Lemma 101.39.4. Assume given a 2-commutative diagram
\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-{x'} \ar[d]_ j & \mathcal{X}' \ar[d]^ p \ar[r]_ q & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-{y'} & \mathcal{Y}' \ar[r]^ g & \mathcal{Y} }
with the right square 2-cartesian. Choose a 2-arrow \gamma ' : y' \circ j \to p \circ x'. Set x = q \circ x', y = g \circ y' and let \gamma : y \circ j \to f \circ x be the composition of \gamma ' with the 2-arrow implicit in the 2-commutativity of the right square. Then the category of dotted arrows for the left square and \gamma ' is equivalent to the category of dotted arrows for the outer rectangle and \gamma .
Proof.
(We do not know how to prove the analogue of this lemma if instead of the category of dotted arrows we look at the set of isomorphism classes of morphisms producing two 2-commutative triangles as in Lemma 101.39.3; in fact this analogue may very well be wrong.) First proof: this lemma is a special case of Categories, Lemma 4.44.2. Second proof: we are allowed to replace \mathcal{X}' by the 2-fibre product \mathcal{Y}' \times _\mathcal {Y} \mathcal{X} as described in Categories, Lemma 4.32.3. Then the object x' becomes the triple (y' \circ j, x, \gamma ). Then we can go from a dotted arrow (a, \alpha , \beta ) for the outer rectangle to a dotted arrow (a', \alpha ', \beta ') for the left square by taking a' = (y', a, \beta ) and \alpha ' = (\text{id}_{y' \circ j}, \alpha ) and \beta ' = \text{id}_{y'}. Details omitted.
\square
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