d] & X \ar [d]^{q} \\ \Delta ^{n+1} \ar [r] \ar@ {-->}[ur] & S. }$

For $0 \leq i < n$, this lifting problem is automatically solvable by virtue of our assumption that $q$ is a left fibration. In the case $i = n$, the edge

$\Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ n, n+1\} ) \hookrightarrow \Lambda ^{n+1}_{n+1} \xrightarrow {f_ i} X$

is an edge of the Kan complex $X_{s}$, and is therefore an isomorphism in the $\infty$-category $X$ (Proposition 1.3.6.10). In this case, the existence of the desired extension follows from Theorem 4.4.2.6. We complete the proof by taking $h = h_{n+1}$. $\square$

Proof of Proposition 4.4.2.14. The implication $(1) \Rightarrow (2)$ is immediate, and the converse follows from Lemma 4.4.2.15. The equivalence $(1) \Leftrightarrow (3)$ follows by a similar argument. $\square$