Kerodon

Proof of Proposition 4.5.6.6. Suppose first that $\mathscr {F}: Q^{\operatorname{op}} \rightarrow \operatorname{Set_{\Delta }}$ is an isofibrant diagram. We will show that, for each element $q \in Q$, the induced map $\theta _{q}: \mathscr {F}(q) \rightarrow \varprojlim _{p < q} \mathscr {F}(p)$ is an isofibration of simplicial sets (for this step, we do not need to assume that $Q$ is well-founded). Fix a simplicial set $B$ and a simplicial subset $A \subseteq B$ for which the inclusion map $A \hookrightarrow B$ is a categorical equivalence; we wish to show that every lifting problem

admits a solution. Define $\mathscr {B}: Q^{\operatorname{op}} \rightarrow \operatorname{Set_{\Delta }}$ by the formula $\mathscr {B}(p) = \left\{ \begin{array}{rl} B & \textnormal{ if } p \leq q \\ \emptyset & \textnormal{ otherwise, } \end{array}\right.$ and let $\mathscr {B}_0 \subseteq \mathscr {B}$ be the subfunctor given by the formula

The lifting problem (4.41) can be identified with a natural transformation of functors $\alpha _0: \mathscr {B}_0 \rightarrow \mathscr {F}$. Since the inclusion $\mathscr {B}_0 \hookrightarrow \mathscr {B}$ is a levelwise categorical equivalence and $\mathscr {F}$ is isofibrant, we can extend $\alpha _0$ to a natural transformation $\alpha : \mathscr {B} \rightarrow \mathscr {F}$, which determines a solution to the lifting problem (4.41).

Now suppose that the partially ordered set $(Q, \leq )$ is well-founded and that for each $q \in Q$, the morphism $\theta _{q}$ is an isofibration of simplicial sets. We wish to show that the diagram $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set_{\Delta }}$ is isofibrant. Let $\mathscr {E}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set_{\Delta }}$ be a functor, let $\mathscr {E}_0 \subseteq \mathscr {E}$ be a subfunctor for which the inclusion $\mathscr {E}_0 \hookrightarrow \mathscr {E}$ is a levelwise categorical equivalence, and let $\alpha _0: \mathscr {E}_0 \rightarrow \mathscr {F}$ be a natural transformation; we wish to show that $\alpha _0$ can be extended to a natural transformation $\alpha : \mathscr {E} \rightarrow \mathscr {F}$.

For every downward-closed subset $P \subseteq Q$, let $\mathscr {E}^{P} \subseteq \mathscr {E}$ denote the subfunctor given by $\mathscr {E}^{P}(q) = \left\{ \begin{array}{rl} \mathscr {E}(q) & \textnormal{ if }q \in P \\ \emptyset & \textnormal{otherwise,} \end{array}\right.$, and set $\mathscr {E}^{P}_{0} = \mathscr {E}^{P} \cap \mathscr {E}_0$. Let $S$ denote the collection of pairs $(P, \alpha ^{P})$, where $P \subseteq Q$ is a downward-closed subset and $\alpha ^{P}: \mathscr {E}^{P} \rightarrow \mathscr {F}$ is a natural transformation satisfying $\alpha ^{P}|_{ \mathscr {E}^{P}_{0} } = \alpha _0|_{ \mathscr {E}^{P}_{0} }$. We regard $S$ as a partially ordered set, where $(P, \alpha ^{P}) \leq (P', \alpha ^{P'} )$ if $P$ is contained in $P'$ and $\alpha ^{P}$ is equal to the restriction $\alpha ^{P'}|_{ \mathscr {E}^{P}}$. The partially ordered set $S$ satisfies the hypotheses of Zorn's lemma, and therefore contains a maximal element $(P, \alpha ^{P})$. To complete the proof, it will suffice to show that $P = Q$, so that $\alpha ^{P}: \mathscr {E} \rightarrow \mathscr {F}$ is an extension of $\alpha _0$. Assume otherwise. Since $Q$ is well-founded, the complement $Q \setminus P$ contains a minimal element $q$. Set $P' = P \cup \{ q\} $. Since $\theta _{q}$ is an isofibration of simplicial sets, the lifting problem

admits a solution in the category of simplicial sets. This solution determines a natural transformation $\alpha ^{P'}: \mathscr {E}^{P'} \rightarrow \mathscr {F}$ satisfying $\alpha ^{P'}|_{ \mathscr {E}^{P} } = \alpha ^{P}$ and $\alpha ^{P'}|_{ \mathscr {E}^{P'}_{0} } = \alpha _{0} |_{ \mathscr {E}^{P'}_{0} }$, contradicting the maximality of the pair $(P, \alpha ^{P})$. $\square$

Proof. Let $B$ be a simplicial set and let $A \subseteq B$ be a simplicial subset. We wish to show that every lifting problem

admits a solution, provided that either the inclusion map $A \hookrightarrow B$ is a categorical equivalence or the inclusion $\mathscr {E}_0 \hookrightarrow \mathscr {E}$ is a levelwise categorical equivalence. Unwinding the definitions, we see that the diagram (4.42) determines a natural transformation

and that solutions to (4.42) can be identified with extensions of $\alpha _0$ to a natural transformation $\alpha : \underline{B} \times \mathscr {E} \rightarrow \mathscr {F}$. By virtue of our assumption that $\mathscr {F}$ is isofibrant, we are reduced to proving that the inclusion map

is a levelwise categorical equivalence, which follows from Corollary 4.5.4.15. $\square$

Proof. Apply Proposition 4.5.6.11 in the special case $\mathscr {E}_0 = \emptyset $. $\square$

Proof. Apply Corollary 4.5.6.12 in the special case $\mathscr {E} = \underline{ \Delta ^{0} }$. $\square$

Proof. Using Exercise 3.1.7.11, we can choose a contractible Kan complex $X$ containing a pair of vertices $x,y \in X$ with $x \neq y$. Evaluation at the vertices $x$ and $y$ determine trivial Kan fibrations of $\infty $-categories

Form a pullback diagram

so that $U$ is also a trivial Kan fibration and therefore an equivalence of $\infty $-categories. It will therefore suffice to show that $\operatorname{ev}_{x} \circ T$ is an equivalence of $\infty $-categories. Since the functors $\operatorname{ev}_{x}$ and $\operatorname{ev}_{y}$ are isomorphic, this is equivalent to the requirement that $\operatorname{ev}_{y} \circ T$ is an equivalence of $\infty $-categories. In fact, the functor $\operatorname{ev}_{y} \circ T$ is a trivial Kan fibration: this follows by applying Proposition 4.5.6.11 to the levelwise categorical equivalence

Proof. Since $\mathscr {E}$ is isofibrant, Proposition 4.5.6.14 guarantees that the functor

is an equivalence of $\infty $-categories. In particular, there exists a natural transformation $\beta : \mathscr {F} \rightarrow \mathscr {E}$ such that $\beta \circ \alpha $ is isomorphic to $\operatorname{id}_{ \mathscr {E} }$ (when viewed as an object of the $\infty $-category $\operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{Set_{\Delta }})} ( \mathscr {E}, \mathscr {E} )_{\bullet }$). To complete the proof, it will suffice to show that $\beta $ is also a right homotopy inverse to $\alpha $: that is, the composition $\alpha \circ \beta $ is isomorphic to $\operatorname{id}_{ \mathscr {F}}$ (when viewed as an object of the $\infty $-category $\operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{Set_{\Delta }})} ( \mathscr {F}, \mathscr {F} )_{\bullet }$).

For each object $C \in \operatorname{\mathcal{C}}$, the functor $\beta _{C}: \mathscr {F}(C) \rightarrow \mathscr {E}(C)$ is a left homotopy inverse of the functor $\alpha _{C}: \mathscr {E}(C) \rightarrow \mathscr {F}(C)$. Since $\alpha _ C$ is an equivalence of $\infty $-categories, it follows that $\beta _{C}$ is also an equivalence of $\infty $-categories. Allowing $C$ to vary, we conclude that $\beta $ is a levelwise categorical equivalence. We can therefore repeat the preceding argument to obtain a natural transformation $\gamma : \mathscr {E} \rightarrow \mathscr {F}$ such that $\gamma \circ \beta $ is isomorphic to $\operatorname{id}_{ \mathscr {F} }$. We then have isomorphisms

in the $\infty $-category $\operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{Set_{\Delta }})} ( \mathscr {E}, \mathscr {F} )_{\bullet }$, so that $\alpha \circ \beta $ is also isomorphic to $\operatorname{id}_{ \mathscr {F}}$. $\square$

Proof. Apply Corollary 4.5.6.16 in the special case $\mathscr {E} = \underline{ \Delta ^{0} }$. $\square$

Proof. By virtue of Corollary 4.5.6.12, the simplicial set $X$ is an $\infty $-category. Define $\mathscr {F}^{\Delta ^{1}}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set_{\Delta }}$ by the formula $\mathscr {F}^{ \Delta ^{1} }(C) = \operatorname{Fun}( \Delta ^1, \mathscr {F}(C) )$. Then $\mathscr {F}^{ \Delta ^1 }$ is also an isofibrant diagram. Moreover, our assumption that each $\mathscr {F}(C)$ is a Kan complex guarantees that the diagonal embedding $\mathscr {F} \hookrightarrow \mathscr {F}^{\Delta ^1}$ is a levelwise categorical equivalence. Applying Corollary 4.5.6.16, we deduce that the diagonal map $X \hookrightarrow \operatorname{Fun}( \Delta ^1, X)$ is an equivalence of $\infty $-categories. In particular, every morphism of $X$ is isomorphic (as an object of the $\infty $-category $\operatorname{Fun}( \Delta ^1, X)$ ) to an identity morphism, and is therefore an isomorphism (Example 4.4.1.14). Applying Proposition 4.4.2.1, we deduce that $X$ is a Kan complex. $\square$

Proof. Apply Proposition 4.5.6.19 in the special case $\mathscr {E} = \underline{ \Delta ^{0} }$. $\square$

Proof. We first show that the diagram $\mathscr {F}^{\simeq }$ is isofibrant. Let $\mathscr {E}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set_{\Delta }}$ be a functor and let $\mathscr {E}_0 \subseteq \mathscr {E}$ be a subfunctor for which the inclusion $\mathscr {E}_0 \hookrightarrow \mathscr {E}$ is a levelwise categorical equivalence. Suppose we are given a natural transformation $\alpha _0: \mathscr {E}_0 \rightarrow \mathscr {F}^{\simeq }$. Our assumption that $\mathscr {F}$ is isofibrant guarantees that $\alpha _0$ can be extended to a natural transformation $\alpha : \mathscr {E} \rightarrow \mathscr {F}$. We claim that $\alpha $ automatically factors through the functor $\mathscr {F}^{\simeq }$: that is, for every object $C \in \operatorname{\mathcal{C}}$, the map $\alpha _{C}: \mathscr {E}(C) \rightarrow \mathscr {F}(C)$ factors through the core of $\mathscr {F}(C)$. This follows from the observation that the lifting problem

has a (unique) solution, since the inclusion $\mathscr {F}(C)^{\simeq } \hookrightarrow \mathscr {F}(C)$ is an isofibration (Proposition 4.4.3.6).

We now prove the second assertion. Let $X$ denote the core of the $\infty $-category $\varprojlim ( \mathscr {F} )$. For every object $C \in \operatorname{\mathcal{C}}$, the projection map $\varprojlim ( \mathscr {F} ) \rightarrow \mathscr {F}(C)$ carries $X$ into the core of $\mathscr {F}(C)$. It follows that $X$ is contained in the inverse limit $\varprojlim ( \mathscr {F}^{\simeq } )$. The reverse inclusion follows from Corollary 4.4.3.18, since the simplicial set $\varprojlim ( \mathscr {F}^{\simeq } )$ is a Kan complex (Corollary 4.5.6.20). $\square$

Proof. Combine Example 4.5.6.8, Corollary 4.5.6.13, and Corollary 4.5.6.21. $\square$

Proof of Proposition 4.5.6.23. Let $\mathscr {F}: \operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{Set_{\Delta }}$ be a diagram which satisfies condition $(\ast _ n)$ of Proposition 4.5.6.23 for every integer $n \geq 0$; we wish to show that $\mathscr {F}$ is isofibrant (the converse follows from Remark 4.5.6.25). Let $\mathscr {E}: \operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{Set_{\Delta }}$ be a diagram of simplicial sets and let $\mathscr {E}_0 \subseteq \mathscr {E}$ be a subfunctor having the property that, for every integer $n \geq 0$, the inclusion map $\mathscr {E}_0( [n] ) \hookrightarrow \mathscr {E}( [n] )$ is a categorical equivalence. We wish to show that every natural transformation $\alpha _0: \mathscr {E}_0 \rightarrow \mathscr {F}$ can be extended to a map $\alpha : \mathscr {E} \rightarrow \mathscr {F}$.

For each integer $n \geq 0$, let $\mathscr {E}_{n} \subseteq \mathscr {E}$ denote the smallest subfunctor which contains $\mathscr {E}_{0}$ and satisfies $\mathscr {E}_{n}( [m] ) = \mathscr {E}( [m] )$ for $m < n$. Then $\mathscr {E}$ can be written as the union of an increasing sequence of subfunctors

To complete the proof, it will suffice to show that every natural transformation $\alpha _{n}: \mathscr {E}_{n} \rightarrow \mathscr {F}$ admits an extension $\alpha _{n+1}: \mathscr {E}_{n+1} \rightarrow \mathscr {F}$. Using Proposition 1.1.4.12, we see that the inclusion $\mathscr {E}_{n} \hookrightarrow \mathscr {E}_{n+1}$ is a pushout of the inclusion map

Consequently, to prove the existence of $\alpha _{n+1}$, we are reduced to solving a lifting problem

By virtue of assumption $(\ast _ n)$, it will suffice to show that the inclusion map $\mathscr {E}_{n}([n] ) \hookrightarrow \mathscr {E}( [n] )$ is a categorical equivalence of simplicial sets.

In fact, we will prove a stronger assertion: the inclusion map $\rho _{n}: \mathscr {E}_{n} \hookrightarrow \mathscr {E}$ is a levelwise categorical equivalence for every integer $n \geq 0$. Our proof proceeds by induction on $n$, the case $n = 0$ being trivial. To carry out the inductive step, let us assume that $\rho _{n}$ is a levelwise categorical equivalence for some $n \geq 0$; in particular, the inclusion map $\mathscr {E}_{n}( [n] ) \hookrightarrow \mathscr {E}( [n] )$ is a categorical equivalence. Since the collection of categorical equivalences is closed under the formation of coproducts (Corollary 4.5.3.10), it follows that the natural transformation $\iota _{n}$ is a levelwise categorical equivalence. The inclusion map $\mathscr {E}_{n} \hookrightarrow \mathscr {E}_{n+1}$ is a pushout of $\iota _ n$, and is therefore also a levelwise categorical equivalence (Remark 4.5.4.13). Since the collection of (levelwise) categorical equivalences satisfies the two-out-of-three property (Remark 4.5.3.5), it follows that $\rho _{n+1}$ is also a categorical equivalence. $\square$