Lemma 100.4.1. 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
\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r]_ b \ar[d]_ a & \mathop{\mathrm{Spec}}(L) \ar[d]^ q & \mathop{\mathrm{Spec}}(\Omega ') \ar[l]^{b'} \ar[d]^{a'} \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X} & \mathop{\mathrm{Spec}}(K') \ar[l]_{p'} } \]
with both squares $2$-commutative we have to show that $p$ is equivalent to $p'$. By the $2$-Yoneda lemma (see Algebraic Stacks, Section 94.5) 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
\[ \xymatrix{ & \mathop{\mathrm{Spec}}(\Omega '') \ar[ld]_ c \ar[d]^{q'} \ar[rd]^{c'} \\ \mathop{\mathrm{Spec}}(\Omega ) \ar[r]_ b \ar[d]_ a & \mathop{\mathrm{Spec}}(L) \ar[d]^ q & \mathop{\mathrm{Spec}}(\Omega ') \ar[l]^{b'} \ar[d]^{a'} \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X} & \mathop{\mathrm{Spec}}(K') \ar[l]_{p'} } \]
with commutative triangles and
\[ (q')^*(\alpha ')^{-1} \circ (q')^*\alpha : (a \circ c)^*x \longrightarrow (a' \circ c')^*x' \]
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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