Lemma 48.5.2. Consider a commutative diagram

$\xymatrix{ X'' \ar[r]_{g'} \ar[d]_{f''} & X' \ar[r]_ g \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{h'} & Y' \ar[r]^ h & Y }$

of quasi-compact and quasi-separated schemes where both diagrams are cartesian and where $f$ and $h$ as well as $f'$ and $h'$ are Tor independent. Then the maps (48.5.0.1) for the two squares compose to give the base change map for the outer rectangle (see proof for a precise statement).

Proof. It follows from the assumptions that $f$ and $h \circ h'$ are Tor independent (details omitted), hence the statement makes sense. In this proof we write $g^*$ in place of $Lg^*$ and $f_*$ instead of $Rf_*$. Let $a$, $a'$, and $a''$ be the right adjoints of Lemma 48.3.1 for $f$, $f'$, and $f''$. The arrow corresponding to the right square is the composition

$\gamma _{right} : g^* \circ a \to g^* \circ a \circ h_* \circ h^* \xleftarrow {\xi _{right}} g^* \circ g_* \circ a' \circ h^* \to a' \circ h^*$

where $\xi _{right} : g_* \circ a' \to a \circ h_*$ is an isomorphism (hence can be inverted) and is the arrow “dual” to the base change map $h^* \circ f_* \to f'_* \circ g^*$. The outer arrows come from the canonical maps $1 \to h_* \circ h^*$ and $g^* \circ g_* \to 1$. Similarly for the left square we have

$\gamma _{left} : (g')^* \circ a' \to (g')^* \circ a' \circ (h')_* \circ (h')^* \xleftarrow {\xi _{left}} (g')^* \circ (g')_* \circ a'' \circ (h')^* \to a'' \circ (h')^*$

For the outer rectangle we get

$\gamma _{rect} : k^* \circ a \to k^* \circ a \circ m_* \circ m^* \xleftarrow {\xi _{rect}} k^* \circ k_* \circ a'' \circ m^* \to a'' \circ m^*$

where $k = g \circ g'$ and $m = h \circ h'$. We have $k^* = (g')^* \circ g^*$ and $m^* = (h')^* \circ h^*$. The statement of the lemma is that $\gamma _{rect}$ is equal to the composition

$k^* \circ a = (g')^* \circ g^* \circ a \xrightarrow {\gamma _{right}} (g')^* \circ a' \circ h^* \xrightarrow {\gamma _{left}} a'' \circ (h')^* \circ h^* = a'' \circ m^*$

To see this we contemplate the following diagram

$\xymatrix{ & (g')^* \circ g^* \circ a \ar[d] \ar[ddl] \\ & (g')^* \circ g^* \circ a \circ h_* \circ h^* \ar[ld] \\ (g')^* \circ g^* \circ a \circ h_* \circ (h')_* \circ (h')^* \circ h^* & (g')^* \circ g^* \circ g_* \circ a' \circ h^* \ar[u]_{\xi _{right}} \ar[d] \ar[ld] \\ (g')^* \circ g^* \circ g_* \circ a' \circ (h')_* \circ (h')^* \circ h^* \ar[u]_{\xi _{right}} \ar[dr] & (g')^* \circ a' \circ h^* \ar[d] \\ (g')^* \circ g^* \circ g_* \circ (g')_* \circ a'' \circ (h')^* \circ h^* \ar[u]_{\xi _{left}} \ar[ddr] \ar[dr] & (g')^* \circ a' \circ (h')_* \circ (h')^* \circ h^* \\ & (g')^*\circ (g')_* \circ a'' \circ (h')^* \circ h^* \ar[u]_{\xi _{left}} \ar[d] \\ & a'' \circ (h')^* \circ h^* }$

Going down the right hand side we have the composition and going down the left hand side we have $\gamma _{rect}$. All the quadrilaterals on the right hand side of this diagram commute by Categories, Lemma 4.28.2 or more simply the discussion preceding Categories, Definition 4.28.1. Hence we see that it suffices to show that

$g_* \circ (g')_* \circ a'' \xrightarrow {\xi _{left}} g_* \circ a' \circ (h')_* \xrightarrow {\xi _{right}} a \circ h_* \circ (h')_*$

is equal to $\xi _{rect}$. This is the statement dual to Cohomology, Remark 20.28.5 and the proof is complete. $\square$

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