Kerodon

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 (8.67) 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 $j$. $\square$