Lemma 99.4.1. The notion above does indeed define an equivalence relation on morphisms from spectra of fields into the algebraic stack $\mathcal{X}$.

**Proof.**
It is clear that the relation is reflexive and symmetric. Hence we have to prove that it is transitive. This comes down to the following: Given a diagram

with both squares $2$-commutative we have to show that $p$ is equivalent to $p'$. By the $2$-Yoneda lemma (see Algebraic Stacks, Section 93.5) the morphisms $p$, $p'$, and $q$ are given by objects $x$, $x'$, and $y$ in the fibre categories of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(K)$, $\mathop{\mathrm{Spec}}(K')$, and $\mathop{\mathrm{Spec}}(L)$. The $2$-commutativity of the squares means that there are isomorphisms $\alpha : a^*x \to b^*y$ and $\alpha ' : (a')^*x' \to (b')^*y$ in the fibre categories of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(\Omega )$ and $\mathop{\mathrm{Spec}}(\Omega ')$. Choose any field $\Omega ''$ and embeddings $\Omega \to \Omega ''$ and $\Omega ' \to \Omega ''$ agreeing on $L$. Then we can extend the diagram above to

with commutative triangles and

is an isomorphism in the fibre category over $\mathop{\mathrm{Spec}}(\Omega '')$. Hence $p$ is equivalent to $p'$ as desired. $\square$

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