Lemma 86.4.1. In diagram (86.4.0.1) the map $a \circ Rg_* \leftarrow Rg'_* \circ a'$ is an isomorphism.
86.4 Right adjoint of pushforward and base change, I
Let us define the base change map between right adjoints of pushforward. Let $S$ be a scheme. Consider a cartesian diagram
86.4.0.1
\begin{equation} \label{spaces-duality-equation-base-change} \vcenter { \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } } \end{equation}
where $Y'$ and $X$ are Tor independent over $Y$. Denote
\[ a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X) \quad \text{and}\quad a' : D_\mathit{QCoh}(\mathcal{O}_{Y'}) \to D_\mathit{QCoh}(\mathcal{O}_{X'}) \]
the right adjoints to $Rf_*$ and $Rf'_*$ (Lemma 86.3.1). The base change map of Cohomology on Sites, Remark 21.19.3 gives a transformation of functors
\[ Lg^* \circ Rf_* \longrightarrow Rf'_* \circ L(g')^* \]
on derived categories of sheaves with quasi-coherent cohomology. Hence a transformation between the right adjoints in the opposite direction
\[ a \circ Rg_* \longleftarrow Rg'_* \circ a' \]
Proof. The base change map $Lg^* \circ Rf_* K \to Rf'_* \circ L(g')^*K$ is an isomorphism for every $K$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Spaces, Lemma 75.20.4 (this uses the assumption of Tor independence). Thus the corresponding transformation between adjoint functors is an isomorphism as well. $\square$
Then we can consider the morphism of functors $D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_{X'})$ given by the composition
86.4.1.1
\begin{equation} \label{spaces-duality-equation-base-change-map} L(g')^* \circ a \to L(g')^* \circ a \circ Rg_* \circ Lg^* \leftarrow L(g')^* \circ Rg'_* \circ a' \circ Lg^* \to a' \circ Lg^* \end{equation}
The first arrow comes from the adjunction map $\text{id} \to Rg_* Lg^*$ and the last arrow from the adjunction map $L(g')^*Rg'_* \to \text{id}$. We need the assumption on Tor independence to invert the arrow in the middle, see Lemma 86.4.1. Alternatively, we can think of (86.4.1.1) by adjointness of $L(g')^*$ and $R(g')_*$ as a natural transformation
\[ a \to a \circ Rg_* \circ Lg^* \leftarrow Rg'_* \circ a' \circ Lg^* \]
were again the second arrow is invertible. If $M \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ then on Yoneda functors this map is given by
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ X(M, a(K)) & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*M, K) \\ & \to \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*M, Rg_* Lg^*K) \\ & = \mathop{\mathrm{Hom}}\nolimits _{Y'}(Lg^*Rf_*M, Lg^*K) \\ & \leftarrow \mathop{\mathrm{Hom}}\nolimits _{Y'}(Rf'_* L(g')^*M, Lg^*K) \\ & = \mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, a'(Lg^*K)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(M, Rg'_*a'(Lg^*K)) \end{align*}
(were the arrow pointing left is invertible by the base change theorem given in Derived Categories of Spaces, Lemma 75.20.4) which makes things a little bit more explicit.
In this section we first prove that the base change map satisfies some natural compatibilities with regards to stacking squares as in Cohomology on Sites, Remarks 21.19.4 and 21.19.5 for the usual base change map. We suggest the reader skip the rest of this section on a first reading.
Lemma 86.4.2. Let $S$ be a scheme. Consider a commutative diagram of quasi-compact and quasi-separated algebraic spaces over $S$ where both diagrams are cartesian and where $f$ and $l$ as well as $g$ and $m$ are Tor independent. Then the maps (86.4.1.1) for the two squares compose to give the base change map for the outer rectangle (see proof for a precise statement).
\[ \xymatrix{ X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z' \ar[r]^ m & Z } \]
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 86.3.1 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 on Sites, Remark 21.19.4 and the proof is complete. $\square$
Lemma 86.4.3. Let $S$ be a scheme. Consider a commutative diagram of quasi-compact and quasi-separated algebraic spaces over $S$ where both diagrams are cartesian and where $f$ and $h$ as well as $f'$ and $h'$ are Tor independent. Then the maps (86.4.1.1) for the two squares compose to give the base change map for the outer rectangle (see proof for a precise statement).
\[ \xymatrix{ X'' \ar[r]_{g'} \ar[d]_{f''} & X' \ar[r]_ g \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{h'} & Y' \ar[r]^ h & Y } \]
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 86.3.1 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. $\square$
Remark 86.4.4. Let $S$ be a scheme. Consider a commutative diagram of quasi-compact and quasi-separated algebraic spaces over $S$ 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 86.3.1 for $f$, $f'$, $f''$, $g$, $g'$, $g''$. Let us label the squares of the diagram $A$, $B$, $C$, $D$ as follows Then the maps (86.4.1.1) for the squares are (where we use $k^* = Lk^*$, etc) For the $2 \times 1$ and $1 \times 2$ rectangles we have four further base change maps By Lemma 86.4.3 we have and by Lemma 86.4.2 we have 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 86.4.2 and 86.4.3 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 namely and The point of this remark is to point out that these transformations are equal. Namely, to see this it suffices to show that commutes. This is true by Categories, Lemma 4.28.2 or more simply the discussion preceding Categories, Definition 4.28.1.
\[ \xymatrix{ X'' \ar[r]_{k'} \ar[d]_{f''} & X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{l'} \ar[d]_{g''} & Y' \ar[r]^ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z'' \ar[r]^{m'} & Z' \ar[r]^ m & Z } \]
\[ \begin{matrix} A
& B
\\ C
& D
\end{matrix} \]
\[ \begin{matrix} \gamma _ A : (k')^* \circ a' \to a'' \circ (l')^*
& \gamma _ B : k^* \circ a \to a' \circ l^*
\\ \gamma _ C : (l')^* \circ b' \to b'' \circ (m')^*
& \gamma _ D : l^* \circ b \to b' \circ m^*
\end{matrix} \]
\[ \begin{matrix} \gamma _{A + B} : (k \circ k')^* \circ a \to a'' \circ (l \circ l')^*
\\ \gamma _{C + D} : (l \circ l')^* \circ b \to b'' \circ (m \circ m')^*
\\ \gamma _{A + C} : (k')^* \circ (a' \circ b') \to (a'' \circ b'') \circ (m')^*
\\ \gamma _{A + C} : k^* \circ (a \circ b) \to (a' \circ b') \circ m^*
\end{matrix} \]
\[ \gamma _{A + B} = \gamma _ A \circ \gamma _ B, \quad \gamma _{C + D} = \gamma _ C \circ \gamma _ D \]
\[ \gamma _{A + C} = \gamma _ C \circ \gamma _ A, \quad \gamma _{B + D} = \gamma _ D \circ \gamma _ B \]
\[ (k')^* \circ k^* \circ a \circ b \longrightarrow a'' \circ b'' \circ (m')^* \circ m^* \]
\[ \gamma _ C \circ \gamma _ A \circ \gamma _ D \circ \gamma _ B = \gamma _{A + C} \circ \gamma _{B + D} \]
\[ \gamma _ C \circ \gamma _ D \circ \gamma _ A \circ \gamma _ B = \gamma _{C + D} \circ \gamma _{A + B} \]
\[ \xymatrix{ (k')^* \circ a' \circ l^* \circ b \ar[r]_{\gamma _ D} \ar[d]_{\gamma _ A} & (k')^* \circ a' \circ b' \circ m^* \ar[d]^{\gamma _ A} \\ a'' \circ (l')^* \circ l^* \circ b \ar[r]^{\gamma _ D} & a'' \circ (l')^* \circ b' \circ m^* } \]
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