Kerodon

Proof. By virtue of Variant 8.5.1.3, we may assume that $\operatorname{\mathcal{C}}$ is (the nerve of) an ordinary category. Choose a commutative diagram

where the horizontal compositions are the identity morphisms $\operatorname{id}_{Y}$ and $\operatorname{id}_{Y'}$, respectively. If $f$ is an isomorphism, then $g$ is also an isomorphism, with inverse given by the composition $Y' \xrightarrow {i'} X' \xrightarrow { f^{-1} } X \xrightarrow {r} Y$. This follows from the calculations

Proof. We will show that $F$ is left Kan extended from $\operatorname{\mathcal{C}}^{0}$; the assertion that $F$ is right Kan extended from $\operatorname{\mathcal{C}}^{0}$ follows by a similar argument. Choose a regular cardinal $\kappa $ for which $\operatorname{\mathcal{C}}$ is essentially $\kappa $-small. Using Corollary 8.3.3.17, we can choose a fully faithful functor $\operatorname{\mathcal{D}}\rightarrow \widehat{\operatorname{\mathcal{D}}}$, where $\widehat{\operatorname{\mathcal{D}}}$ admits $\kappa $-small colimits. By virtue of Remark 7.3.1.14, we can replace $\operatorname{\mathcal{D}}$ by $\widehat{\operatorname{\mathcal{D}}}$ and thereby reduce to proving Proposition 8.5.1.8 in the special case where $\operatorname{\mathcal{D}}$ admits $\kappa $-small colimits.

Set $F^{0} = F|_{ \operatorname{\mathcal{C}}^{0} }$. Using Proposition 7.6.6.13, we can extend $F^{0}$ to a functor $F': \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ which is left Kan extended from $\operatorname{\mathcal{C}}^{0}$. Invoking the universal mapping property of Corollary 7.3.6.9, we deduce that there is a natural transformation $\alpha : F' \rightarrow F$ which restricts to the identity transformation from $F^{0}$ to itself. The natural transformation $\alpha $ carries each object $Y \in \operatorname{\mathcal{C}}$ to a morphism $\alpha _{Y}: F'(Y) \rightarrow F(Y)$ in the $\infty $-category $\operatorname{\mathcal{D}}$. By assumption, the object $Y$ is a retract of some object $X \in \operatorname{\mathcal{C}}^{0}$. It follows that $\alpha _ Y$ is a retract of the morphism $\alpha _{X} = \operatorname{id}_{ F(X) }$, and is therefore an isomorphism (Proposition 8.5.1.7). Invoking Theorem 4.4.4.4, we deduce that the natural transformation $\alpha $ is an isomorphism, so that $F$ is left Kan extended from $\operatorname{\mathcal{C}}^{0}$ by virtue of Remark 7.3.3.17. $\square$

Proof. Without loss of generality, we may assume that $\operatorname{\mathcal{C}}$ is spanned by $\operatorname{\mathcal{C}}^{0}$ together with the objects $X$ and $Y$. By virtue of Proposition 7.3.8.6, we can further assume that $\operatorname{\mathcal{C}}^{0}$ contains the object $X$. In this case, Proposition 8.5.1.8 implies that the functors $F$ and $U \circ F$ are left Kan extended from $\operatorname{\mathcal{C}}^{0}$, so that $F$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ by virtue of Remark 7.3.3.20. $\square$

Proof. Let $\operatorname{ev}: \operatorname{\mathcal{C}}\times \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{\mathcal{D}}$ denote the evaluation functor. By virtue of Remark 7.3.3.3, the functor $F$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ at an object $C \in \operatorname{\mathcal{C}}$ if and only if the functor $\operatorname{ev}$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0} \times \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$ at the object $(C,F)$. If this condition is satisfied, then Corollary 8.5.1.9 guarantees that $\operatorname{ev}$ is also $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0} \times \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$ at the object $(C,G)$, so that $G$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ at $C$. The desired result now follows by allowing the object $C \in \operatorname{\mathcal{C}}$ to vary. $\square$

Proof. Using Corollary 4.1.3.3, we can choose an inner anodyne morphism $K \hookrightarrow \operatorname{\mathcal{K}}$, where $\operatorname{\mathcal{K}}$ is an $\infty $-category. Using Remark 4.3.6.7, we see that the induced map $K^{\triangleright } \hookrightarrow \operatorname{\mathcal{K}}^{\triangleright }$ is also inner anodyne. We may therefore extend $f$ and $g$ to functors $F,G: \operatorname{\mathcal{K}}^{\triangleright } \rightarrow \operatorname{\mathcal{D}}$. Since the restriction functor $\operatorname{Fun}( \operatorname{\mathcal{K}}^{\triangleright }, \operatorname{\mathcal{D}}) \rightarrow \operatorname{Fun}(K^{\triangleright }, \operatorname{\mathcal{D}})$ is a trivial Kan fibration (Proposition 1.5.7.6), it follows that $G$ is a retract of $F$. By virtue of Corollary 7.2.2.2, we can replace $K$ by $\operatorname{\mathcal{K}}$ and thereby reduce to proving Corollary 8.5.1.11 in the special case where $K$ is an $\infty $-category. In this case, the desired result is a special case of Corollary 8.5.1.10 (see Example 7.3.3.9). $\square$

Proof. Apply Corollary 8.5.1.11 in the special case $\operatorname{\mathcal{E}}= \Delta ^0$. $\square$

Proof. Apply Corollary 8.5.1.11 in the special case $K = \Delta ^0$ (see Example 7.1.6.9). $\square$

Proof. By virtue of Proposition 4.7.6.15, we may assume that the $\infty $-categories $\operatorname{\mathcal{X}}$ and $\operatorname{\mathcal{Y}}$ are minimal, so that $\operatorname{\mathcal{X}}$ is a $\kappa $-small simplicial set (Corollary 4.7.6.12). Choose functors $i: \operatorname{\mathcal{Y}}\rightarrow \operatorname{\mathcal{X}}$ and $r: \operatorname{\mathcal{X}}\rightarrow \operatorname{\mathcal{Y}}$ such that the composition $(r \circ i): \operatorname{\mathcal{Y}}\rightarrow \operatorname{\mathcal{Y}}$ is isomorphic to the identity functor. Then $r \circ i$ is an equivalence of $\infty $-categories. Since $\operatorname{\mathcal{Y}}$ is minimal, it follows that $r \circ i$ is an isomorphism of simplicial sets (Proposition 4.7.6.13). In particular, the functor $i: \operatorname{\mathcal{Y}}\rightarrow \operatorname{\mathcal{X}}$ is a monomorphism of simplicial sets. It follows that $\operatorname{\mathcal{Y}}$ is $\kappa $-small (Remark 4.7.4.8), and therefore essentially $\kappa $-small. $\square$

Proof. Combine Propositions 8.5.1.27 and 1.5.7.6. $\square$

Proof. Combine Corollary 8.5.1.28 with Remark 8.5.1.26. $\square$

Proof. The morphism $\theta $ fits into a commutative diagram

where the vertical maps are trivial Kan fibrations (Corollary 8.5.1.30). It will therefore suffice to show that $\theta '$ is an equivalence of $\infty $-categories. In fact, $\theta '$ is an isomorphism of simplicial sets, since the simplicial set $\mathcal{R}$ is finite (Corollary 3.6.1.10). $\square$

Proof of Lemma 8.5.1.33. Let $P$ denote the collection of all subsets $S' \subseteq [n]$ which contain $S \cup \{ j\} $. Choose a linear ordering

of $P$ with the property that if $S(a) \subseteq S(b)$, then $a \leq b$. Let For $0 \leq b \leq c$, let $K(b) \subseteq \Delta ^ n$ denote the union of $K$ with the faces $\{ \operatorname{N}_{\bullet }( S(a) ) \subseteq \Delta ^ n \} _{ 1 \leq a \leq b}$. We then have inclusion maps

It will therefore suffice to show that, for every positive integer $b \leq c$, the inclusion map $K(b-1) \hookrightarrow K(b)$ is inner anodyne.

Let us identify $\operatorname{N}_{\bullet }( S(b) )$ with the image of a nondegenerate simplex $\sigma : \Delta ^ m \hookrightarrow \Delta ^ n$. Let $L \subseteq \Delta ^ m$ be the inverse image $\sigma ^{-1}( S(b-1) )$, so that we have a pushout diagram of simplicial sets

It will therefore suffice to show that the inclusion map $L \subseteq \Delta ^ m$ is inner anodyne. Because $S(b)$ contains the integers $i < j < k$, we can write $j = \sigma (\overline{j})$ for some $0 < \overline{j} < n$. We conclude by observing that $L$ can be identified with the inner horn $\Lambda ^{m}_{\overline{j} } \subseteq \Delta ^ m$. $\square$

Proof of Proposition 8.5.1.27. Let $\tau $ be a nondegenerate $m$-simplex of the simplicial set $\operatorname{N}_{\bullet }( \operatorname{Ret})$. We define the weight $w(\tau )$. to be the cardinality of the set $\{ i \in [m]: \tau (i) = \widetilde{X} \} $. Note that, if $\tau '$ is any nondegenerate facet of $\tau $, then $w( \tau ' ) \leq w( \tau )$. For $n \geq 1$, the collection of nondegenerate simplices of weight $\leq n$ span a simplicial subset $\mathcal{R}(n) \subseteq \operatorname{N}_{\bullet }( \operatorname{Ret})$. It follows that we can write $\operatorname{N}_{\bullet }( \operatorname{Ret})$ as the union of an increasing sequence

where $\mathcal{R}(1)$ coincides with the simplicial set $\mathcal{R}$ introduced in Notation 8.5.1.24. We will complete the proof by showing that, for each $n \geq 2$, the inclusion map $\mathcal{R}(n-1) \hookrightarrow \mathcal{R}(n)$ is inner anodyne.

Let $\sigma _{n}: \Delta ^{2n} \rightarrow \operatorname{N}_{\bullet }( \operatorname{Ret})$ denote the simplex corresponding to the diagram

Note that $\sigma _{n}$ is a nondegenerate simplex of weight $n$, and therefore factors through $\mathcal{R}(n)$. Let $K \subseteq \Delta ^{2n}$ denote the inverse image $\sigma _{n}^{-1} \mathcal{R}(n-1)$, so that we have a commutative diagram of simplicial sets

Note that a nondegenerate simplex of $\Delta ^{2n}$ belongs to $K$ if and only if it does not contain $\operatorname{N}_{\bullet }( \{ 1 < 3 < \cdots < 2n-1 \} ) \subseteq \Delta ^{2n}$. Applying Lemma 8.5.1.33, we deduce that the inclusion map $K \hookrightarrow \Delta ^{2n}$ is inner anodyne. It will therefore suffice to show that the diagram (8.64) is a pushout square.

Let $\tau $ be an $m$-simplex of $\operatorname{N}_{\bullet }(\operatorname{Ret})$ which belongs to $\mathcal{R}(n)$, but does not belong to $\mathcal{R}(n-1)$. We wish to show that $\tau $ factors uniquely through $\sigma _{n}$. We first prove the existence of the desired factorization. For this, we may assume without loss of generality that $\tau $ is nondegenerate. Then $\tau $ has weight $n$, so we can write

Let $\alpha : \Delta ^ m \rightarrow \Delta ^{2n}$ be the unique morphism of simplices which is given on vertices by the formula $\alpha ( d_ i ) = 2i - 1$ for $1 \leq i \leq n$. We claim that $\tau = \sigma _{n} \circ \alpha $. Note that $\tau $ and $\sigma _ n \circ \alpha $ can both be regarded as functors from the linearly ordered set $[m]$ to the category $\operatorname{Ret}$. By construction, these functors coincide on objects. It will therefore suffice to show that, for $0 \leq j < j' < m$, the functors $\tau $ and $\sigma _{n} \circ \alpha $ determine the same element of $\operatorname{Hom}_{\operatorname{Ret}}( \tau (j), \tau (j') )$. If $\tau (j) = \widetilde{Y}$ or $\tau (j') = \widetilde{Y}$, this condition is automatic (since the set $\operatorname{Hom}_{\operatorname{Ret}}( \tau (j), \tau (j') )$ has only one element). We may therefore assume without loss of generality that $\tau (j) = \widetilde{X} = \tau (j')$: that is, we have $j = d_ i$ and $j' = d_{i'}$ for some $i < i'$. In this case, the functors $\tau $ and $\sigma _{n} \circ \alpha $ both carry the pair $(j < j')$ to the element $e \in \operatorname{Hom}_{\operatorname{Ret}}(X,X)$.

We now prove uniqueness. Suppose we are given a pair of maps $\alpha , \beta : \Delta ^{m} \rightarrow \Delta ^{2n}$ satisfying $\sigma _{n} \circ \alpha = \tau = \sigma _{n} \circ \beta $; we wish to show that $\alpha = \beta $. Suppose otherwise. Then there is some smallest integer $j \in [m]$ such that $\alpha (j) \neq \beta (j)$. Without loss of generality, we may assume that $\alpha (j) < \beta (j)$. Assume first that $\alpha (j)$ is odd. Since $\tau $ does not belong to $K$, $\alpha (j)$ is contained in the image of $\beta $; that is, we can write $\alpha (j) = \beta (i)$ for some $i < j$. Then minimality of $j$ then guarantees that $\alpha (i) = \alpha (j)$, so that $\sigma _{n} \circ \alpha $ carries the pair $(i < j)$ to the identity morphism $\operatorname{id}_{\widetilde{X}}$ in the category $\operatorname{Ret}$. Since $\sigma _{n} \circ \beta = \tau = \sigma _{n} \circ \alpha $, the morphism $\sigma _{n} \circ \beta $ also carries $(i < j)$ to the identity morphism $\operatorname{id}_{\widetilde{X}}$. It follows that $\beta (i) = \beta (j)$, contradicting our assumption that $\beta (i) = \alpha (j) < \beta (j)$.

We now treat the case where $\alpha (j)$ is even, so that $\tau (j) = (\sigma _{n} \circ \alpha )(j) = Y$. Using the equality $\sigma _{n} \circ \beta = \tau $, we deduce that $\beta (j)$ is also even. Since $\tau $ does not belong to $K$, the odd number $\beta (j)-1$ belongs to the image of $\beta $. We therefore have $\beta (j) - 1 = \beta (i)$ for some integer $i < j$. We then have

contradicting the minimality of $j$. $\square$