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 $(P, \alpha ^{P})$. $\square$