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.
$\square$

## Comments (0)

There are also: