Proposition 4.3.3.1. Let $q: X \rightarrow S$ be a Kan fibration of simplicial sets and let $e: s \rightarrow s'$ be an edge of $S$. Then the morphisms $e_{!}: X_{s} \rightarrow X_{s'}$ and $e^{\ast }: X_{s'} \rightarrow X_{s}$ are homotopy inverse to one another.

### 4.3.3 A Characterization of Kan Fibrations

Let $q: X \rightarrow S$ be a Kan fibration of simplicial sets. Then $q$ is both a left fibration and a right fibration (Example 4.1.0.5). Consequently, to each edge $e: s \rightarrow s'$ of the simplicial set $S$, we can associate both a covariant transport morphism $e_{!}: X_{s} \rightarrow X_{s'}$ (Notation 4.3.2.5) and a contravariant transport morphism $e^{\ast }: X_{s'} \rightarrow X_{s}$ (Notation 4.3.2.11).

**Proof.**
We will show that the composition $e^{\ast } \circ e_{!}$ is homotopic to the identity map $\operatorname{id}_{ X_ s }$; a similar argument will show that $e_{!} \circ e^{\ast }$ is homotopic to the identity map $\operatorname{id}_{ X_{s'} }$. As in the proof of Proposition 4.3.2.4, we let $\underline{e}$ denote the image of $e$ along the diagonal map $S \rightarrow \operatorname{Fun}( X_ s, S)$, and let $Q: \operatorname{Fun}( X_ s, X) \rightarrow \operatorname{Fun}(X_ s, S)$ be the morphism given by postcomposition with $q$. Let $h: \Delta ^1 \times X_ s \rightarrow X$ be a morphism of simplicial sets which exhibits $e_{!}$ as given by covariant transport along $e$, so that $h$ can be identified with an edge $\overline{e}: \operatorname{id}_{X_ s} \rightarrow e_{!}$ of the simplicial set $\operatorname{Fun}(X_ s, X)$ satisfying $Q( \overline{e} ) = \underline{e}$. Let $h': \Delta ^1 \times X_{s'} \rightarrow X$ be a morphism of simplicial sets which exhibits $e^{\ast }$ as given by contravariant transport along $e$. Then the composite map

can be identified with an edge $\overline{e}': (e^{\ast } \circ e_{!}) \rightarrow e_{!}$ of $\operatorname{Fun}( X_ s, X)$ satisfying $Q(\overline{e}' ) = \underline{e}$. Since $Q$ is a Kan fibration (Corollary 3.1.3.2), the lifting problem

admits a solution. Here $\sigma $ is a $2$-simplex of the simplicial set $\operatorname{Fun}(X_ s, X)$ which we can display as a diagram

By construction, the edge $h''$ can be regarded as a homotopy from $\operatorname{id}_{X_ s}$ to the composite map $e^{\ast } \circ e_{!}$. $\square$

It follows from Proposition 4.3.3.1 that if $q: X \rightarrow S$ is a Kan fibration and $e: s \rightarrow s'$ is an edge of the simplicial set $S$, then the transport morphisms $e_{!}: X_{s} \rightarrow X_{s'}$ and $e^{\ast }: X_{s'} \rightarrow X_{s}$ are homotopy equivalences. Our goal in this section is to prove the converse:

Theorem 4.3.3.2. Let $q: X \rightarrow S$ be a morphism of simplicial sets. The following conditions are equivalent:

- $(1)$
The morphism $q$ is a Kan fibration.

- $(2)$
The morphism $q$ is a left fibration and, for every edge $e: s \rightarrow s'$ of the simplicial set $S$, the covariant transport morphism $e_{!}: X_{s} \rightarrow X_{s'}$ is a homotopy equivalence.

- $(3)$
The morphism $q$ is a right fibration and, for every edge $e: s \rightarrow s'$ of the simplicial set $S$, the contravariant transport morphism $e^{\ast }: X_{s'} \rightarrow X_{s}$ is a homotopy equivalence.

The proof of Theorem 4.3.3.2 will require some preliminaries.

Lemma 4.3.3.3. Let $q: X \rightarrow S$ be a left fibration of simplicial sets, where $S$ is a Kan complex. Suppose that $q$ induces a surjection $\pi _0(q): \pi _0(X) \rightarrow \pi _0(S)$. Then $q$ is surjective on vertices.

**Proof.**
Fix a vertex $s \in S$. Since $\pi _0(q)$ is surjective, there exists a vertex $x \in X$ for which $q(x)$ and $s$ belong to the same connected component of $\pi _0(S)$. Since $S$ is a Kan complex, we can choose an edge $e: q(x) \rightarrow s$ in the simplicial set $S$. Our assumption that $q$ is a left fibration guarantees that we can write $e = q(\overline{e})$ for some edge $\overline{e}: x \rightarrow \overline{s}$ of the simplicial set $X$. In particular, there exists a vertex $\overline{s} \in X$ satisfying $q( \overline{s} ) = s$.
$\square$

Lemma 4.3.3.4. Let $q: X \rightarrow S$ be a morphism of simplicial sets, where $S$ is a Kan complex. Then the following conditions are equivalent:

- $(1)$
The morphism $q$ is a Kan fibration.

- $(2)$
The morphism $q$ is a left fibration.

- $(3)$
The morphism $q$ is a right fibration.

**Proof.**
We will show that $(1)$ and $(2)$ are equivalent (the equivalence of $(1)$ and $(3)$ follows from a similar argument). The implication $(1) \Rightarrow (2)$ is contained in Example 4.1.0.5 (and does not require the assumption that $S$ is a Kan complex). Conversely, suppose that $q$ is a left fibration. Then the projection map $X \rightarrow \Delta ^0$ is a left fibration (since it can be written as a composition of $q$ with the Kan fibration $S \rightarrow \Delta ^0$). Invoking Proposition 4.3.1.1, we deduce that $X$ is a Kan complex. To prove that $q$ is a Kan fibration, we must show that for every anodyne morphism of simplicial sets $i: A \hookrightarrow B$, the induced map

is surjective on vertices (Remark 3.1.2.6). Note that we have a commutative diagram of simplicial sets

where the vertical maps are trivial Kan fibrations (since $i$ is anodyne and $X$ and $S$ are Kan complexes; see Corollary 3.1.3.6). Invoking Corollary 3.4.1.3, we see that the diagram (4.14) is homotopy Cartesian. It follows that $\theta $ is a weak homotopy equivalence of simplicial sets (Example 3.4.1.5), and is therefore bijective on connected components (Remark 3.1.5.4). Since $\theta $ is a left fibration of simplicial sets (Proposition 4.1.2.1), it is surjective on vertices by virtue of Lemma 4.3.3.3. $\square$

Remark 4.3.3.5. Assuming Theorem 4.3.3.2, one can give a more direct proof of Lemma 4.3.3.4. Let $q: X \rightarrow S$ be a left fibration of simplicial sets, where $S$ is a Kan complex. To show that $q$ is a Kan fibration, it will suffice (by virtue of Theorem 4.3.3.2) to show that the covariant transport functor $\mathrm{h} \mathit{S} \rightarrow \mathrm{h} \mathit{\operatorname{Kan}}$ carries every morphism in $\mathrm{h} \mathit{S}$ to an invertible morphism in $\mathrm{h} \mathit{\operatorname{Kan}}$. This is clear, since the homotopy category $\mathrm{h} \mathit{S}$ is a groupoid (Proposition 1.3.6.11).

Lemma 4.3.3.6. 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

admits a solution.

**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 4.1.0.4).

Let $\overline{h}: \Delta ^{1} \times \Delta ^{n} \rightarrow \Delta ^ n$ be the morphism given on vertices by $h(i,j) = \begin{cases} j & \text{ if }i=0 \\ n & \text{ if }i=1. \end{cases}$ Since the inclusion $\{ 0\} \times \operatorname{\partial \Delta }^ n \hookrightarrow \Delta ^1 \times \operatorname{\partial \Delta }^{n}$ is left anodyne (Proposition 4.1.2.3), 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 3.1.2.9. Then $Y(0)$ can be described as the pushout

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

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

is an edge of the Kan complex $X_{s}$, and is therefore an isomorphism in the $\infty $-category $X$ (Proposition 1.3.6.11). In this case, the existence of the desired extension follows from Theorem 4.3.1.3. We complete the proof by taking $h = h_{n+1}$. $\square$

Lemma 4.3.3.7. 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.

**Proof.**
The implication $(1) \Rightarrow (2)$ is immediate, and the converse follows from Lemma 4.3.3.6. The equivalence $(1) \Leftrightarrow (3)$ follows by a similar argument.
$\square$

Remark 4.3.3.8. Assuming Theorem 4.3.3.2, one can give a more direct proof of Lemma 4.3.3.7. Let $q: X \rightarrow S$ be a left fibration of simplicial sets with the property that, for every vertex $s \in S$, the fiber $X_{s}$ is a contractible Kan complex. For each edge $e: s \rightarrow s'$ of $S$, the covariant transport $e_{!}: X_{s} \rightarrow X_{s'}$ is a morphism between contractible Kan complexes, and is therefore automatically a homotopy equivalence. Applying Theorem 4.3.3.2, we deduce that $q$ is a Kan fibration. Since the fibers of $q$ are contractible, Proposition 3.3.7.4 guarantees that $q$ is a trivial Kan fibration.

**Proof of Theorem 4.3.3.2.**
We will show that $(1) \Leftrightarrow (2)$; the proof of the equivalence $(1) \Leftrightarrow (3)$ is similar. The implication $(1) \Rightarrow (2)$ is immediate from Proposition 4.3.3.1. For the converse, assume that $q: X \rightarrow S$ is a left fibration of simplicial sets and that, for every edge $e: s \rightarrow s'$ of $S$, the covariant transport morphism $e_{!}: X_{s} \rightarrow X_{s'}$ is a homotopy equivalence. We wish to show that $q$ is a Kan fibration. By virtue of Example 4.1.0.5, it will suffice to show that $q$ is a right fibration. By virtue of Proposition 4.1.3.1, this is equivalent to the assertion that the induced map

is a trivial Kan fibration. Note that our assumption that $q$ is a left fibration guarantees that $\theta $ is also a left fibration (Proposition 4.1.2.1). It will therefore suffice to show that the fibers of $\theta $ are contractible Kan complexes (Lemma 4.3.3.7).

Fix an edge $e: s \rightarrow s'$ of the simplicial set $S$ and set $Y = \operatorname{Fun}( \Delta ^1, X) \times _{ \operatorname{Fun}( \Delta ^1, S) } \{ e\} $. Then evaluation at the vertex $1 \in \Delta ^1$ induces a morphism $\theta _{e}: Y \rightarrow X_{s'}$. Note that $\theta _{e}$ is a pullback of $\theta $, and is therefore also a left fibration. Since $X_{s'}$ is a Kan complex (Corollary 4.3.1.2), Lemma 4.3.3.4 guarantees that $\theta _{e}$ is Kan fibration (so $Y$ is also a Kan complex). Evaluation at the vertex $0 \in \Delta ^1$ induces another morphism of simplicial sets $u: Y \rightarrow X_{s}$. Since $q$ is a left fibration, the morphism $u$ is a trivial Kan fibration. By construction, the covariant transport morphism $e_{!}: X_{s} \rightarrow X_{s'}$ can be realized as a composition

where $v$ is a section of $u$ (and therefore a homotopy equivalence). Consequently, our assumption that $e_{!}$ is a homotopy equivalence of Kan complexes guarantees that $\theta _{e}$ is a homotopy equivalence of Kan complexes (Remark 3.1.5.16). Applying Corollary 3.2.6.9, we deduce that the fibers of $\theta _{e}$ are contractible Kan complexes. Since every fiber of $\theta $ can also be viewed as a fiber of $\theta _{e}$ for some edge $e$ of the simplicial set $S$, it follows that the fibers of $\theta $ are also contractible Kan complexes, as desired. $\square$