Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 8.5.9.11. The homomorphism $\varphi : E_{\bullet } \rightarrow \operatorname{N}_{\bullet }( \mathrm{Dy} )$ is a weak homotopy equivalence of simplicial sets.

Proof of Proposition 8.5.9.11 from Proposition 8.5.9.15. By virtue of Proposition 8.5.9.15, it will suffice to show that the inclusion of categories $\mathrm{Dy}_{+} \hookrightarrow \mathrm{Dy}$ induces a weak homotopy equivalence of simplicial sets $U: \operatorname{N}_{\bullet }( \mathrm{Dy}_{+} ) \hookrightarrow \operatorname{N}_{\bullet }( \mathrm{Dy} )$. Using Quillen's Theorem A (Example 7.2.3.3), we are reduced to proving the following: for every object $[0,s] \in \mathrm{Dy}$, the category $\operatorname{\mathcal{A}}= \mathrm{Dy}_{+} \times _{ \mathrm{Dy} } \mathrm{Dy}_{ [0,s] / }$ has weakly contractible nerve. We can describe the category $\operatorname{\mathcal{A}}$ more concretely as follows:

  • The objects of $\operatorname{\mathcal{A}}$ are dyadic homeomorphisms $f: [0,s] \xrightarrow {\sim } [0,m]$, where $m$ is an integer.

  • Let $f: [0,s] \xrightarrow {\sim } [0,m]$ and $g: [0,s] \xrightarrow {\sim } [0,n]$ be dyadic homeomorphisms. Then there is a morphism from $f$ to $g$ (in the category $\operatorname{\mathcal{A}}$) if and only if the homeomorphism $(g \circ f^{-1}): [0,m] \rightarrow [0,n]$ is a dyadic contraction. If this condition is satisfied, then the morphism is unique.

It follows that the category $\operatorname{\mathcal{A}}$ can be viewed as a partially ordered set. Moreover, every finite subset of $\operatorname{\mathcal{A}}$ has a lower bound, given by the dyadic homeomorphism

\[ [0,s] \xrightarrow {\sim } [0, 2^ k s] \quad \quad x \mapsto 2^{k} x \]

for some integer $k \gg 0$. Consequently, the simplicial set $\operatorname{N}_{\bullet }( \operatorname{\mathcal{A}}^{\operatorname{op}} )$ is a filtered $\infty $-category (Definition 9.1.1.1 and Example 9.1.1.2), and therefore weakly contractible (see Proposition 9.1.1.13). $\square$