# Kerodon

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

Lemma 3.1.2.11. Let $X$ be a simplicial set which is the union of a simplicial subset $Y \subseteq X$ with the image of an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$, where $n > 0$. Suppose that the inverse image $\sigma ^{-1}(Y) \subseteq \Delta ^ n$ is equal to the horn $\Lambda ^{n}_{i}$ for some $0 \leq i \leq n$. Then pullback diagram of simplicial sets

3.3
$$\begin{gathered}\label{equation:build-using-horn} \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [r] \ar [d] & \Delta ^ n \ar [d]^{\sigma } \\ Y \ar [r] & X } \end{gathered}$$

is also a pushout square.

Proof. Fix an integer $m \geq 0$. We wish to show that $\sigma$ induces a bijection from the set of $m$-simplices of $\Delta ^{n}$ which are not contained in $\Lambda ^{n}_{i}$ to the set of $m$-simplices of $X$ which are not contained in $Y$. Surjectivity follows from our assumption that $X$ is the union of $Y$ with the image of $\sigma$. To prove injectivity, we proceed by induction on $m$. Let $\alpha ,\beta : \Delta ^{m} \rightarrow \Delta ^{n}$ be morphisms which do not factor through $\Lambda ^{n}_{i}$, and suppose that $\sigma \circ \alpha = \tau = \sigma \circ \beta$ for some simplex $\tau : \Delta ^ m \rightarrow X$; we wish to show that $\alpha = \beta$.

Suppose first that the simplex $\alpha$ is degenerate: that is, we have $\alpha (j) = \alpha (k)$ for some $0 \leq j < k \leq m$. Then $d^{m}_ j(\alpha )$ is an $(m-1)$-simplex of $\Delta ^ n$ which is not contained $\Lambda ^{n}_{i}$. It follows that $\sigma \circ d^{m}_ j(\alpha ) = d^{m}_ j(\tau ) = \sigma \circ d^{m}_ j(\beta )$ is an $(m-1)$-simplex of $X$ which is not contained in $Y$, so that $d^{m}_ j(\beta )$ is not contained in $\Lambda ^{n}_{i}$. Applying our inductive hypothesis, we deduce that $d^{m}_ j(\alpha ) = d^{m}_ j(\beta )$. The same argument shows that $d^{m}_ k(\alpha ) = d^{m}_ k(\beta )$, so that $\alpha = \beta$.

We may therefore assume without loss of generality that the simplex $\alpha$ is nondegenerate. By a similar argument, we may assume that $\beta$ is nondegenerate. The equality $\alpha = \beta$ now follows from the observation that $\Delta ^ n$ contains at most one nondegenerate $m$-simplex which is not contained in $\Lambda ^{n}_{i}$. $\square$