### 5.2.4 Cartesian Fibrations

We now introduce an $\infty $-categorical counterpart of Definition 5.1.4.8.

Definition 5.2.4.1. Let $q: X \rightarrow S$ be a morphism of simplicial sets. We say that $q$ is a *cartesian fibration* if the following conditions are satisfied:

- $(1)$
The morphism $q$ is an inner fibration.

- $(2)$
For every edge $\overline{e}: s \rightarrow t$ of the simplicial set $S$ and every vertex $z \in X$ satisfying $q(z) = t$, there exists a $q$-cartesian edge $e: y \rightarrow z$ of $X$ satisfying $q(e) = \overline{e}$.

We say that $q$ is a *cocartesian fibration* if it satisfies condition $(1)$ together with the following dual version of $(2)$:

- $(2')$
For every edge $\overline{e}: s \rightarrow t$ of the simplicial set $S$ and every vertex $y \in X$ satisfying $q(y) = s$, there exists a $q$-cocartesian edge $e: y \rightarrow z$ of $X$ satisfying $q( e) = \overline{e}$.

Example 5.2.4.2. Let $q: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between ordinary categories. Then $q$ is a cartesian fibration (in the sense of Definition 5.1.4.8) if and only if the induced morphism of simplicial sets $\operatorname{N}_{\bullet }(q): \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ is a cartesian fibration (in the sense of Definition 5.2.4.1). Similarly, $q$ is a cocartesian fibration if and only if $\operatorname{N}_{\bullet }(q)$ is a cocartesian fibration of simplicial sets. See Corollary 5.2.2.2.

Example 5.2.4.3. Let $X$ be a simplicial set and let $q: X \rightarrow \Delta ^0$ denote the projection map. The following conditions are equivalent:

The simplicial set $X$ is an $\infty $-category.

The morphism $q$ is a cartesian fibration.

The morphism $q$ is a cocartesian fibration.

Proposition 5.2.4.7. Let $q: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a cartesian fibration of $\infty $-categories. Then $q$ is an isofibration.

**Proof.**
Suppose we are given an object $Y \in \operatorname{\mathcal{C}}$ and an isomorphism $\overline{e}: \overline{X} \rightarrow q(Y)$ in the $\infty $-category $\operatorname{\mathcal{D}}$. We wish to show that there exists an isomorphism $e: X \rightarrow Y$ in the $\infty $-category $\operatorname{\mathcal{C}}$ satisfying $q(e) = \overline{e}$. Our assumption that $q$ is a cartesian fibration guarantees that we can write $\overline{e} = q(e)$, where $e: X \rightarrow Y$ is a $q$-cartesian morphism of $\operatorname{\mathcal{C}}$. Since $\overline{e} = q(e)$ is an isomorphism, Proposition 5.2.1.7 guarantees that $e$ is an isomorphism.
$\square$

Proposition 5.2.4.9. Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets and let $e$ be an edge of $X$ such that $q(e) = \operatorname{id}_{s}$ is a degenerate edge of $S$. Then $e$ is $q$-cartesian if and only if it is an isomorphism in the $\infty $-category $X_{s} = \{ s\} \times _{S} X$.

**Proof.**
Combine Example 5.2.3.7 with Remark 5.2.4.5.
$\square$

Proposition 5.2.4.10. Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets and let $\sigma : \Delta ^2 \rightarrow X$ be a $2$-simplex of $X$, which we will depict as a diagram

\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{g} & \\ x \ar [ur]^{f} \ar [rr]^{h} & & z. } \]

Suppose that $g$ is $q$-cartesian. Then $f$ is $q$-cartesian if and only if $h$ is $q$-cartesian.

**Proof.**
Combine Proposition 5.2.3.8 with Remark 5.2.4.5.
$\square$

Proposition 5.2.4.11. Let $p: X \rightarrow Y$ and $q: Y \rightarrow Z$ be cartesian fibrations of simplicial sets. Then:

The composite morphism $(q \circ p): X \rightarrow Z$ is a cartesian fibration of simplicial sets.

An edge $e$ of $X$ is $(q \circ p)$-cartesian if and only if $e$ is $p$-cartesian and $p(e)$ is $q$-cartesian.

**Proof.**
It follows from Remark 4.1.1.8 that $q \circ p$ is an inner fibration. Let us say that an edge $e$ of $X$ is *special* if $e$ is $p$-cartesian and $p(e)$ is $q$-cartesian. Remark 5.2.1.6 guarantees that every special edge of $X$ is $(q \circ p)$-cartesian. Consequently, to prove the first assertion, it will suffice to verify the following:

To prove $(\ast )$, set $y = p(x)$. Using our assumption that $q$ is a cartesian fibration, we can choose a $q$-cartesian edge $\widetilde{e}: y' \rightarrow y$ of the simplicial set $Y$ satisfying $q( \widetilde{e} ) = \overline{e}$. Using our assumption that $p$ is a cartesian fibration, we can choose a $p$-cartesian edge $e: x' \rightarrow x$ of $X$ satisfying $p(e) = \widetilde{e}$. Then $e$ is a special edge of $X$ satisfying $(q \circ p)(e) = q( \widetilde{e} ) = \overline{e}$.

To complete the proof, it will suffice to show that every $(q \circ p)$-cartesian edge $f: x'' \rightarrow x$ of $X$ is special. Let $\overline{f}: z'' \rightarrow z$ be the image of $f$ under $(q \circ p): X \rightarrow Z$. Using $(\ast )$, we can choose a special edge $e: x' \rightarrow x$ satisfying $(q \circ p)(e) = \overline{f}$. Since $e$ is $(q \circ p)$-cartesian, we can choose a $2$-simplex $\sigma $ of $X$ satisfying

\[ d_0(\sigma ) = e \quad \quad d_1(\sigma ) = f \quad \quad (q \circ p)(\sigma ) = s_0( \overline{e} ). \]

Set $g = d_2(\sigma )$, so that we can view $\sigma $ informally as a diagram

\[ \xymatrix@R =50pt@C=50pt{ & x' \ar [dr]^{ e } & \\ x'' \ar [ur]^{g} \ar [rr]^{f} & & x. } \]

Set $y' = p(x') \in Y$ and $z' = q(y') \in Z$. Since $f$ is $(q \circ p)$-cartesian, the edge $g$ is an isomorphism in the $\infty $-category $X_{z'}$ (Remark 5.2.3.9). Then $g$ is $q'$-cartesian, where $q': X_{z'} \rightarrow Y_{z'}$ is the projection map (Proposition 5.2.1.7). Applying Remark 5.2.3.6, we conclude that $g$ is locally $q$-cartesian. Since $q$ is a cartesian fibration, it follows that $g$ is $q$-cartesian (Remark 5.2.4.5). Invoking Proposition 5.2.4.10, we deduce that $f$ is also $q$-cartesian. Since $p(g)$ is an isomorphism in the $\infty $-category $Y_{z'}$, it is $q$-cartesian (Proposition 5.2.4.9). Applying Proposition 5.2.4.10, we conclude that $p(f)$ is also $q$-cartesian.
$\square$

Recall that an $\infty $-category $\operatorname{\mathcal{C}}$ is a Kan complex if and only if every morphism in $\operatorname{\mathcal{C}}$ is an isomorphism (Proposition 4.4.2.1). We now establish a relative version of this assertion:

Proposition 5.2.4.12. Let $q: X \rightarrow S$ be a morphism of simplicial sets. The following conditions are equivalent:

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

- $(2)$
The morphism $q$ is a cartesian fibration and every edge of $X$ is $q$-cartesian.

- $(3)$
The morphism $q$ is a cartesian fibration and, for every vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a Kan complex.

**Proof.**
The equivalence $(1) \Leftrightarrow (2)$ is immediate from the definitions. The implication $(2) \Rightarrow (3)$ follows from Propositions 5.2.4.9 and 4.4.2.1. We will complete the proof by showing that $(3)$ implies $(2)$. Assume that $q$ is a cartesian fibration and that each fiber of $q$ is a Kan complex. Let $h: x \rightarrow z$ be an edge of $X$; we wish to show that $h$ is $q$-cartesian. Since $q$ is a cartesian fibration, we can choose a $q$-cartesian edge $g: y \rightarrow z$ of $X$ satisfying $q(g) = q(h)$. The assumption that $g$ is $q$-cartesian then guarantees the existence of a $2$-simplex $\sigma $ of $X$ satisfying

\[ d_0(\sigma ) = g \quad \quad d_1(\sigma ) = h \quad \quad q(\sigma ) = s_0( q(h) ), \]

as depicted in the diagram

\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{g} & \\ x \ar [ur]^{f} \ar [rr]^{h} & & z. } \]

Set $s = q(x)$, so that $f$ is a morphism in the $\infty $-category $X_{s} = \{ s\} \times _{S} X$. Since $X_{s}$ is a Kan complex, $f$ is an isomorphism (Proposition 1.3.6.10). Applying Remark 5.2.3.9 and Corollary 5.2.3.10, we deduce that $h$ is $q$-cartesian.
$\square$

Recall that every $\infty $-category $\operatorname{\mathcal{C}}$ has an underlying Kan complex $\operatorname{\mathcal{C}}^{\simeq }$, obtained by discarding the noninvertible morphisms of $\operatorname{\mathcal{C}}$ (Construction 4.4.3.1). Using Proposition 5.2.4.12, we can establish a relative version of this result.

Corollary 5.2.4.13. Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets, and let $X' \subseteq X$ be the simplicial subset spanned by those simplices $\sigma : \Delta ^ n \rightarrow X$ which carry each edge of $\Delta ^ n$ to a $q$-cartesian edge of $X$. Then the morphism $q|_{X'}: X' \rightarrow S$ is a right fibration of simplicial sets.

**Proof.**
Choose integers $0 < i \leq n$; we wish to show that every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [r]^-{ \sigma _0 } \ar [d] & X' \ar [d]^{ q|_{X'} } \\ \Delta ^ n \ar@ {-->}[ur]^{\sigma } \ar [r]^-{\overline{\sigma } } & S } \]

admits a solution. In the special case $i = n = 1$, this follows immediately from our assumption that $q$ is a cartesian fibration. Assume therefore that $n \geq 2$. We first show that $\sigma _0$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$ satisfying $q \circ \sigma = \overline{\sigma }$. For $i < n$, this follows from the assumption that $q$ is an inner fibration. For $i = n$, it follows from the assumption that the edge

\[ \Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ n-1 < n \} ) \hookrightarrow \Lambda ^ n_ i \xrightarrow { \sigma _0 } X' \hookrightarrow X \]

is $q$-cartesian. We now complete the proof by showing that $\sigma $ factors through the simplicial subset $X'$; that is, it carries each edge of $\Delta ^ n$ to a $q$-cartesian edge of $X$. For $n \geq 3$, this is immediate (since every edge of $\Delta ^ n$ is contained in $\Lambda ^{n}_{i}$). The case $n=2$ follows from Proposition 5.2.4.10.
$\square$

We now study the behavior of cartesian fibrations with respect to the slice and coslice constructions of §4.3.

Proposition 5.2.4.14. Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets and let $f: K \rightarrow X$ be any morphism of simplicial sets. Then:

- $(1)$
The induced map $q': X_{/f} \rightarrow S_{/(q \circ f)}$ is a cartesian fibration of simplicial sets.

- $(2)$
An edge $e$ of $X_{/f}$ is $q'$-cartesian if and only if its image in $X$ is $q$-cartesian.

**Proof.**
The morphism $q'$ factors as a composition

\[ X_{/f} \xrightarrow {u} X \times _{S} S_{/(q \circ f)} \xrightarrow {v} S_{ /(q \circ f)}. \]

Since $q$ is an inner fibration, the morphism $u$ is a right fibration (Proposition 4.3.6.5). In particular, $u$ is a cartesian fibration and every edge of $X_{/f}$ is $u$-cartesian (Proposition 5.2.4.12). The morphism $v$ is a pullback of $q$, and is therefore a cartesian fibration (Remark 5.2.4.6). Moreover, an edge of $X \times _{S} S_{/(q \circ f)}$ is $v$-cartesian if and only if its image in $X$ is $q$-cartesian. Assertions $(1)$ and $(2)$ now follow from Proposition 5.2.4.11.
$\square$

Lemma 5.2.4.15. Let $q: X \rightarrow S$ be an inner fibration of simplicial sets, let $f: B \rightarrow X$ be a morphism of simplicial sets, let $A$ be a simplicial subset of $B$, and let

\[ q': X_{f/} \rightarrow X_{ f|_{A} / } \times _{ S_{ ( q \circ f|_{A} )/}} S_{ (q \circ f)/} \]

be the induced map. Let $\overline{e}$ be an edge of the simplicial set $X_{f/}$, and let $e$ be its image in $X$. If $e$ is $q$-cartesian, then $\overline{e}$ is $q'$-cartesian.

**Proof.**
Let $q'': X_{ f|_{A} / } \times _{ S_{ ( q \circ f|_{A} )/}} S_{ (q \circ f)/} \rightarrow S_{ (q \circ f)/ }$ be the projection map onto the second factor. Then $q''$ is a pullback of the map $u: X_{ f|_{A} / } \rightarrow S_{ ( q \circ f|_{A} )/}$, and is therefore an inner fibration (Corollary 4.3.6.7). By virtue of Corollary 5.2.1.12, the edge $\overline{e}$ is $(q'' \circ q')$-cartesian and the image of $\overline{e}$ in $X_{ f_{A}/}$ is $u$-cartesian. It follows from Remark 5.2.1.9 that $q'( \overline{e} )$ is $q''$-cartesian, so that $\overline{e}$ is $q'$-cartesian by virtue of Corollary 5.2.2.7.
$\square$

Proposition 5.2.4.16. Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets and let $f: K \rightarrow X$ be any morphism of simplicial sets. Then:

- $(1)$
The induced map $q': X_{f/} \rightarrow S_{(q \circ f)/}$ is a cartesian fibration of simplicial sets.

- $(2)$
An edge $e$ of $X_{f/}$ is $q'$-cartesian if and only if its image in $X$ is $q$-cartesian.

**Proof.**
The morphism $q'$ factors as a composition

\[ X_{f/} \xrightarrow {u} X \times _{S} S_{(q \circ f)/} \xrightarrow {v} S_{(q \circ f)/}, \]

where $v$ is a pullback of $q$ and is therefore a cartesian fibration by virtue of Remark 5.2.4.6. The morphism $u$ is a left fibration (Proposition 4.3.6.5), and therefore an inner fibration. It follows that $q'$ is an inner fibration (Remark 4.1.1.8).

Let us say that an edge $e$ of $X_{f/}$ is *special* if its image in $X$ is $q$-cartesian. If this condition is satisfied, then $u(e)$ is $v$-cartesian (Remark 5.2.4.6) and $e$ is $v$-cartesian (Lemma 5.2.4.15), so that $e$ is $q'$-cartesian by virtue of Remark 5.2.1.6. This proves the “if” direction of assertion $(2)$.

To prove $(1)$, it will suffice to show that for every vertex $y$ of the simplicial set $X_{f/}$ and every edge $\overline{e}: \overline{x} \rightarrow q'(y)$ of the simplicial set $S_{ (q \circ f)/}$, there exists a special edge $e: x \rightarrow y$ of $X_{f/}$ satisfying $q'(e) = \overline{e}$. Since $v$ is a cartesian fibration, we can choose a $v$-cartesian edge $\widetilde{e}: \widetilde{x} \rightarrow u(y)$ of the simplicial set $X \times _{S} S_{(q \circ f)/}$. In this case, the image $\widetilde{e}$ in $X$ is $q$-cartesian (Remark 5.2.4.6). Corollary 5.2.1.13 then guarantees that there exists an edge $e: x \rightarrow y$ of $X_{f/}$ satisfying $u(e) = \widetilde{e}$. The edge $e$ is automatically special and satisfies $q'(e) = (v \circ u)(e) = v( \widetilde{e} ) = \overline{e}$, as desired.

To complete the proof of $(2)$, it will suffice to show that every $q'$-cartesian edge $e: x \rightarrow z$ of $X_{f/}$ is special. It follows from the preceding argument that there exists a special edge $e': y \rightarrow z$ satisfying $q'(e') = q'(e)$, which is also $q'$-cartesian. Applying Remark 5.2.3.9, we can choose a $2$-simplex $\sigma $ of $X_{f/}$ as indicated in the diagram

\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{ e' } & \\ x \ar [ur]^{e''} \ar [rr]^{e} & & z, } \]

where $e''$ is an isomorphism in the $\infty $-category $\{ q'(x) \} \times _{ S_{ (q \circ f)/} } X_{f/}$. Using Proposition 5.2.4.9, we deduce that the image of $e''$ in $X$ is $q$-cartesian. Applying Proposition 5.2.4.10, we conclude the image of $e$ in $X$ is also $q$-cartesian, as desired.
$\square$