# Kerodon

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### 5.1.5 Fiberwise Equivalence

Let $\operatorname{\mathcal{D}}$ be an $\infty$-category. Our primary goal in this section is to show that, when studying $\infty$-categories $\operatorname{\mathcal{C}}$ equipped with a cartesian fibration $\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$, equivalence can be detected fiberwise. More precisely, we have the following result:

Theorem 5.1.5.1. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}' }$

where $U$ is a cartesian fibration of $\infty$-categories, $U'$ is an isofibration of $\infty$-categories, and $\overline{F}$ is an equivalence of $\infty$-categories. Then the functor $F$ is an equivalence of $\infty$-categories if and only if it satisfies the following conditions:

$(1)$

For every object $D \in \operatorname{\mathcal{D}}$ having image $D' = \overline{F}(D)$ in $\operatorname{\mathcal{D}}'$, the induced functor

$F_{D}: \operatorname{\mathcal{C}}_{D} = \{ D\} \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}\rightarrow \{ D' \} \times _{\operatorname{\mathcal{D}}'} \operatorname{\mathcal{C}}' = \operatorname{\mathcal{C}}'_{D'}$

is an equivalence of $\infty$-categories.

$(2)$

The functor $F$ carries $U$-cartesian morphisms of $\operatorname{\mathcal{C}}$ to $U'$-cartesian morphisms of $\operatorname{\mathcal{C}}'$.

Moreover, if these conditions are satisfied, then $U'$ is also a cartesian fibration of $\infty$-categories.

We will give the proof of Theorem 5.1.5.1 at the end of this section. First, let us collect some of its consequences.

Corollary 5.1.5.2. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}'. }$

Assume that $U$ and $U'$ are isofibrations of $\infty$-categories and that $F$ and $\overline{F}$ are equivalences of $\infty$-categories. Then:

• The functor $U$ is a cartesian fibration if and only if $U'$ is a cartesian fibration.

• The functor $U$ is a cocartesian fibration if and only if $U'$ is a cocartesian fibration.

Proof. We will prove the first assertion; the second follows by a similar argument. It follows from Theorem 5.1.5.1 that if $U$ is a cartesian fibration, then $U'$ is also a cartesian fibration. To prove the converse, choose functors $G': \operatorname{\mathcal{C}}' \rightarrow \operatorname{\mathcal{C}}$ and $\overline{G}: \operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$ which are homotopy inverse to the equivalences $F$ and $\overline{F}$, respectively. We then have isomorphisms

$U \circ G' \circ F \simeq U \simeq \overline{G} \circ \overline{F} \circ U = \overline{G} \circ U' \circ F$

in the functor $\infty$-category $\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$. Since $F$ is an equivalence of $\infty$-categories, it follows that there exists an isomorphism $\overline{\alpha }: U \circ G' \rightarrow \overline{G} \circ U'$ in the functor $\infty$-category $\operatorname{Fun}(\operatorname{\mathcal{C}}', \operatorname{\mathcal{D}})$. Using our assumption that $U$ is an isofibration, we can lift $\overline{\alpha }$ to an isomorphism of functors $\alpha : G' \rightarrow G$ (Proposition 4.4.5.8). Applying Theorem 5.1.5.1 to the commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [r]^-{G} \ar [d]^-{U'} & \operatorname{\mathcal{C}}\ar [d]^-{U} \\ \operatorname{\mathcal{D}}' \ar [r]^-{\overline{G}} & \operatorname{\mathcal{D}}, }$

we conclude that if $U'$ is a cartesian fibration, then $U$ is also a cartesian fibration. $\square$

Corollary 5.1.5.3. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}'. }$

Assume that $U$ and $U'$ are isofibrations of $\infty$-categories and that $F$ and $\overline{F}$ are equivalences of $\infty$-categories. Then:

• The functor $U$ is a right fibration if and only if $U'$ is a right fibration.

• The functor $U$ is a left fibration if and only if $U'$ is a left fibration.

Proof. We will prove the first assertion; the second follows by a similar argument. Assume first that $U'$ is a right fibration of $\infty$-categories. Then $U'$ is a cartesian fibration (Proposition 5.1.4.14), so Corollary 5.1.5.2 implies that $U$ is a cartesian fibration. To prove that $U$ is a right fibration, it will suffice to show that for every object $D \in \operatorname{\mathcal{D}}$, the fiber $\operatorname{\mathcal{C}}_{D} = \{ D\} \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$ is a Kan complex (Proposition 5.1.4.14). This follows from Remark 4.5.1.21, since the functor $F$ induces an equivalence of $\infty$-categories $F_{D}: \operatorname{\mathcal{C}}_{D} \rightarrow \operatorname{\mathcal{C}}'_{ \overline{F}(D)}$ (Corollary 4.5.2.26).

We now prove the reverse implication. Arguing as in the proof of Corollary 5.1.5.2, we can construct a commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [r]^-{G} \ar [d]^-{U'} & \operatorname{\mathcal{C}}\ar [d]^-{U} \\ \operatorname{\mathcal{D}}' \ar [r]^-{\overline{G}} & \operatorname{\mathcal{D}}, }$

where $G$ and $\overline{G}$ are homotopy inverses of the equivalences $F$ and $\overline{F}$, respectively. It then follows from the preceding argument that if $U$ is a right fibration of $\infty$-categories, then $U'$ is also a right fibration of $\infty$-categories. $\square$

Corollary 5.1.5.4. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{\overline{F}} & \operatorname{\mathcal{D}}'. }$

where $U$ and $U'$ are right fibrations and the functor $\overline{F}$ is an equivalence of $\infty$-categories. Then $F$ is an equivalence of $\infty$-categories if and only if, for every object $D \in \operatorname{\mathcal{D}}$ having image $D' = \overline{F}(D) \in \operatorname{\mathcal{D}}'$, the induced map of fibers $F_{D}: \operatorname{\mathcal{C}}_{D} \rightarrow \operatorname{\mathcal{C}}'_{D'}$ is a homotopy equivalence of Kan complexes.

The proof of Theorem 5.1.5.1 will require some preliminaries. Our first step is to show that if $U: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is an isofibration of $\infty$-categories, then the collection of $U$-cartesian morphisms of $\operatorname{\mathcal{C}}$ is invariant under categorical equivalence.

Lemma 5.1.5.5. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}', }$

where the functors $U$ and $U'$ are inner fibrations and the functors $F$ and $\overline{F}$ are fully faithful. Let $g: Y \rightarrow Z$ be a morphism in $\operatorname{\mathcal{C}}$. If $F(g)$ is a $U'$-cartesian morphism of $\operatorname{\mathcal{C}}'$, then $g$ is a $U$-cartesian morphism of $\operatorname{\mathcal{C}}$.

Proof. By virtue of Proposition 5.1.2.1, it will suffice to show that for every object $X \in \operatorname{\mathcal{C}}$, the diagram of Kan complexes

5.9
$$\begin{gathered}\label{equation:fully-faithful-check-cartesian} \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y,Z) \times _{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y,Z) } \{ g\} \ar [r] \ar [d] & \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z) \ar [d] \\ \operatorname{Hom}_{\operatorname{\mathcal{D}}}( U(X), U(Y), U(Z) ) \times _{ \operatorname{Hom}_{\operatorname{\mathcal{D}}}( U(Y), U(Z)) } \{ U(g) \} \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{D}}}( U(X), U(Z) ) } \end{gathered}$$

is a homotopy pullback square. Set $X' = F(X)$, $Y' = F(Y)$, $Z' = F(Z)$, and $g' = F(g)$. Since the functors $F$ and $\overline{F}$ are fully faithful, (5.9) is homotopy equivalent to the diagram

5.10
$$\begin{gathered}\label{equation:fully-faithful-check-cartesian2} \xymatrix@R =50pt@C=25pt{ \operatorname{Hom}_{\operatorname{\mathcal{C}}'}(X',Y',Z') \times _{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y',Z') } \{ g'\} \ar [r] \ar [d] & \operatorname{Hom}_{\operatorname{\mathcal{C}}'}(X',Z') \ar [d] \\ \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( U'(X'), U'(Y'), U'(Z') ) \times _{ \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( U'(Y), U'(Z)) } \{ U'(g') \} \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( U'(X'), U'(Z') ). } \end{gathered}$$

Our assumption that $g'$ is $U'$-cartesian guarantees that (5.10) is a homotopy pullback square of Kan complexes (Proposition 5.1.2.1), so that (5.9) is also a homotopy pullback square (Corollary 3.4.1.12). $\square$

Proposition 5.1.5.6. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}', }$

where the functors $U$ and $U'$ are isofibrations and the functors $F$ and $\overline{F}$ are equivalences of $\infty$-categories. Let $g: Y \rightarrow Z$ be a morphism in $\operatorname{\mathcal{C}}$. Then $g$ is $U$-cartesian if and only if $F(g)$ is $U'$-cartesian.

Proof. It follows from Lemma 5.1.5.5 that if $F(g)$ is $U'$-cartesian, then $g$ is $U$-cartesian. For the converse, suppose that $g$ is $U$-cartesian. Arguing as in the proof of Corollary 5.1.5.2, we can construct a commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [r]^-{G} \ar [d]^-{U'} & \operatorname{\mathcal{C}}\ar [d]^-{U} \\ \operatorname{\mathcal{D}}' \ar [r]^-{\overline{G}} & \operatorname{\mathcal{D}}, }$

where $G$ and $\overline{G}$ are homotopy inverses of the equivalences $F$ and $\overline{F}$, respectively. Then $G( F(g) )$ is isomorphic to $g$ as an object of the arrow $\infty$-category $\operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}})$. Invoking Corollary 5.1.2.5, we conclude that $G( F(g) )$ is $U$-cartesian, so that $F(g)$ is $U'$-cartesian by virtue of Lemma 5.1.5.5. $\square$

Proposition 5.1.5.7. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{q} & \operatorname{\mathcal{C}}' \ar [d]^-{q'} \\ \operatorname{\mathcal{D}}\ar [r]^-{\overline{F}} & \operatorname{\mathcal{D}}'. }$

Assume that:

$(1)$

The functors $q$ and $q'$ are inner fibrations.

$(2)$

The inner fibration $q$ is cartesian and the functor $F$ carries $q$-cartesian morphisms of $\operatorname{\mathcal{C}}$ to locally $q'$-cartesian morphisms of $\operatorname{\mathcal{C}}'$.

$(3)$

The functor $\overline{F}: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}'$ is fully faithful.

Then $F$ is fully faithful if and only if, for every object $D \in \operatorname{\mathcal{D}}$ having image $D' = \overline{F}(D) \in \operatorname{\mathcal{D}}'$, the induced map of fibers $F_{D}: \operatorname{\mathcal{C}}_{D} \rightarrow \operatorname{\mathcal{C}}'_{D'}$ is fully faithful.

Proof. The “only if” direction follows from Proposition 4.6.2.6. For the converse, assume that each of the functors $F_{D}$ is fully faithful; we will show that $F$ is fully faithful. Let $X$ and $Z$ be objects of $\operatorname{\mathcal{C}}$ having images $\overline{X}, \overline{Z} \in \operatorname{\mathcal{D}}$; we wish to show that the upper horizontal map in the diagram of Kan complexes

$\xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z) \ar [r] \ar [d] & \operatorname{Hom}_{\operatorname{\mathcal{C}}'}( F(X), F(Z) ) \ar [d] \\ \operatorname{Hom}_{\operatorname{\mathcal{D}}}( \overline{X}, \overline{Z} ) \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( \overline{F}( \overline{X} ), \overline{F}( \overline{Z} )) }$

is a homotopy equivalence. Since $q$ and $q'$ are inner fibrations, the vertical maps are Kan fibrations (Proposition 4.6.1.19). Assumption $(3)$ guarantees that the lower horizontal map is a homotopy equivalence. By virtue of Proposition 3.2.8.1, it will suffice to show that for every morphism $\overline{e}: \overline{X} \rightarrow \overline{Z}$ in $\operatorname{\mathcal{D}}$, the induced map of fibers

$\theta : \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z)_{\overline{e}} \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}'}( F(X), F(Z) )_{ \overline{F}(\overline{e} ) }$

is a homotopy equivalence.

Let $[\theta ]$ denote the homotopy class of $\theta$, regarded as a morphism in the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}$. Since $q$ is a cartesian fibration, there exists a $q$-cartesian morphism $g: Y \rightarrow Z$ of $\operatorname{\mathcal{C}}$ satisfying $q(g) = \overline{e}$. We then have a commutative diagram

$\xymatrix@C =50pt@R=50pt{ \operatorname{Hom}_{ \operatorname{\mathcal{C}}_{ \overline{X} }}(X,Y) \ar [d]^-{ [g] \circ } \ar [r] & \operatorname{Hom}_{ \operatorname{\mathcal{C}}'_{ \overline{F}(\overline{X})} }( F(X), F(Y) ) \ar [d]^-{ [F(g)] \circ } \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z)_{ \overline{e} } \ar [r]^-{ [ \theta ] } & \operatorname{Hom}_{\operatorname{\mathcal{C}}'}( F(X), F(Z) )_{ \overline{F}( \overline{e} ) } }$

in $\mathrm{h} \mathit{\operatorname{Kan}}$, where the vertical maps are given by the composition law of Notation 5.1.3.10. Assumption $(2)$ guarantees that $F(g)$ is locally $q'$-cartesian, so that the vertical maps in this diagram are isomorphisms in $\mathrm{h} \mathit{\operatorname{Kan}}$ (Proposition 5.1.3.11). It will therefore suffice to show that the functor $F_{\overline{X}}$ induces a homotopy equivalence of mapping spaces $\operatorname{Hom}_{ \operatorname{\mathcal{C}}_{ \overline{X} }}(X,Y) \rightarrow \operatorname{Hom}_{ \operatorname{\mathcal{C}}'_{ \overline{F}(\overline{X})} }( F(X), F(Y) )$, which follows from our assumption that $F_{ \overline{X} }$ is fully faithful. $\square$

Remark 5.1.5.8. In the situation of Proposition 5.1.5.7, we can replace $(2)$ with the following a priori weaker assumption:

$(2')$

For every object $Z \in \operatorname{\mathcal{C}}$ and every morphism $\overline{u}: \overline{Y} \rightarrow q(Z)$ in $\operatorname{\mathcal{D}}$, there exists a $q$-cartesian $u: Y \rightarrow Z$ of $\operatorname{\mathcal{C}}$ satisfying $q(u) = \overline{u}$ and for which $F(u)$ is locally $q'$-cartesian.

Assume that $(2)$ is satisfied and let $v: X \rightarrow Z$ be any $q$-cartesian morphism in $\operatorname{\mathcal{C}}$; we wish to show that $F(v)$ is locally $q'$-cartesian. To prove this, we can assume without loss of generality that $\operatorname{\mathcal{D}}= \Delta ^1 = \operatorname{\mathcal{D}}'$ and that $\overline{F}$ is the identity map. Using $(2')$, we can choose another $q$-cartesian morphism $u: Y \rightarrow Z$ satisfying $q(u) = q(v)$ for which $F(u)$ is $q'$-cartesian. Applying Remark 5.1.3.8, we see that $v$ can be obtained as a composition of $u$ with an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}$. Then $F(v)$ can be obtained as the composition of $F(u)$ with an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}'$, and is therefore $q$-cartesian by virtue of Corollary 5.1.2.4 (and Proposition 5.1.1.8).

Proof of Theorem 5.1.5.1. Suppose we are given a commutative diagram of $\infty$-categories

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}' }$

where $U$ is a cartesian fibration of $\infty$-categories, $U'$ is an isofibration of $\infty$-categories, and $\overline{F}$ is an equivalence of $\infty$-categories. If $F$ satisfies conditions $(1)$ and $(2)$ of Theorem 5.1.5.1, then it is fully faithful (Proposition 5.1.5.7) and essentially surjective (Remark 4.6.2.16), hence an equivalence of $\infty$-categories by virtue of Theorem 4.6.2.17. Conversely, if $F$ is an equivalence of $\infty$-categories, then it satisfies conditions $(1)$ and $(2)$ by virtue of Corollary 4.5.2.26 and Proposition 5.1.5.7, respectively. To complete the proof, we must show that if these conditions are satisfied, then $U'$ is also a cartesian fibration of $\infty$-categories.

Let $Z'$ be an object of $\operatorname{\mathcal{C}}'$ and let $\overline{g}': \overline{Y}' \rightarrow U'(Z')$ be a morphism in $\operatorname{\mathcal{D}}'$; we wish to show that $\overline{g}'$ can be lifted to a $U'$-cartesian morphism $Y' \rightarrow Z'$ in $\operatorname{\mathcal{C}}'$. Since $F$ is essentially surjective, we can choose an object $Z \in \operatorname{\mathcal{C}}$ and an isomorphism $v: F(Z) \rightarrow Z'$ in the $\infty$-category $\operatorname{\mathcal{C}}'$. Since $\overline{F}$ is essentially surjective, we can choose an object $\overline{Y} \in \operatorname{\mathcal{D}}$ and an isomorphism $\overline{u}: \overline{F}(\overline{Y}) \rightarrow \overline{Y}'$ in the $\infty$-category $\operatorname{\mathcal{D}}'$. Since $\overline{F}$ is fully faithful at the level of homotopy categories, we can choose a morphism $\overline{g}: \overline{Y} \rightarrow U(Z)$ in $\operatorname{\mathcal{D}}$ for which the diagram

$\xymatrix@R =50pt@C=50pt{ \overline{F}( \overline{Y} ) \ar [r]^-{ \overline{F}( \overline{g} ) } \ar [d]^{ \overline{u} } & \overline{F}( U(Z) ) \ar [d]^-{ U'(v) } \\ \overline{Y}' \ar [r]^-{ \overline{g}' } & \overline{Z}', }$

commutes in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{D}}'}$, and can therefore be lifted to a commutative diagram $\overline{\sigma }$ in $\infty$-category $\operatorname{\mathcal{D}}'$ (see Exercise 1.4.2.10). Using our assumption that $U$ is a cartesian fibration, we can lift $\overline{g}$ to a $U$-cartesian morphism $g: Y \rightarrow Z$ of $\operatorname{\mathcal{C}}$. Since $U'$ is an isofibration, Corollary 4.4.5.9 guarantees that we can lift $\overline{\sigma }$ to a commutative diagram $\sigma :$

$\xymatrix@R =50pt@C=50pt{ F(Y) \ar [r]^-{ F(g) } \ar [d] & F(Z) \ar [d]^-{ v } \\ Y' \ar [r]^-{ g' } & Z' }$

in the $\infty$-category $\operatorname{\mathcal{C}}'$, where the vertical maps are isomorphisms. To complete the proof, it will suffice to show that the morphism $g'$ is $U'$-cartesian. This follows from Corollary 5.1.2.5, since the morphism $F(g)$ is $U'$-cartesian (Proposition 5.1.5.6). $\square$