Kerodon

Proof. The implication $(1) \Rightarrow (2)$ follows from Remark 3.1.2.7. To deduce the converse, we first apply Proposition 3.1.7.1 to write $f$ as a composition $X \xrightarrow {f'} Q \xrightarrow {f''} Y$, where $f'$ is anodyne and $f''$ is a Kan fibration. If $f$ satisfies condition $(2)$, then the lifting problem

admits a solution. It follows that $f$ is a retract of $f'$ (in the arrow category $\operatorname{Fun}([1], \operatorname{Set_{\Delta }})$). Since the collection of anodyne morphisms is closed under retracts, it follows that $f$ is anodyne. $\square$

Proof. The implication $(\ast ) \Rightarrow (\ast ')$ is immediate (note that a morphism of Kan complexes $u: X \rightarrow Y$ is a homotopy equivalence if and only if its homotopy class $[u]$ is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}$). For the converse, assume that $(\ast ')$ is satisfied, let $X$ and $Y$ be Kan complexes, and let $u_0, u_1: X \rightarrow Y$ be a pair of homotopic morphisms. Let us regard $u_0$ and $u_1$ as vertices of the Kan complex $\operatorname{Fun}(X,Y)$. Since $u_0$ and $u_1$ are homotopic, there exists an edge $e: \Delta ^1 \rightarrow \operatorname{Fun}(X,Y)$ satisfying $e(0) = u_0$ and $e(1) = u_1$. By virtue of Proposition 3.1.7.1, this morphism factors as a composition $\Delta ^1 \xrightarrow {e'} Q \xrightarrow {e''} \operatorname{Fun}(X,Y)$, where $e'$ is anodyne and $e''$ is a Kan fibration. Since $\operatorname{Fun}(X,Y)$ is a Kan complex (Corollary 3.1.3.4), it follows that $Q$ is also a Kan complex. Let us identify $e''$ with a morphism of Kan complexes $h: Q \times X \rightarrow Y$. Let $i_0: X \hookrightarrow Q \times X$ be the product of the identity map $\operatorname{id}_{X}$ with the inclusion $\{ e'(0) \} \hookrightarrow Q$, and define $i_1: X \hookrightarrow Q \times X$ similarly. Since $e'$ is anodyne, the restrictions $e'|_{\{ 0\} }$ and $e'|_{\{ 1\} }$ are anodyne. In particular, they are weak homotopy equivalences (Proposition 3.1.6.14) and therefore homotopy equivalences (Proposition 3.1.6.13), since $Q$ is a Kan complex. It follows that $i_0$ and $i_1$ are also homotopy equivalences, so that $F(i_0)$ and $F(i_1)$ are isomorphisms (by virtue of assumption $(\ast ')$). Using the fact that $i_0$ and $i_1$ are left inverse to the projection map $\pi : Q \times X \rightarrow X$, we see that $F(\pi )$ is an isomorphism in $\operatorname{\mathcal{C}}$ and that we have

as desired. $\square$

Proof. Combine Remark 3.1.5.11 with Proposition 3.1.7.6. $\square$

Proof of Variant 3.1.7.8. Let $\operatorname{\mathcal{E}}\subseteq \operatorname{Fun}( \operatorname{Kan}, \operatorname{\mathcal{C}})$ be the full subcategory spanned by those functors $F: \operatorname{Kan}\rightarrow \operatorname{\mathcal{C}}$ which carry homotopy equivalences of Kan complexes to isomorphisms in $\operatorname{\mathcal{C}}$. By virtue of Corollary 3.1.7.7, it will suffice to show that the restriction functor $F \mapsto F|_{\operatorname{Kan}}$ induces an equivalence of categories $\operatorname{\mathcal{E}}' \rightarrow \operatorname{\mathcal{E}}$. Using Proposition 3.1.7.1, we can choose a functor $Q: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Kan}$ and a natural transformation $u: \operatorname{id}_{\operatorname{Set_{\Delta }}} \rightarrow Q$ with the property that, for every simplicial set $X$, the induced map $u_{X}: X \rightarrow Q(X)$ is anodyne. For every morphism of simplicial sets $f: X \rightarrow Y$, we have a commutative diagram

where the vertical maps are weak homotopy equivalences (Proposition 3.1.6.14). It follows that if $f$ is a weak homotopy equivalence, then $Q(f)$ is also a weak homotopy equivalence (Remark 3.1.6.16) and therefore a homotopy equivalence (Proposition 3.1.6.13). In other words, the functor $Q$ carries weak homotopy equivalences of simplicial sets to homotopy equivalences of Kan complexes. It follows that precomposition with $Q$ induces a functor $\theta : \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{E}}'$. We claim that $\theta $ is homotopy inverse to the restriction functor $\operatorname{\mathcal{E}}' \rightarrow \operatorname{\mathcal{E}}$. This follows from the following pair of observations:

• For every functor $F: \operatorname{Set_{\Delta }}\rightarrow \operatorname{\mathcal{C}}$, $u$ induces a natural transformation $F \rightarrow F|_{\operatorname{Kan}} \circ Q$, which depends functorially on $F$ and is an isomorphism for $F \in \operatorname{\mathcal{E}}'$.

• For every functor $F_0: \operatorname{Kan}\rightarrow \operatorname{\mathcal{C}}$, $u$ induces a natural transformation $F_0 \rightarrow (F_0 \circ Q)|_{\operatorname{Kan}}$, which depends functorially on $F_0$ and is an isomorphism for $F_0 \in \operatorname{\mathcal{E}}$.

Proof of Proposition 3.1.7.1. Let $f: X_{} \rightarrow Y_{}$ be a morphism of simplicial sets. We construct a sequence of simplicial sets $\{ X(m)_{} \} _{m \geq 0}$ and morphisms $f(m): X(m)_{} \rightarrow Y_{}$ by recursion. Set $X(0)_{} = X_{}$ and $f(0) = f$. Assuming that $f(m): X(m)_{} \rightarrow Y_{}$ has been defined, let $S(m)$ denote the set of all commutative diagrams $\sigma :$

where $0 \leq i \leq n$, $n > 0$, and the left vertical map is the inclusion. For every such commutative diagram $\sigma $, let $C_{\sigma } = \Lambda ^ n_{i}$ denote the upper left hand corner of the diagram $\sigma $, and $D_{\sigma } = \Delta ^ n$ the lower left hand corner. Form a pushout diagram

and let $f(m+1): X(m+1)_{} \rightarrow Y_{}$ be the unique map whose restriction to $X(m)_{}$ is equal to $f(m)$ and whose restriction to each $D_{\sigma }$ is equal to $u_{\sigma }$. By construction, we have a direct system of anodyne morphisms

Set $Q_{}(f) = \varinjlim _{m} X(m)_{}$. Then the natural map $f': X_{} \rightarrow Q_{}(f)$ is anodyne (since the collection of anodyne maps is closed under transfinite composition), and the system of morphisms $\{ f(m) \} _{m \geq 0}$ can be amalgamated to a single map $f'': Q_{}(f) \rightarrow Y_{}$ satisfying $f = f'' \circ f'$. It is clear from the definition that the construction $f \mapsto Q_{}(f)$ is functorial and commutes with filtered colimits. To complete the proof, it will suffice to show that $f''$ is a Kan fibration: that is, that every lifting problem $\sigma :$

admits a solution (provided that $n > 0$). Let us abuse notation by identifying each $X(m)_{}$ with its image in $Q_{}(f)$. Since $\Lambda ^{n}_{i}$ is a finite simplicial set, its image under $v$ is contained in $X(m)_{}$ for some $m \gg 0$. In this case, we can identify $\sigma $ with an element of the set $S(m)$, so that the lifting problem

admits a solution by construction. $\square$