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

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

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 (48.4.1.1) 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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