Lemma 106.3.3. Let $U_1 \to \mathcal{X}$ and $U_2 \to \mathcal{X}$ be two smooth morphisms from schemes to a locally Noetherian algebraic stack $\mathcal{X}$. Let $T_1'$ and $T_2'$ be irreducible components of $|U_1|$ and $|U_2|$ respectively. Assume the closures of the images of $T_1'$ and $T_2'$ are the same irreducible component $T$ of $|\mathcal{X}|$. Then $m_{T_1', U_1} = m_{T_2', U_2}$.

**Proof.**
Let $V_1$ and $V_2$ be dense subsets of $T_1'$ and $T'_2$, respectively, that are open in $U_1$ and $U_2$ respectively (see proof of Lemma 106.3.1). The images of $|V_1|$ and $|V_2|$ in $|\mathcal{X}|$ are non-empty open subsets of the irreducible subset $T$, and therefore have non-empty intersection. By Properties of Stacks, Lemma 99.4.3, the map $|V_1 \times _\mathcal {X} V_2| \to |V_1| \times _{|\mathcal{X}|} |V_2|$ is surjective. Consequently $V_1 \times _\mathcal {X} V_2$ is a non-empty algebraic space; we may therefore choose an étale surjection $V \to V_1 \times _\mathcal {X} V_2$ whose source is a (non-empty) scheme. If we let $T'$ be any irreducible component of $V$, then Lemma 106.3.1 shows that the closure of the image of $T'$ in $U_1$ (respectively $U_2$) is equal to $T'_1$ (respectively $T'_2$).

Applying Lemma 106.3.2 twice we find that

as required. $\square$

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