Kerodon

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 $\Lambda ^{n}_{i}$. $\square$