Lemma 48.5.1. Consider a commutative diagram

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

of quasi-compact and quasi-separated schemes where both diagrams are cartesian and where $f$ and $l$ as well as $g$ and $m$ 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 $g \circ f$ and $m$ are Tor independent (details omitted), hence the statement makes sense. In this proof we write $k^*$ in place of $Lk^*$ and $f_*$ instead of $Rf_*$. Let $a$, $b$, and $c$ be the right adjoints of Lemma 48.3.1 for $f$, $g$, and $g \circ f$ and similarly for the primed versions. The arrow corresponding to the top square is the composition

$\gamma _{top} : k^* \circ a \to k^* \circ a \circ l_* \circ l^* \xleftarrow {\xi _{top}} k^* \circ k_* \circ a' \circ l^* \to a' \circ l^*$

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

$\gamma _{bot} : l^* \circ b \to l^* \circ b \circ m_* \circ m^* \xleftarrow {\xi _{bot}} l^* \circ l_* \circ b' \circ m^* \to b' \circ m^*$

For the outer rectangle we get

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

We have $(g \circ f)_* = g_* \circ f_*$ and hence $c = a \circ b$ and similarly $c' = a' \circ b'$. The statement of the lemma is that $\gamma _{rect}$ is equal to the composition

$k^* \circ c = k^* \circ a \circ b \xrightarrow {\gamma _{top}} a' \circ l^* \circ b \xrightarrow {\gamma _{bot}} a' \circ b' \circ m^* = c' \circ m^*$

To see this we contemplate the following diagram:

$\xymatrix{ & & k^* \circ a \circ b \ar[d] \ar[lldd] \\ & & k^* \circ a \circ l_* \circ l^* \circ b \ar[ld] \\ k^* \circ a \circ b \circ m_* \circ m^* \ar[r] & k^* \circ a \circ l_* \circ l^* \circ b \circ m_* \circ m^* & k^* \circ k_* \circ a' \circ l^* \circ b \ar[u]_{\xi _{top}} \ar[d] \ar[ld] \\ & k^*\circ k_* \circ a' \circ l^* \circ b \circ m_* \circ m^* \ar[u]_{\xi _{top}} \ar[rd] & a' \circ l^* \circ b \ar[d] \\ k^* \circ k_* \circ a' \circ b' \circ m^* \ar[uu]_{\xi _{rect}} \ar[ddrr] & k^*\circ k_* \circ a' \circ l^* \circ l_* \circ b' \circ m^* \ar[u]_{\xi _{bot}} \ar[l] \ar[dr] & a' \circ l^* \circ b \circ m_* \circ m^* \\ & & a' \circ l^* \circ l_* \circ b' \circ m^* \ar[u]_{\xi _{bot}} \ar[d] \\ & & a' \circ b' \circ m^* }$

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 the diagram

$\xymatrix{ a \circ l_* \circ l^* \circ b \circ m_* & a \circ b \circ m_* \ar[l] \\ k_* \circ a' \circ l^* \circ b \circ m_* \ar[u]_{\xi _{top}} & \\ k_* \circ a' \circ l^* \circ l_* \circ b' \ar[u]_{\xi _{bot}} \ar[r] & k_* \circ a' \circ b' \ar[uu]_{\xi _{rect}} }$

becomes commutative if we invert the arrows $\xi _{top}$, $\xi _{bot}$, and $\xi _{rect}$ (note that this is different from asking the diagram to be commutative). However, the diagram

$\xymatrix{ & a \circ l_* \circ l^* \circ b \circ m_* \\ a \circ l_* \circ l^* \circ l_* \circ b' \ar[ru]^{\xi _{bot}} & & k_* \circ a' \circ l^* \circ b \circ m_* \ar[ul]_{\xi _{top}} \\ & k_* \circ a' \circ l^* \circ l_* \circ b' \ar[ul]^{\xi _{top}} \ar[ur]_{\xi _{bot}} }$

commutes by Categories, Lemma 4.28.2. Since the diagrams

$\vcenter { \xymatrix{ a \circ l_* \circ l^* \circ b \circ m_* & a \circ b \circ m \ar[l] \\ a \circ l_* \circ l^* \circ l_* \circ b' \ar[u] & a \circ l_* \circ b' \ar[l] \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ a \circ l_* \circ l^* \circ l_* \circ b' \ar[r] & a \circ l_* \circ b' \\ k_* \circ a' \circ l^* \circ l_* \circ b' \ar[u] \ar[r] & k_* \circ a' \circ b' \ar[u] } }$

commute (see references cited) and since the composition of $l_* \to l_* \circ l^* \circ l_* \to l_*$ is the identity, we find that it suffices to prove that

$k \circ a' \circ b' \xrightarrow {\xi _{bot}} a \circ l_* \circ b \xrightarrow {\xi _{top}} a \circ b \circ m_*$

is equal to $\xi _{rect}$ (via the identifications $a \circ b = c$ and $a' \circ b' = c'$). This is the statement dual to Cohomology, Remark 20.28.4 and the proof is complete. $\square$

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