Processing math: 100%

The Stacks project

Lemma 63.5.4. Let f : X \to Y and g : Y \to Z be locally quasi-finite morphisms. Let w_ f : X \to \mathbf{Z} be a weighting of f and let w_ g : Y \to \mathbf{Z} be a weighting of g. For K \in D(Z_{\acute{e}tale}, \Lambda ) the composition

(g \circ f)_!(g \circ f)^{-1}K = g_! f_! f^{-1} g^{-1}K \xrightarrow {g_! \text{Tr}_{f, w_ f, g^{-1}K}} g_!g^{-1}K \xrightarrow {\text{Tr}_{g, w_ g, K}} K

is equal to \text{Tr}_{g \circ f, w_{g \circ f}, K} where w_{g \circ f}(x) = w_ f(x) w_ g(f(x)).

Proof. We have (g \circ f)_! = g_! \circ f_! by Lemma 63.4.12. In More on Morphisms, Lemma 37.75.5 we have seen that w_{g \circ f} is a weighting for g \circ f so the statement makes sense. To check equality compute on stalks. Details omitted. \square


Comments (0)


Your email address will not be published.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.