86.6 Right adjoint of pushforward and trace maps
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated algebraic spaces over $S$. Let $a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ be the right adjoint as in Lemma 86.3.1. By Categories, Section 4.24 we obtain a transformation of functors
\[ \text{Tr}_ f : Rf_* \circ a \longrightarrow \text{id} \]
The corresponding map $\text{Tr}_{f, K} : Rf_*a(K) \longrightarrow K$ for $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ is sometimes called the trace map. This is the map which has the property that the bijection
\[ \mathop{\mathrm{Hom}}\nolimits _ X(L, a(K)) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*L, K) \]
for $L \in D_\mathit{QCoh}(\mathcal{O}_ X)$ which characterizes the right adjoint is given by
\[ \varphi \longmapsto \text{Tr}_{f, K} \circ Rf_*\varphi \]
The canonical map (86.3.2.1)
\[ Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \]
comes about by composition with $\text{Tr}_{f, K}$. Every trace map we are going to consider in this section will be a special case of this trace map. Before we discuss some special cases we show that formation of the trace map commutes with base change.
Lemma 86.6.1 (Trace map and base change). Suppose we have a diagram (86.4.0.1). Then the maps $1 \star \text{Tr}_ f : Lg^* \circ Rf_* \circ a \to Lg^*$ and $\text{Tr}_{f'} \star 1 : Rf'_* \circ a' \circ Lg^* \to Lg^*$ agree via the base change maps $\beta : Lg^* \circ Rf_* \to Rf'_* \circ L(g')^*$ (Cohomology on Sites, Remark 21.19.3) and $\alpha : L(g')^* \circ a \to a' \circ Lg^*$ (86.4.1.1). More precisely, the diagram
\[ \xymatrix{ Lg^* \circ Rf_* \circ a \ar[d]_{\beta \star 1} \ar[r]_-{1 \star \text{Tr}_ f} & Lg^* \\ Rf'_* \circ L(g')^* \circ a \ar[r]^{1 \star \alpha } & Rf'_* \circ a' \circ Lg^* \ar[u]_{\text{Tr}_{f'} \star 1} } \]
of transformations of functors commutes.
Proof.
In this proof we write $f_*$ for $Rf_*$ and $g^*$ for $Lg^*$ and we drop $\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. Recall that $\beta : g^* \circ f_* \to f'_* \circ (g')^*$ is an isomorphism and that $\alpha $ is defined using the isomorphism $\beta ^\vee : g'_* \circ a' \to a \circ g_*$ which is the adjoint of $\beta $, see Lemma 86.4.1 and its proof. First we note that the top horizontal arrow of the diagram in the lemma is equal to the composition
\[ g^* \circ f_* \circ a \to g^* \circ f_* \circ a \circ g_* \circ g^* \to g^* \circ g_* \circ g^* \to g^* \]
where the first arrow is the unit for $(g^*, g_*)$, the second arrow is $\text{Tr}_ f$, and the third arrow is the counit for $(g^*, g_*)$. This is a simple consequence of the fact that the composition $g^* \to g^* \circ g_* \circ g^* \to g^*$ of unit and counit is the identity. Consider the diagram
\[ \xymatrix{ & g^* \circ f_* \circ a \ar[ld]_\beta \ar[d] \ar[r]_{\text{Tr}_ f} & g^* \\ f'_* \circ (g')^* \circ a \ar[dr] & g^* \circ f_* \circ a \circ g_* \circ g^* \ar[d]_\beta \ar[ru] & g^* \circ f_* \circ g'_* \circ a' \circ g^* \ar[l]_{\beta ^\vee } \ar[d]_\beta & f'_* \circ a' \circ g^* \ar[lu]_{\text{Tr}_{f'}} \\ & f'_* \circ (g')^* \circ a \circ g_* \circ g^* & f'_* \circ (g')^* \circ g'_* \circ a' \circ g^* \ar[ru] \ar[l]_{\beta ^\vee } } \]
In this diagram the two squares commute Categories, Lemma 4.28.2 or more simply the discussion preceding Categories, Definition 4.28.1. The triangle commutes by the discussion above. By Categories, Lemma 4.24.8 the square
\[ \xymatrix{ g^* \circ f_* \circ g'_* \circ a' \ar[d]_{\beta ^\vee } \ar[r]_-\beta & f'_* \circ (g')^* \circ g'_* \circ a' \ar[d] \\ g^* \circ f_* \circ a \circ g_* \ar[r] & \text{id} } \]
commutes which implies the pentagon in the big diagram commutes. Since $\beta $ and $\beta ^\vee $ are isomorphisms, and since going on the outside of the big diagram equals $\text{Tr}_ f \circ \alpha \circ \beta $ by definition this proves the lemma.
$\square$
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated algebraic spaces over $S$. Let $a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ be the right adjoint of $Rf_*$ as in Lemma 86.3.1. By Categories, Section 4.24 we obtain a transformation of functors
\[ \eta _ f : \text{id} \to a \circ Rf_* \]
which is called the unit of the adjunction.
Lemma 86.6.2. Suppose we have a diagram (86.4.0.1). Then the maps $1 \star \eta _ f : L(g')^* \to L(g')^* \circ a \circ Rf_*$ and $\eta _{f'} \star 1 : L(g')^* \to a' \circ Rf'_* \circ L(g')^*$ agree via the base change maps $\beta : Lg^* \circ Rf_* \to Rf'_* \circ L(g')^*$ (Cohomology on Sites, Remark 21.19.3) and $\alpha : L(g')^* \circ a \to a' \circ Lg^*$ (86.4.1.1). More precisely, the diagram
\[ \xymatrix{ L(g')^* \ar[r]_-{1 \star \eta _ f} \ar[d]_{\eta _{f'} \star 1} & L(g')^* \circ a \circ Rf_* \ar[d]^\alpha \\ a' \circ Rf'_* \circ L(g')^* & a' \circ Lg^* \circ Rf_* \ar[l]_-\beta } \]
of transformations of functors commutes.
Proof.
This proof is dual to the proof of Lemma 86.6.1. In this proof we write $f_*$ for $Rf_*$ and $g^*$ for $Lg^*$ and we drop $\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. Recall that $\beta : g^* \circ f_* \to f'_* \circ (g')^*$ is an isomorphism and that $\alpha $ is defined using the isomorphism $\beta ^\vee : g'_* \circ a' \to a \circ g_*$ which is the adjoint of $\beta $, see Lemma 86.4.1 and its proof. First we note that the left vertical arrow of the diagram in the lemma is equal to the composition
\[ (g')^* \to (g')^* \circ g'_* \circ (g')^* \to (g')^* \circ g'_* \circ a' \circ f'_* \circ (g')^* \to a' \circ f'_* \circ (g')^* \]
where the first arrow is the unit for $((g')^*, g'_*)$, the second arrow is $\eta _{f'}$, and the third arrow is the counit for $((g')^*, g'_*)$. This is a simple consequence of the fact that the composition $(g')^* \to (g')^* \circ (g')_* \circ (g')^* \to (g')^*$ of unit and counit is the identity. Consider the diagram
\[ \xymatrix{ & (g')^* \circ a \circ f_* \ar[r] & (g')^* \circ a \circ g_* \circ g^* \circ f_* \ar[ld]_\beta \\ (g')^* \ar[ru]^{\eta _ f} \ar[dd]_{\eta _{f'}} \ar[rd] & (g')^* \circ a \circ g_* \circ f'_* \circ (g')^* & (g')^* \circ g'_* \circ a' \circ g^* \circ f_* \ar[u]_{\beta ^\vee } \ar[ld]_\beta \ar[d] \\ & (g')^* \circ g'_* \circ a' \circ f'_* \circ (g')^* \ar[ld] \ar[u]_{\beta ^\vee } & a' \circ g^* \circ f_* \ar[lld]^\beta \\ a' \circ f'_* \circ (g')^* } \]
In this diagram the two squares commute Categories, Lemma 4.28.2 or more simply the discussion preceding Categories, Definition 4.28.1. The triangle commutes by the discussion above. By the dual of Categories, Lemma 4.24.8 the square
\[ \xymatrix{ \text{id} \ar[r] \ar[d] & g'_* \circ a' \circ g^* \circ f_* \ar[d]^\beta \\ g'_* \circ a' \circ g^* \circ f_* \ar[r]^{\beta ^\vee } & a \circ g_* \circ f'_* \circ (g')^* } \]
commutes which implies the pentagon in the big diagram commutes. Since $\beta $ and $\beta ^\vee $ are isomorphisms, and since going on the outside of the big diagram equals $\beta \circ \alpha \circ \eta _ f$ by definition this proves the lemma.
$\square$
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