ScalingStacks

Proof. Assume first $\tilde{\xi}$ is a homeomorphism and $Z_{o}=\emptyset$. Let $T$ and $T^{\prime}$ be two finite subsets of ${\mathbf{R}}$ with same cardinality $m$. Let $a:\{1,\ldots,m\}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}T$ and $a^{\prime}:\{1,\ldots,m\}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}T^{\prime}$ be the increasing bijections. There is an isomorphism of differential modules (Proposition 7.4.33) $\phi:\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z)}(T,T^{\prime})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}H_{m}$: given $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(T,T^{\prime})$ non-zero and given $i\in\{1,\ldots,m\}$, we put $\phi(\theta)(i)=a^{\prime-1}(\theta_{a(i)}(1))$.

Assume in addition that $\{-n,\ldots,-1\}\subset T$ and $T\setminus\{-n,\ldots,-1\}\subset(-1,\infty)$ and $\{1,\ldots,n\}\subset T^{\prime}$ and $T^{\prime}\setminus\{1,\ldots,n\}\subset(-\infty,1)$. There is a commutative diagram

The lemma follows now from §6.1.1.

Assume now $Z$ is smooth. If $Z(\xi^{+})$ is unoriented, then the lemma holds by the discussion above, using §7.4.10. In general, we consider the morphism of curves $f:Z\to\bar{Z}$ that is an isomorphism outside $Z(\xi^{+})$ and the identity on $Z(\xi^{+})$, with $f(Z(\xi^{+}))_{o}=\emptyset$. The vertical maps of the commutative diagram (8.1.6) are injective, hence the lemma holds for $Z$ since it holds for $\bar{Z}$.

Consider now a general $Z$. Let $f:\hat{Z}\to Z$ be a non-singular cover. The vertical maps of the commutative diagram (8.1.7) are injective, hence the lemma holds for $Z$ since it holds for $\hat{Z}$. ∎