**Proof.**
Proof of (1). Choose a scheme $Y$ and a surjective smooth morphism $Y \to \mathcal{Y}$. Set $\mathcal{X}' = \mathcal{X} \times _\mathcal {Y} Y$. By Lemma 101.5.5 we obtain cartesian squares

\[ \xymatrix{ \mathcal{I}_{\mathcal{X}'} \ar[r] \ar[d] & \mathcal{X}' \ar[r] \ar[d] & Y \ar[d] \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r] & \mathcal{Y} } \]

By Lemmas 101.25.4 and 101.27.11 it suffices to prove that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is flat and locally of finite presentation. This follows from Proposition 101.28.9 (because $\mathcal{X}'$ is a gerbe over $Y$ by Lemma 101.28.3).

Proof of (2). With notation as above, note that we may assume that $\mathcal{X}' = [Y/G]$ for some group algebraic space $G$ flat and locally of finite presentation over $Y$, see Lemma 101.28.7. The base change of the morphism $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ over $\mathcal{Y}$ by the morphism $Y \to \mathcal{Y}$ is the morphism $\Delta ' : \mathcal{X}' \to \mathcal{X}' \times _ Y \mathcal{X}'$. Hence it suffices to show that $\Delta '$ is surjective, flat, and locally of finite presentation (see Lemmas 101.25.4 and 101.27.11). In other words, we have to show that

\[ [Y/G] \longrightarrow [Y/G \times _ Y G] \]

is surjective, flat, and locally of finite presentation. This is true because the base change by the surjective, flat, locally finitely presented morphism $Y \to [Y/G \times _ Y G]$ is the morphism $G \to Y$.

Proof of (3). Observe that the diagram

\[ \xymatrix{ T_1 \times _{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times _{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

is cartesian. Hence (3) follows from (2).

Proof of (4). This is true because

\[ \mathit{Isom}_\mathcal {X}(x_1, x_2) = (T \times _{x_1, \mathcal{X}, x_2} T) \times _{T \times T, \Delta _ T} T \]

hence the morphism in (4) is a base change of the morphism in (3).
$\square$

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