Example 99.11.6. Here is an example where the morphism constructed in Lemma 99.11.5 isn't an isomorphism. This example shows that imposing that residual gerbes are locally Noetherian is necessary in Definition 99.11.8. In fact, the example is even an algebraic space! Let $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group of $\mathbf{Q}$ with the pro-finite topology. Let

$U = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}}) \times _{\mathop{\mathrm{Spec}}(\mathbf{Q})} \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}}) = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \times \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$

(we omit a precise explanation of the meaning of the last equal sign). Let $G$ denote the absolute Galois group $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with the discrete topology viewed as a constant group scheme over $\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$, see Groupoids, Example 39.5.6. Then $G$ acts freely and transitively on $U$. Let $X = U/G$, see Spaces, Definition 64.14.4. Then $X$ is a non-noetherian reduced algebraic space with exactly one point. Furthermore, $X$ has a (locally) finite type point:

$x : \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}}) \longrightarrow U \longrightarrow X$

Indeed, every point of $U$ is actually closed! As $X$ is an algebraic space over $\overline{\mathbf{Q}}$ it follows that $x$ is a monomorphism. So $x$ is the morphism constructed in Lemma 99.11.5 but $x$ is not an isomorphism. In fact $\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}}) \to X$ is the residual gerbe of $X$ at $x$.

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