Lemma 14.21.7. Let $U \subset V$ be simplicial sets. Suppose $n \geq 0$ and $x \in V_ n$, $x \not\in U_ n$ are such that

$V_ i = U_ i$ for $i < n$,

$V_ n = U_ n \cup \{ x\} $,

any $z \in V_ j$, $z \not\in U_ j$ for $j > n$ is degenerate.

Let $\Delta [n] \to V$ be the unique morphism mapping the nondegenerate $n$-simplex of $\Delta [n]$ to $x$. In this case the diagram

\[ \xymatrix{ \Delta [n] \ar[r] & V \\ i_{(n - 1)!} \text{sk}_{n - 1} \Delta [n] \ar[r] \ar[u] & U \ar[u] } \]

is a pushout diagram.

**Proof.**
Let us denote $\partial \Delta [n] = i_{(n - 1)!} \text{sk}_{n - 1} \Delta [n]$ for convenience. There is a natural map $U \amalg _{\partial \Delta [n]} \Delta [n] \to V$. We have to show that it is bijective in degree $j$ for all $j$. This is clear for $j \leq n$. Let $j > n$. The third condition means that any $z \in V_ j$, $z \not\in U_ j$ is a degenerate simplex, say $z = s^{j - 1}_ i(z')$. Of course $z' \not\in U_{j - 1}$. By induction it follows that $z'$ is a degeneracy of $x$. Thus we conclude that all $j$-simplices of $V$ are either in $U$ or degeneracies of $x$. This implies that the map $U \amalg _{\partial \Delta [n]} \Delta [n] \to V$ is surjective. Note that a nondegenerate simplex of $U \amalg _{\partial \Delta [n]} \Delta [n]$ is either the image of a nondegenerate simplex of $U$, or the image of the (unique) nondegenerate $n$-simplex of $\Delta [n]$. Since clearly $x$ is nondegenerate we deduce that $U \amalg _{\partial \Delta [n]} \Delta [n] \to V$ maps nondegenerate simplices to nondegenerate simplices and is injective on nondegenerate simplices. Hence it is injective, by Lemma 14.18.3.
$\square$

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