ScalingStacks

Proof. Lemma 8.1.14 shows that $\hat{\kappa}$ commutes with differentials.

Let $S$ be a finite subset of $M$ of cardinality $n$.

Assume $\tilde{\xi}$ is a homeomorphism and $Z_{o}=\emptyset$. There is a commutative diagram (see the proof of Lemma 8.1.14 with $(T,T^{\prime})=(S,\xi^{+}(\{1,\ldots,n\})$ and $(T,T^{\prime})=(\xi^{-}(\{-n,\ldots,-1\}),S)$)

The bottom horizontal map is bijective by Corollary 3.1.2, hence $\hat{\kappa}(\emptyset,S)$ is bijective.

Assume now $Z(\xi^{+})$ is smooth and unoriented. The map $\hat{\kappa}(\emptyset,S)$ is the same for $Z$ and for $Z(\xi^{+})$, so $\hat{\kappa}(\emptyset,S)$ is still bijective.

Assume $Z(\xi^{+})$ is smooth. There is a 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 map $\hat{\kappa}(\emptyset,S)$ is the same for $Z$ and for $\bar{Z}$, so $\hat{\kappa}(\emptyset,S)$ is still bijective.

Consider now a general $Z$ and let $f:\hat{Z}\to Z$ be a non-singular cover. Let $\tilde{\hat{\xi}}:Z^{\prime}\to\hat{Z}$ be the morphism of curves such that $\tilde{\xi}=f\circ\tilde{\hat{\xi}}$. The functors $f$ and $f^{\#}$ are inverse bijections between $\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z)}(S,\xi^{+}(\{1,\ldots,n\})$ and $\bigoplus_{S^{\prime}}\operatorname{Hom}\nolimits_{{\mathcal{S}}(\hat{Z})}(S^{\prime},\tilde{\hat{\xi}}(\{1,\ldots,n\})$ (resp. $\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z)}(\xi^{-}(\{-n,\ldots,-1\},S)$ and $\bigoplus_{S^{\prime}}\operatorname{Hom}\nolimits_{{\mathcal{S}}(\hat{Z})}(\tilde{\hat{\xi}}(\{-n,\ldots,-1\},S^{\prime})$), where $S^{\prime}$ runs over $n$-elements subsets of $\hat{Z}$ such that $f(S^{\prime})=S$. Furthermore, $\hat{\kappa}(\emptyset,S)$ is compatible with these bijections (see the proof of Lemma 8.1.14). It follows that $\hat{\kappa}(\emptyset,S)$ is bijective.

We consider now two arbitrary subsets $S$ and $T$ of $M$. The canonical isomorphisms of Lemma 8.1.2 and of §8.1.5 fit in a commutative diagram of ${\mathbf{F}}_{2}$-modules

where $S^{\prime}$ runs over $n$ elements subsets of $S$. The discussion above shows that the left vertical arrow is an isomorphism, hence $\hat{\kappa}(T,S)$ is an isomorphism. ∎