# Kerodon

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### 5.1.4 Cartesian Fibrations

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:

$(1)$

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 5.1.4.2. 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 5.1.3.10). 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

$\xymatrix@R =50pt@C=50pt{ X' \ar [r]^-{f} \ar [d]^-{q'} & X \ar [d]^-{q} \\ S' \ar [r] & S. }$

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 5.1.3.6). 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 $X$).

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 $X$, 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 5.1.1.11, 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 5.1.1.7 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 5.6.6.4).

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:

$(1)$

The morphism $q$ is an isofibration.

$(2)$

The morphism $q$ is a cartesian fibration.

$(3)$

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 $(2) \Leftrightarrow (1)$ 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 1.3.6.10). 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 5.1.1.9. $\square$

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 $X_{s} = \{ s\} \times _{S} X$.

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

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

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

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 $\overline{e}: z' \rightarrow z$ is an edge 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

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

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 5.1.3.9). Then $g$ is $q'$-cartesian, where $q': X_{z'} \rightarrow Y_{z'}$ is the projection map (Proposition 5.1.1.7). Applying Remark 5.1.3.6, we conclude that $g$ is locally $q$-cartesian. Since $q$ is a cartesian fibration, it follows that $g$ is $q$-cartesian (Remark 5.1.4.5). Invoking Proposition 5.1.4.12, 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 5.1.4.11). 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:

$(1)$

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 5.1.4.11 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

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

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 5.1.3.9 and Corollary 5.1.3.10, 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 5.1.4.14, 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

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

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 5.1.4.12. $\square$

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

Proposition 5.1.4.16. 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

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

Since $q$ is an inner fibration, the morphism $u$ is a right fibration (Proposition 4.3.6.6). 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 5.1.4.17. 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 4.3.6.8). By virtue of Corollary 5.1.1.13, 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 5.1.1.10 that $q'( \overline{e} )$ is $q''$-cartesian, so that $\overline{e}$ is $q'$-cartesian by virtue of Corollary 5.1.2.7. $\square$

Proposition 5.1.4.18. 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 5.1.4.6. The morphism $u$ is a left fibration (Proposition 4.3.6.6), 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 $v$-cartesian (Lemma 5.1.4.17), 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 5.1.1.14 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 5.1.3.9, 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 5.1.4.11, 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. $\square$