# Kerodon

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Proposition 1.2.3.10. Let $n$ be a nonnegative integer and let $\operatorname{\mathcal{U}}$ be a downward closed collection of nonempty subsets of $[n]$. Then the canonical map $\Delta ^{n} \rightarrow \operatorname{Sing}_{\bullet }( | \Delta ^{n} | )$ restricts to a map of simplicial sets $f_{\operatorname{\mathcal{U}}}: \Delta ^ n_{\operatorname{\mathcal{U}}} \rightarrow \operatorname{Sing}_{\bullet } ( | \Delta ^{n} |_{\operatorname{\mathcal{U}}} )$, which exhibits the topological space $| \Delta ^{n} |_{\operatorname{\mathcal{U}}}$ as a geometric realization of $\Delta ^{n}_{\operatorname{\mathcal{U}}}$.

Proof. We proceed by induction on the cardinality of $\operatorname{\mathcal{U}}$. If $\operatorname{\mathcal{U}}$ is empty, then the simplicial set $\Delta ^{n}_{\operatorname{\mathcal{U}}}$ and the topological space $| \Delta ^{n} |_{\operatorname{\mathcal{U}}}$ are both empty, in which case there is nothing to prove. We may therefore assume that $\operatorname{\mathcal{U}}$ is nonempty. Choose some $S \in \operatorname{\mathcal{U}}$ whose cardinality is as large as possible. Set

$\operatorname{\mathcal{U}}_0 = \operatorname{\mathcal{U}}\setminus \{ S \} \quad \quad \operatorname{\mathcal{U}}_1 = \{ T \subseteq S: T \neq \emptyset \} \quad \quad \operatorname{\mathcal{U}}_{01} = \operatorname{\mathcal{U}}_0 \cap \operatorname{\mathcal{U}}_1.$

Our inductive hypothesis implies that the maps $f_{\operatorname{\mathcal{U}}_0}$ and $f_{ \operatorname{\mathcal{U}}_{01} }$ exhibit $| \Delta ^{n} |_{\operatorname{\mathcal{U}}_0}$ and $| \Delta ^{n} |_{\operatorname{\mathcal{U}}_{01}}$ as geometric realizations of $\Delta ^{n}_{\operatorname{\mathcal{U}}_0}$ and $\Delta ^{n}_{ \operatorname{\mathcal{U}}_{01} }$, respectively. Moreover, if $S = \{ i_0 < i_1 < \cdots < i_ m \} \subseteq [n]$, then we can identify $f_{\operatorname{\mathcal{U}}_1}$ with the tautological map $\Delta ^{m} \rightarrow \operatorname{Sing}_{\bullet } ( | \Delta ^{m} | )$, so that $f_{ \operatorname{\mathcal{U}}_1}$ exhibits $| \Delta ^{n} |_{\operatorname{\mathcal{U}}_1}$ as a geometric realization of $\Delta ^{n}_{\operatorname{\mathcal{U}}_1}$ by virtue of Example 1.2.3.2. It follows immediately from the definitions that the diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \Delta ^{n}_{\operatorname{\mathcal{U}}_{01}} \ar [r] \ar [d] & \Delta ^{n}_{\operatorname{\mathcal{U}}_0} \ar [d] \\ \Delta ^{n}_{\operatorname{\mathcal{U}}_1} \ar [r] & \Delta ^{n}_{\operatorname{\mathcal{U}}} }$

is a pushout square. By virtue of Lemma 1.2.3.6, we are reduced to proving that the diagram of topological spaces

$\xymatrix@R =50pt@C=50pt{ |\Delta ^{n}|_{\operatorname{\mathcal{U}}_{01}} \ar [r] \ar [d] & |\Delta ^{n}|_{\operatorname{\mathcal{U}}_0} \ar [d] \\ |\Delta ^{n}|_{\operatorname{\mathcal{U}}_1} \ar [r] & |\Delta ^{n}|_{\operatorname{\mathcal{U}}} }$

is also a pushout square. This is clear, since $| \Delta ^{n} |_{\operatorname{\mathcal{U}}_0}$ and $| \Delta ^{n} |_{\operatorname{\mathcal{U}}_1}$ are closed subsets of $| \Delta ^{n} |$ whose union is $| \Delta ^{n} |_{\operatorname{\mathcal{U}}}$ and whose intersection is $| \Delta ^{n} |_{\operatorname{\mathcal{U}}_{01} }$. $\square$