The Stacks project

Lemma 48.7.1 (Trace map and base change). Suppose we have a diagram ( 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 $\alpha : L(g')^* \circ a \to a' \circ Lg^*$ ( 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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B6J. Beware of the difference between the letter 'O' and the digit '0'.