## 107.6 Open substacks of the stack of curves

Below we will often characterize an open substack of $\mathcal{C}\! \mathit{urves}$ by a propery $P$ of morphisms of algebraic spaces. To see that $P$ defines an open substack it suffices to check

1. given a family of curves $f : X \to S$ there exists a largest open subscheme $S' \subset S$ such that $f|_{f^{-1}(S')} : f^{-1}(S') \to S'$ has $P$ and such that formation of $S'$ commutes with arbitrary base change.

Namely, suppose (o) holds. Choose a scheme $U$ and a surjective smooth morphism $m : U \to \mathcal{C}\! \mathit{urves}$. Let $R = U \times _{\mathcal{C}\! \mathit{urves}} U$ and denote $t, s : R \to U$ the projections. Recall that $\mathcal{C}\! \mathit{urves}= [U/R]$ is a presentation, see Algebraic Stacks, Lemma 92.16.2 and Definition 92.16.5. By construction of $\mathcal{C}\! \mathit{urves}$ as the stack of curves, the morphism $m$ is the classifying morphism for a family of curves $C \to U$. The $2$-commutativity of the diagram

$\xymatrix{ R \ar[r]_ s \ar[d]_ t & U \ar[d] \\ U \ar[r] & \mathcal{C}\! \mathit{urves}}$

implies that $C \times _{U, s} R \cong C \times _{U, t} R$ (isomorphism of families of curves over $R$). Let $W \subset U$ be the largest open subscheme such that $f|_{f^{-1}(W)} : f^{-1}(W) \to W$ has $P$ as in (o). Since formation of $W$ commutes with base change according to (o) and by the isomorphism above we find that $s^{-1}(W) = t^{-1}(W)$. Thus $W \subset U$ corresponds to an open substack

$\mathcal{C}\! \mathit{urves}^ P \subset \mathcal{C}\! \mathit{urves}$

according to Properties of Stacks, Lemma 98.9.7.

Continuing with the setup of the previous paragrpah, we claim the open substack $\mathcal{C}\! \mathit{urves}^ P$ has the following two universal properties:

1. given a family of curves $X \to S$ the following are equivalent

1. the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^ P$,

2. the morphism $X \to S$ has $P$,

2. given $X$ a proper scheme over a field $k$ of dimension $\leq 1$ the following are equivalent

1. the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^ P$,

2. the morphism $X \to \mathop{\mathrm{Spec}}(k)$ has $P$.

This follows by considering the $2$-fibre product

$\xymatrix{ T \ar[r]_ p \ar[d]_ q & U \ar[d] \\ S \ar[r] & \mathcal{C}\! \mathit{urves}}$

Observe that $T \to S$ is surjective and smooth as the base change of $U \to \mathcal{C}\! \mathit{urves}$. Thus the open $S' \subset S$ given by (o) is determined by its inverse image in $T$. However, by the invariance under base change of these opens in (o) and because $X \times _ S T \cong C \times _ U T$ by the $2$-commutativity, we find $q^{-1}(S') = p^{-1}(W)$ as opens of $T$. This immediately implies (1). Part (2) is a special case of (1).

Given two properties $P$ and $Q$ of morphisms of algebraic spaces, supposing we already have esthablished $\mathcal{C}\! \mathit{urves}^ Q$ is an open substack of $\mathcal{C}\! \mathit{urves}$, then we can use exactly the same method to prove openness of $\mathcal{C}\! \mathit{urves}^{Q, P} \subset \mathcal{C}\! \mathit{urves}^ Q$. We omit a precise explanation.

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