Lemma 48.7.1 (Trace map and base change). Suppose we have a diagram (48.4.0.1) where $f$ and $g$ are tor independent. 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, Remark 20.28.3) and $\alpha : L(g')^* \circ a \to a' \circ Lg^*$ (48.5.0.1). 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 48.4.1 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.
$\square$

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