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Tag 0CLE

Chapter 91: Morphisms of Algebraic Stacks > Section 91.38: Valuative criteria

Lemma 91.38.4. Assume given a $2$-commutative diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-{x'} \ar[d]_j & \mathcal{X}' \ar[d]^p \ar[r]_q & \mathcal{X} \ar[d]^f \\ \mathop{\rm Spec}(A) \ar[r]^-{y'} & \mathcal{Y}' \ar[r]^g & \mathcal{Y} } $$ 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. This lemma, although a bit of a brain teaser, is straightforward. (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 91.38.3; in fact this analogue may very well be wrong.) To prove the lemma we are allowed to replace $\mathcal{X}'$ by the $2$-fibre product $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$ as described in Categories, Lemma 4.31.3. 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. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 8709–8730 (see updates for more information).

    \begin{lemma}
    \label{lemma-cat-dotted-arrows-base-change}
    Assume given a $2$-commutative diagram
    $$
    \xymatrix{
    \Spec(K) \ar[r]_-{x'} \ar[d]_j &
    \mathcal{X}' \ar[d]^p \ar[r]_q &
    \mathcal{X} \ar[d]^f \\
    \Spec(A) \ar[r]^-{y'} &
    \mathcal{Y}' \ar[r]^g &
    \mathcal{Y}
    }
    $$
    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 $\gamma$.
    \end{lemma}
    
    \begin{proof}
    This lemma, although a bit of a brain teaser, is straightforward.
    (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 \ref{lemma-cat-dotted-arrows-independent};
    in fact this analogue may very well be wrong.)
    To prove the lemma we are allowed to replace
    $\mathcal{X}'$ by the $2$-fibre product
    $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$
    as described in Categories, Lemma
    \ref{categories-lemma-2-product-categories-over-C}.
    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.
    \end{proof}

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