# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

Lemma 3.1.2.9. Let $n$ be a nonnegative integer. Then there exists a chain of simplicial subsets

$X(0) \subset X(1) \subset \cdots \subset X(n) \subset X(n+1) = \Delta ^1 \times \Delta ^ n$

with the following properties:

$(a)$

The simplicial $X(0)$ is given by the union of $\Delta ^1 \times \operatorname{\partial \Delta }^ n$ with $\{ 1\} \times \Delta ^ n$ (and can therefore be described abstractly as the pushout $(\Delta ^1 \times \operatorname{\partial \Delta }^ n) \coprod _{ \{ 1\} \times \operatorname{\partial \Delta }^ n} ( \{ 1 \} \times \Delta ^ n )$).

$(b)$

For $0 \leq i \leq n$, the inclusion map $X(i) \hookrightarrow X(i+1)$ fits into a pushout diagram

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{n+1}_{i+1} \ar [r] \ar [d] & X(i) \ar [d] \\ \Delta ^{n+1} \ar [r] & X(i+1). }$

Proof. For $0 \leq i \leq n$, let $\sigma _ i: \Delta ^{n+1} \rightarrow \Delta ^1 \times \Delta ^ n$ denote the map of simplicial sets given on vertices by the formula $\sigma _ i(j) = \begin{cases} (0,j) & \text{ if } j \leq i \\ (1,j-1) & \text{ if } j > i. \end{cases}$ We define simplicial subsets $X(i) \subseteq \Delta ^1 \times \Delta ^ n$ inductively by the formulae

$X(0) = (\Delta ^1 \times \operatorname{\partial \Delta }^ n) \cup (\{ 1\} \times \Delta ^ n) \quad \quad X(i+1) = X(i) \cup \operatorname{im}( \sigma _{i} ),$

where $\operatorname{im}( \sigma _{i} )$ denotes the image of the morphism $\sigma _{i}$. Note that $\Delta ^1 \times \Delta ^ n$ is the union of the simplicial subsets $\{ \operatorname{im}(\sigma _ i) \} _{0 \leq i \leq n}$, and is therefore equal to $X(n+1)$. This definition satisfies condition $(a)$ by construction. To verify $(b)$, it will suffice to show that for $0 \leq i \leq n$, the inverse image $A_{} = \sigma _{i}^{-1} X(i)$ is equal to $\Lambda ^{n+1}_{i+1}$ (as a simplicial subset of $\Delta ^{n+1}$). Regarding $\sigma _{i}$ as an $(n+1)$-simplex of $\Delta ^1 \times \Delta ^ n$, we are reduced to showing that the faces $d_{j}( \sigma _{i} )$ belong to $X(i)$ if and only if $j \neq i+1$. One direction is clear: the face $d_{j}( \sigma _{i} )$ is contained in $\Delta ^1 \times \operatorname{\partial \Delta }^ n$ for $j \notin \{ i, i+1 \}$, the face $d_{i}(\sigma _ i) = d_{i}( \sigma _{i-1} )$ is contained in $\operatorname{im}( \sigma _{i-1} ) \subseteq X(i)$ for $i > 0$, and $d_0( \sigma _0)$ is contained in $\{ 1\} \times \Delta ^ n$. To complete the proof, it suffices to show that the face $d_{i+1}( \sigma _ i )$ is not contained in $X(i)$, which follows by inspection. $\square$