Lemma 62.5.4. 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 62.4.12. In More on Morphisms, Lemma 37.73.5 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.
$\square$

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