**Proof.**
Part (1) either follows from the construction of the trace map in the proof of Lemma 62.5.2 or more simply because the characterization of the map forces it to be true on all stalks. Let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

be a cartesian diagram of schemes. Then the function $w' = w \circ g' : X' \to \mathbf{Z}$ is a weighting of $f'$ by More on Morphisms, Lemma 37.73.3. Statement (2) means that the diagram

\[ \xymatrix{ g^{-1}f_!f^{-1}\mathcal{F} \ar[rr]_-{g^{-1}\text{Tr}_{f, w, \mathcal{F}}} \ar@{=}[d] & & g^{-1}\mathcal{F} \ar@{=}[d] \\ f'_!(f')^{-1}g^{-1}\mathcal{F} \ar[rr]^-{\text{Tr}_{f', w', g^{-1}\mathcal{F}}} & & g^{-1}\mathcal{F} } \]

is commutative where the left vertical equality is given by

\[ g^{-1}f_!f^{-1}\mathcal{F} = f'_!(g')^{-1}f^{-1}\mathcal{F} = f'_!(f')^{-1}g^{-1}\mathcal{F} \]

with first equality sign given by Lemma 62.4.10 (base change for lower shriek). The commutativity of this diagram follows from the characterization of the action of our trace maps on stalks and the fact that the base change map of Lemma 62.4.10 respects the descriptions of stalks.

Given parts (1) and (2), part (3) follows as the functors $f^{-1} : D(Y_{\acute{e}tale}, \Lambda ) \to D(X_{\acute{e}tale}, \Lambda )$ and $f_! : D(X_{\acute{e}tale}, \Lambda ) \to D(Y_{\acute{e}tale}, \Lambda )$ are obtained by applying $f^{-1}$ and $f_!$ to any complexes of modules representing the objects in question.
$\square$

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