We now introduce an $\infty $-categorical counterpart of Definition 5.0.0.3.
Definition 5.1.4.1. 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:
The morphism $q$ is an inner fibration.
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 $(2)$:
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 5.1.4.2. 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 5.0.0.3) 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 5.1.4.1). Similarly, $q$ is a cocartesian fibration if and only if $\operatorname{N}_{\bullet }(q)$ is a cocartesian fibration of simplicial sets. See Corollary 5.1.2.2.
Example 5.1.4.3. 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.
Remark 5.1.4.4. Let $q: X \rightarrow S$ be a morphism of simplicial sets. Then $q$ is a cartesian fibration if and only if the opposite morphism $q^{\operatorname{op}}: X^{\operatorname{op}} \rightarrow S^{\operatorname{op}}$ is a cocartesian fibration.
Remark 5.1.4.5. Let $q: X \rightarrow S$ be an inner fibration of simplicial sets and let $e$ be an edge of $X$. If $q$ is a cartesian fibration, then $e$ is $q$-cartesian if and only if it is locally $q$-cartesian (see Corollary 5.1.3.9). Similarly, if $q$ is a cocartesian fibration, then $e$ is $q$-cocartesian if and only if it is locally $q$-cocartesian.
Remark 5.1.4.6. Suppose we are given a pullback diagram of simplicial sets If $q$ is a cartesian fibration, then $q'$ is also a cartesian fibration. Moreover, an edge $e'$ of $X'$ is $q'$-cartesian if and only if $e = f(e')$ is a $q$-cartesian edge of $X$ (this follows from Remarks 5.1.4.5 and 5.1.3.5). Similarly, if $q$ is a cocartesian fibration, then $q'$ is also a cocartesian fibration (and an edge $e'$ of $X'$ is $q'$-cocartesian if and only if $e = f(e')$ is a $q$-cocartesian edge of $X$).
\[ \xymatrix@R =50pt@C=50pt{ X' \ar [r]^-{f} \ar [d]^-{q'} & X \ar [d]^-{q} \\ S' \ar [r] & S. } \]
Proposition 5.1.4.7. Let $q: X \rightarrow S$ be a morphism of simplicial sets. Then $q$ is a cartesian fibration if and only if, for every simplex $\sigma : \Delta ^ n \rightarrow S$, the projection map $q_{\sigma }: \Delta ^ n \times _{S} X \rightarrow \Delta ^ n$ is a cartesian fibration.
Proof. If $q$ is a cartesian fibration, then Remark 5.1.4.6 guarantees that every pullback of $q$ is a cartesian fibration. Conversely, suppose that for every $n$-simplex $\sigma : \Delta ^ n \rightarrow S$, the projection map $q_{\sigma }: \Delta ^ n \times _{S} X \rightarrow \Delta ^ n$ is a cartesian fibration. Applying this assumption in the case $n=1$, we conclude that for every vertex $y \in X$ and every edge $\overline{e}: \overline{x} \rightarrow q(y)$ of $S$, there exists a locally $q$-cartesian edge $e: x \rightarrow y$ satisfying $q(e) = \overline{e}$. Moreover, Remark 4.1.1.13 guarantees that $q$ is an inner fibration. It will therefore suffice to show that every locally $q$-cartesian edge $e$ of $X$ is $q$-cartesian. By virtue of Remark 5.1.1.12, it suffices to verify the analogous assertion for each of the projection maps $q_{\sigma }: \Delta ^ n \times _{S} X \rightarrow \Delta ^ n$, which follows from Remark 5.1.4.5 (since $q_{\sigma }$ is assumed to be a cartesian fibration). $\square$
Proposition 5.1.4.8. 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 5.1.1.8 guarantees that $e$ is an isomorphism. $\square$
Remark 5.1.4.9. In the statement of Proposition 5.1.4.8, the hypothesis that $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are $\infty $-categories is superfluous: we will later show that every cartesian fibration of simplicial sets is an isofibration (Corollary 5.6.7.5).
Corollary 5.1.4.10. Let $q: X \rightarrow S$ be a morphism of simplicial sets, where $S$ is a Kan complex. The following conditions are equivalent:
The morphism $q$ is an isofibration.
The morphism $q$ is a cartesian fibration.
The morphism $q$ is a cocartesian fibration.
Proof. We will prove the equivalence $(1) \Leftrightarrow (2)$; the equivalence $(1) \Leftrightarrow (3)$ follows by a similar argument. The implication $(2) \Rightarrow (1)$ is a special case of Proposition 5.1.4.8. For the converse, suppose that $q$ is an isofibration. Then $q$ is an inner fibration. To complete the proof, we must show that for every vertex $y \in X$ and every edge $\overline{e}: \overline{x} \rightarrow q(y)$ of $S$, we can write $\overline{e} = q(e)$ for some $q$-cartesian edge $e$ of $X$. Since $S$ is a Kan complex, $\overline{e}$ is an isomorphism (Proposition 1.4.6.10). Our assumption that $q$ is an isofibration then guarantees that we can write $\overline{e} = q(e)$ for some isomorphism $e: x \rightarrow y$ of $X$. The edge $e$ is automatically $q$-cartesian by virtue of Corollary 5.1.1.10. $\square$
Proposition 5.1.4.11. 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 5.1.3.6 with Remark 5.1.4.5. $\square$
Proposition 5.1.4.12. 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 Suppose that $g$ is $q$-cartesian. Then $f$ is $q$-cartesian if and only if $h$ is $q$-cartesian.
\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{g} & \\ x \ar [ur]^{f} \ar [rr]^-{h} & & z. } \]
Proof. Combine Proposition 5.1.3.7 with Remark 5.1.4.5. $\square$
Proposition 5.1.4.13. 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 5.1.1.6 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:
For every edge $\overline{e}: z' \rightarrow z$ of $Z$ and every vertex $x \in X$ satisfying $z = (q \circ p)(x)$, there exists a special edge $e: x' \rightarrow x$ of $X$ satisfying $\overline{e} = (q \circ p)(e)$.
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
Set $g = d^{2}_2(\sigma )$, so that we can view $\sigma $ informally as a diagram
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 5.1.3.8). Then $g$ is $p'$-cartesian, where $p': X_{z'} \rightarrow Y_{z'}$ is the projection map (Proposition 5.1.1.8). Applying Remark 5.1.3.5, we conclude that $g$ is locally $p$-cartesian. Since $p$ is a cartesian fibration, it follows that $g$ is $p$-cartesian (Remark 5.1.4.5). Invoking Proposition 5.1.4.12, we deduce that $f$ is also $p$-cartesian. Since $p(g)$ is an isomorphism in the $\infty $-category $Y_{z'}$, it is $q$-cartesian (Proposition 5.1.4.11). Applying Proposition 5.1.4.12, 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 5.1.4.14. Let $q: X \rightarrow S$ be a morphism of simplicial sets. The following conditions are equivalent:
The morphism $q$ is a right fibration.
The morphism $q$ is a cartesian fibration and every edge of $X$ is $q$-cartesian.
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 5.1.4.11 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
as depicted in the diagram
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 1.4.6.10). Applying Remark 5.1.3.8 and Corollary 5.1.3.9, 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 5.1.4.14, we can establish a relative version of this result.
Corollary 5.1.4.15. 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
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
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 5.1.4.12. $\square$
Proposition 5.1.4.16. Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets, and let $X' \subseteq X$ be a full simplicial subset with the following property:
For every vertex $y \in X'$ and every edge $\overline{e}: \overline{x} \rightarrow q(y)$ in $S$, there exists a vertex $x \in X'$ and a $q$-cartesian edge $e: x \rightarrow y$ of $X$ satisfying $q(e) = \overline{e}$.
Then $q' = q|_{X'}$ is a cartesian fibration from $X'$ to $S$. Moreover, an edge $e$ of $X'$ is $q'$-cartesian if and only if it is $q$-cartesian.
Proof. Since the inclusion map $X' \hookrightarrow X$ is an inner fibration (see Definition 4.1.2.15), the restriction $q|_{X'}$ is also an inner fibration. Remark 5.1.1.7 guarantees that every edge of $X'$ which is $q$-cartesian is also $q'$-cartesian, so that $(\ast )$ immediately guarantees that $q'$ is a cartesian fibration. To complete the proof, we must show that if $e: x \rightarrow z$ is a $q'$-cartesian edge of $X'$, then $e$ is $q$-cartesian when viewed as an edge of $X$. Applying $(\ast )$, we can choose a $q$-cartesian edge $e': y \rightarrow z$ satisfying $q(e') = q(e)$, where $y$ belongs to $X'$. Then $e'$ is also $q'$-cartesian, so Remark 5.1.3.8 guarantees that there exists a $2$-simplex
of $X'$, where $u$ is an isomorphism in the $\infty $-category $\{ q(x) \} \times _{S} X'$. It follows that $u$ is also an isomorphism in the $\infty $-category $\{ q(x) \} \times _{S} X$, and is therefore $q$-cartesian (Proposition 5.1.4.11). Applying Proposition 5.1.4.12, we see that the edge $e$ is also $q$-cartesian. $\square$
We now study the behavior of cartesian fibrations with respect to the slice and coslice constructions of §4.3.
Proposition 5.1.4.17. 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:
The induced map $q': X_{/f} \rightarrow S_{/(q \circ f)}$ is a cartesian fibration of simplicial sets.
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
Since $q$ is an inner fibration, the morphism $u$ is a right fibration (Proposition 4.3.6.8). In particular, $u$ is a cartesian fibration and every edge of $X_{/f}$ is $u$-cartesian (Proposition 5.1.4.14). The morphism $v$ is a pullback of $q$, and is therefore a cartesian fibration (Remark 5.1.4.6). 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 5.1.4.13. $\square$
Lemma 5.1.4.18. 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 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.
\[ q': X_{f/} \rightarrow X_{ f|_{A} / } \times _{ S_{ ( q \circ f|_{A} )/}} S_{ (q \circ f)/} \]
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 4.3.6.10). By virtue of Corollary 5.1.1.14, 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 5.1.1.11 that $q'( \overline{e} )$ is $q''$-cartesian, so that $\overline{e}$ is $q'$-cartesian by virtue of Corollary 5.1.2.6. $\square$
Proposition 5.1.4.19. 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:
The induced map $q': X_{f/} \rightarrow S_{(q \circ f)/}$ is a cartesian fibration of simplicial sets.
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
where $v$ is a pullback of $q$ and is therefore a cartesian fibration by virtue of Remark 5.1.4.6. The morphism $u$ is a left fibration (Proposition 4.3.6.8), 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 5.1.4.6) and $e$ is $u$-cartesian (Lemma 5.1.4.18), so that $e$ is $q'$-cartesian by virtue of Remark 5.1.1.6. 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 5.1.4.6). Corollary 5.1.1.15 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 5.1.3.8, we can choose a $2$-simplex $\sigma $ of $X_{f/}$ as indicated in the diagram
where $e''$ is an isomorphism in the $\infty $-category $\{ q'(x) \} \times _{ S_{ (q \circ f)/} } X_{f/}$. Using Proposition 5.1.4.11, we deduce that the image of $e''$ in $X$ is $q$-cartesian. Applying Proposition 5.1.4.12, we conclude the image of $e$ in $X$ is also $q$-cartesian, as desired. $\square$
Proposition 5.1.4.20. Suppose we are given a commutative diagram of simplicial sets satisfying the following conditions:
The morphisms $U_0$ and $U_1$ are cartesian fibrations.
The morphism $F_0$ carries $U_0$-cartesian edges of $X_0$ to $U$-cartesian edges of $X$.
The morphism $F_1$ carries $U_1$-cartesian edges of $X_1$ to $U$-cartesian edges of $X$.
The morphism $F_1$ is an isofibration.
\[ \xymatrix@R =50pt@C=50pt{ X_0 \ar [r]^-{F_0} \ar [dr]_{U_0} & X \ar [d]^{U} & X_1 \ar [dl]^{ U_1 } \ar [l]_{F_1} \\ & S & } \]
Then the induced map $U_{01}: X_0 \times _{X} X_1 \rightarrow S$ is also a cartesian fibration. Moreover, an edge $e = (e_0, e_1)$ of $X_0 \times _{X} X_1$ is $U_{01}$-cartesian if and only if $e_0$ is $U_0$-cartesian and $e_1$ is $U_1$-cartesian.
Proof. Let $\pi : X_0 \times _{X} X_1 \rightarrow X_0$ and $\pi ': X_0 \times _{X} X_1 \rightarrow X_1$ be the projection maps. Since $F_1$ is an isofibration, $\pi $ is also an isofibration. In particular, $\pi $ is an inner fibration, so the composition $U_{01} = U_0 \circ \pi $ is also an inner fibration. Let us say that an edge $e = (e_0, e_1)$ of $X_0 \times _{X} X_1$ is special if $e_0$ is $U_0$-cartesian and $e_1$ is $U_1$-cartesian. The second assumption guarantees that $e$ is $\pi $-cartesian (Remark 5.1.1.11) and the first guarantees that $\pi (e)$ is $U_0$-cartesian. Applying Corollary 5.1.2.4, we deduce that every special edge of $X_0 \times _{X} X_1$ is $U_{01}$-cartesian.
To prove that $U_{01}$ is a cartesian fibration, it will suffice to show that for every vertex $x = (x_0, x_1)$ of $X_{0} \times _{X} X_1$ and every edge $\overline{e}: s \rightarrow U_{01}(x)$ in $S$, there exists a special edge $e: y \rightarrow x$ of $X_{0} \times _{X} X_1$ satisfying $U_{01}(e) = \overline{e}$. Since $U_0$ is a cartesian fibration, we can choose a $U_0$-cartesian edge $e_0: y_0 \rightarrow x_0$ of $X_0$ satisfying $U_0(e_0) = \overline{e}$. Similarly, we can choose a $U_1$-cartesian edge $e_1: y_1 \rightarrow x_1$ of $X_1$ satisfying $U_1(e_1) = \overline{e}$. We now observe that $F_0(e_0)$ and $F_1(e_1)$ are $U$-cartesian edges of $X$ having the same target and the same image in the simplicial set $S$. Applying Remark 5.1.3.8, we can choose a $2$-simplex $\sigma $ of $X$ as indicated in the diagram
where $v$ is an isomorphism in the $\infty $-category $X_{s} = \{ s\} \times _{S} X$. Our assumption that $F_1$ is an isofibration guarantees that we can lift $v$ to an isomorphism $\widetilde{v}: y'_1 \rightarrow y_1$ in the $\infty $-category $\{ s\} \times _{S} X_1$. Since $F_1$ is an inner fibration, we can lift $\sigma $ to a $2$-simplex $\widetilde{\sigma }$ of $X_1$ with boundary indicated in the diagram
It follows from Propositions 5.1.4.11 and 5.1.4.12 that $e'_1$ is a $U_1$-cartesian edge of $X_1$, so that $e = (e_0, e'_1)$ is a special edge of $X_0 \times _{X} X_1$ with target $x = (x_0, x_1)$ which satisfies $U_{01}(e) = \overline{e}$.
To complete the proof of Proposition 5.1.4.20, it will suffice to show that if $f: z \rightarrow x$ is a $U_{01}$-cartesian edge of the fiber product $X_{0} \times _{X} X_{1}$, then $f$ is special. Set $s = U_{01}(z)$. Applying the above argument, we can choose a special edge $e: y \rightarrow x$ satisfying $U_{01}(e) = U_{01}(f)$. Using Remark 5.1.3.8, we can choose a $2$-simplex $\tau $ of $X_0 \times _{X} X_1$ with boundary indicated in the diagram
where $v$ is an isomorphism in the $\infty $-category $\{ s\} \times _{S} (X_0 \times _{X} X_1)$. Applying Propositions 5.1.4.11 and 5.1.4.12 to the $2$-simplices $\pi (\tau )$ and $\pi '(\tau )$, we conclude that the edges $\pi (f)$ and $\pi '(f)$ are $U_0$-cartesian and $U_1$-cartesian, as desired. $\square$
As an application of Proposition 5.1.4.20, we record a generalization of Proposition 5.1.4.19 which will be useful later.
Corollary 5.1.4.21. Suppose we are given a commutative diagram of simplicial sets where $q$ and $q'$ are cartesian fibrations and the morphism $u$ carries $q'$-cartesian edges of $X'$ to $q$-cartesian edges of $X$. Let $f: K \rightarrow X$ be a morphism of simplicial sets. Then $q'$ induces a cartesian fibration $\widetilde{q}': X' \times _{X} X_{f/} \rightarrow S_{ (q \circ f)/ }$. Moreover, an edge of $X' \times _{X} X_{f/}$ is $\widetilde{q}'$-cartesian if and only if its image in $X'$ is $q'$-cartesian.
\[ \xymatrix@R =50pt@C=50pt{ X' \ar [dr]^{q'} \ar [rr]^{u} & & X \ar [dl]_{q} \\ & S & } \]
Proof. Let $\widetilde{u}: X' \times _{S} S_{ (q \circ f) / } \rightarrow X \times _{S} S_{ (q \circ f) /}$ denote the pullback of $u$, let $\widetilde{q}: X \times _{S} S_{ (q \circ f) /} \rightarrow S_{ (q \circ f) / }$ be given by projection onto the second factor, and let $v: X_{ f/} \rightarrow X \times _{S} S_{ (q \circ f)/ }$ be the left fibration of Proposition 4.3.6.8. Note that $\widetilde{q}$ is a pullback of $q$, and therefore a cartesian fibration (Remark 5.1.4.6). Moreover, an edge of $X \times _{S} S_{ (q \circ f) /}$ is $\widetilde{q}$-cartesian if and only if its image in $X$ is $q$-cartesian. Similarly, the composite map $\widetilde{q} \circ \widetilde{u}$ is a pullback of $q'$. It follows that $\widetilde{q} \circ \widetilde{u}$ is a cartesian fibration, and that an edge of $X' \times _{S} S_{ (q \circ f) / }$ is $(\widetilde{q} \circ \widetilde{u})$-cartesian if and only if its image in $X'$ is $q'$-cartesian. Applying Proposition 5.1.4.19, we deduce that the composition $\widetilde{q} \circ v$ is a cartesian fibration, and that an edge of $X_{ f/}$ is $(\widetilde{q} \circ v)$-cartesian if and only if its image in $X$ is $q$-cartesian. The desired result now follows by applying Proposition 5.1.4.20 to the diagram