Lemma 100.45.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism, then $f$ is representable by algebraic spaces.

## 100.45 Stabilizer preserving morphisms

In the literature a morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is said to be *stabilizer preserving* or *fixed-point reflecting* if the induced morphism $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism. Such a morphism induces an isomorphism between automorphism groups (Remark 100.19.5) in every point of $\mathcal{X}$. In this section we prove some simple lemmas around this concept.

**Proof.**
Immediate from Lemma 100.6.2.
$\square$

Remark 100.45.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $U \to \mathcal{X}$ be a morphism whose source is an algebraic space. Let $G \to H$ be the pullback of the morphism $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ to $U$. If $\Delta _ f$ is unramified, étale, etc, so is $G \to H$. This is true because

is cartesian and the morphism $G \to H$ is the base change of the left vertical arrow by the diagonal $U \to U \times U$. Compare with the proof of Lemma 100.6.6.

Lemma 100.45.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be an unramified morphism of algebraic stacks. The following are equivalent

$\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism, and

$f$ induces an isomorphism between automorphism groups at $x$ and $f(x)$ (Remark 100.19.5) for all $x \in |\mathcal{X}|$.

**Proof.**
Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Denote $G \to H$ the pullback of the morphism $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ to $U$. By Remark 100.45.2 and Lemma 100.36.9 the morphism $G \to H$ is étale. Condition (1) is equivalent to the condition that $G \to H$ is an isomorphism (this follows for example by applying Properties of Stacks, Lemma 99.3.3). Condition (2) is equivalent to the condition that for every $u \in U$ the morphism $G_ u \to H_ u$ of fibres is an isomorphism. Thus (1) $\Rightarrow $ (2) is trivial. If (2) holds, then $G \to H$ is a surjective, universally injective, étale morphism of algebraic spaces. Such a morphism is an isomorphism by Morphisms of Spaces, Lemma 66.51.2.
$\square$

Lemma 100.45.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume

$f$ is representable by algebraic spaces and unramified, and

$\mathcal{I}_\mathcal {Y} \to \mathcal{Y}$ is proper.

Then the set of $x \in |\mathcal{X}|$ such that $f$ induces an isomorphism between automorphism groups at $x$ and $f(x)$ (Remark 100.19.5) is open. Letting $\mathcal{U} \subset \mathcal{X}$ be the corresponding open substack, the morphism $\mathcal{I}_\mathcal {U} \to \mathcal{U} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism.

**Proof.**
Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Denote $G \to H$ the pullback of the morphism $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ to $U$. By Remark 100.45.2 and Lemma 100.36.9 the morphism $G \to H$ is étale. Since $f$ is representable by algebraic spaces, we see that $G \to H$ is a monomorphism. Hence $G \to H$ is an open immersion, see Morphisms of Spaces, Lemma 66.51.2. By assumption $H \to U$ is proper.

With these preparations out of the way, we can prove the lemma as follows. The inverse image of the subset of $|\mathcal{X}|$ of the lemma is clearly the set of $u \in U$ such that $G_ u \to H_ u$ is an isomorphism (since after all $G_ u$ is an open sub group algebraic space of $H_ u$). This is an open subset because the complement is the image of the closed subset $|H| \setminus |G|$ and $|H| \to |U|$ is closed. By Properties of Stacks, Lemma 99.9.12 we can consider the corresponding open substack $\mathcal{U}$ of $\mathcal{X}$. The final statement of the lemma follows from applying Lemma 100.45.3 to $\mathcal{U} \to \mathcal{Y}$. $\square$

Lemma 100.45.5. Let

be a cartesian diagram of algebraic stacks.

Let $x' \in |\mathcal{X}'|$ with image $x \in |\mathcal{X}|$. If $f$ induces an isomorphism between automorphism groups at $x$ and $f(x)$ (Remark 100.19.5), then $f'$ induces an isomorphism between automorphism groups at $x'$ and $f(x')$.

If $\mathcal{I}_\mathcal {X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{I}_\mathcal {Y}$ is an isomorphism, then $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{I}_{\mathcal{Y}'}$ is an isomorphism.

**Proof.**
Omitted.
$\square$

Lemma 100.45.6. Let

be a cartesian diagram of algebraic stacks. If $f$ induces an isomorphism between automorphism groups at points (Remark 100.19.5), then

is injective on isomorphism classes for any field $k$.

**Proof.**
We have to show that given $(y', x)$ there is at most one $x'$ mapping to it. By our construction of $2$-fibre products, a morphism $x'$ is given by a triple $(x, y', \alpha )$ where $\alpha : g \circ y' \to f \circ x$ is a $2$-morphism. Now, suppose we have a second such triple $(x, y', \beta )$. Then $\alpha $ and $\beta $ differ by a $k$-valued point $\epsilon $ of the automorphism group algebraic space $G_{f(x)}$. Since $f$ induces an isomorphism $G_ x \to G_{f(x)}$ by assumption, this means we can lift $\epsilon $ to a $k$-valued point $\gamma $ of $G_ x$. Then $(\gamma , \text{id}) : (x, y', \alpha ) \to (x, y', \beta )$ is an isomorphism as desired.
$\square$

Lemma 100.45.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $f$ is étale, $f$ induces an isomorphism between automorphism groups at points (Remark 100.19.5), and for every algebraically closed field $k$ the functor

is an equivalence. Then $f$ is an isomorphism.

**Proof.**
By Lemma 100.14.5 we see that $f$ is universally injective. Combining Lemmas 100.45.1 and 100.45.3 we see that $f$ is representable by algebraic spaces. Hence $f$ is an open immersion by Morphisms of Spaces, Lemma 66.51.2. To finish we remark that the condition in the lemma also guarantees that $f$ is surjective.
$\square$

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