$\xymatrix{ \mathcal{X}' \ar[r] \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Y}' \ar[r]^ g & \mathcal{Y} }$

be a cartesian diagram of algebraic stacks. If $f$ induces an isomorphism between automorphism groups at points (Remark 101.19.5), then

$\mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{X}') \longrightarrow \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{Y}') \times \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{X})$

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. $\square$

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