The Stacks project

Lemma 48.16.4. In Situation 48.16.1 the constructions of Lemmas 48.16.2 and 48.16.3 define a pseudo functor from the category $\textit{FTS}_ S$ into the $2$-category of categories (see Categories, Definition 4.29.5).

Proof. To show this we have to prove given morphisms $f : X \to Y$, $g : Y \to Z$, $h : Z \to T$ that

\[ \xymatrix{ (h \circ g \circ f)^! \ar[r]_{\gamma _{A + B}} \ar[d]_{\gamma _{B + C}} & f^! \circ (h \circ g)^! \ar[d]^{\gamma _ C} \\ (g \circ f)^! \circ h^! \ar[r]^{\gamma _ A} & f^! \circ g^! \circ h^! } \]

is commutative (for the meaning of the $\gamma $'s, see below). To do this we choose a compactification $\overline{Z}$ of $Z$ over $T$, then a compactification $\overline{Y}$ of $Y$ over $\overline{Z}$, and then a compactification $\overline{X}$ of $X$ over $\overline{Y}$. This uses More on Flatness, Theorem 38.33.8 and Lemma 38.32.2. Let $W \subset \overline{Y}$ be the inverse image of $Z$ under $\overline{Y} \to \overline{Z}$ and let $U \subset V \subset \overline{X}$ be the inverse images of $Y \subset W$ under $\overline{X} \to \overline{Y}$. This produces the following diagram

\[ \xymatrix{ X \ar[d]_ f \ar[r] & U \ar[r] \ar[d] \ar@{}[dr]|A & V \ar[d] \ar[r] \ar@{}[rd]|B & \overline{X} \ar[d] \\ Y \ar[d]_ g \ar[r] & Y \ar[r] \ar[d] & W \ar[r] \ar[d] \ar@{}[rd]|C & \overline{Y} \ar[d] \\ Z \ar[d]_ h \ar[r] & Z \ar[d] \ar[r] & Z \ar[d] \ar[r] & \overline{Z} \ar[d] \\ T \ar[r] & T \ar[r] & T \ar[r] & T } \]

Without introducing tons of notation but arguing exactly as in the proof of Lemma 48.16.3 we see that the maps in the first displayed diagram use the maps ( for the rectangles $A + B$, $B + C$, $A$, and $C$ as indicated. Since by Lemmas 48.5.1 and 48.5.2 we have $\gamma _{A + B} = \gamma _ A \circ \gamma _ B$ and $\gamma _{B + C} = \gamma _ C \circ \gamma _ B$ we conclude that the desired equality holds provided $\gamma _ A \circ \gamma _ C = \gamma _ C \circ \gamma _ A$. This is true because the two squares $A$ and $C$ only intersect in one point (similar to the last argument in Remark 48.5.3). $\square$

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