Remark 48.5.3. Consider a commutative diagram

$\xymatrix{ X'' \ar[r]_{k'} \ar[d]_{f''} & X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{l'} \ar[d]_{g''} & Y' \ar[r]^ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z'' \ar[r]^{m'} & Z' \ar[r]^ m & Z }$

of quasi-compact and quasi-separated schemes where all squares are cartesian and where $(f, l)$, $(g, m)$, $(f', l')$, $(g', m')$ are Tor independent pairs of maps. Let $a$, $a'$, $a''$, $b$, $b'$, $b''$ be the right adjoints of Lemma 48.3.1 for $f$, $f'$, $f''$, $g$, $g'$, $g''$. Let us label the squares of the diagram $A$, $B$, $C$, $D$ as follows

$\begin{matrix} A & B \\ C & D \end{matrix}$

Then the maps (48.5.0.1) for the squares are (where we use $k^* = Lk^*$, etc)

$\begin{matrix} \gamma _ A : (k')^* \circ a' \to a'' \circ (l')^* & \gamma _ B : k^* \circ a \to a' \circ l^* \\ \gamma _ C : (l')^* \circ b' \to b'' \circ (m')^* & \gamma _ D : l^* \circ b \to b' \circ m^* \end{matrix}$

For the $2 \times 1$ and $1 \times 2$ rectangles we have four further base change maps

$\begin{matrix} \gamma _{A + B} : (k \circ k')^* \circ a \to a'' \circ (l \circ l')^* \\ \gamma _{C + D} : (l \circ l')^* \circ b \to b'' \circ (m \circ m')^* \\ \gamma _{A + C} : (k')^* \circ (a' \circ b') \to (a'' \circ b'') \circ (m')^* \\ \gamma _{B + D} : k^* \circ (a \circ b) \to (a' \circ b') \circ m^* \end{matrix}$

By Lemma 48.5.2 we have

$\gamma _{A + B} = \gamma _ A \circ \gamma _ B, \quad \gamma _{C + D} = \gamma _ C \circ \gamma _ D$

and by Lemma 48.5.1 we have

$\gamma _{A + C} = \gamma _ C \circ \gamma _ A, \quad \gamma _{B + D} = \gamma _ D \circ \gamma _ B$

Here it would be more correct to write $\gamma _{A + B} = (\gamma _ A \star \text{id}_{l^*}) \circ (\text{id}_{(k')^*} \star \gamma _ B)$ with notation as in Categories, Section 4.28 and similarly for the others. However, we continue the abuse of notation used in the proofs of Lemmas 48.5.1 and 48.5.2 of dropping $\star$ products with identities as one can figure out which ones to add as long as the source and target of the transformation is known. Having said all of this we find (a priori) two transformations

$(k')^* \circ k^* \circ a \circ b \longrightarrow a'' \circ b'' \circ (m')^* \circ m^*$

namely

$\gamma _ C \circ \gamma _ A \circ \gamma _ D \circ \gamma _ B = \gamma _{A + C} \circ \gamma _{B + D}$

and

$\gamma _ C \circ \gamma _ D \circ \gamma _ A \circ \gamma _ B = \gamma _{C + D} \circ \gamma _{A + B}$

The point of this remark is to point out that these transformations are equal. Namely, to see this it suffices to show that

$\xymatrix{ (k')^* \circ a' \circ l^* \circ b \ar[r]_{\gamma _ D} \ar[d]_{\gamma _ A} & (k')^* \circ a' \circ b' \circ m^* \ar[d]^{\gamma _ A} \\ a'' \circ (l')^* \circ l^* \circ b \ar[r]^{\gamma _ D} & a'' \circ (l')^* \circ b' \circ m^* }$

commutes. This is true by Categories, Lemma 4.28.2 or more simply the discussion preceding Categories, Definition 4.28.1.

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