ScalingStacks

Remark 8.1.1. Assume $\xi$ is not outgoing for $Z$ and let $z_{0}\in Z$ such that $\overline{\xi({\mathbf{R}}_{\geq 1})}\setminus\xi({\mathbf{R}}_{\geq 1})=\{z_{0}\}$. Note that $\xi$ is outgoing for $Z\setminus\{z_{0}\}$. The map $\xi$ is terminal for $(Z,M)$ if and only if $z_{0}{\not\in M}$ and the inclusion induces an isomorphism $\operatorname{Hom}\nolimits_{{\mathcal{A}}^{\bullet}(Z\setminus\{z_{0}\},1)}(m,z)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{A}}^{\bullet}(Z,1)}(m,z)$ for all $m\in M$ and $z\in M\cup\{\xi(1)\}$.