Lemma 62.5.2. Let $f : X \to Y$ be a locally quasi-finite morphism of schemes. Let $w : X \to \mathbf{Z}$ be a weighting of $f$. For any abelian sheaf $\mathcal{F}$ on $Y$ there exists a unique trace map $\text{Tr}_{f, w, \mathcal{F}} : f_!f^{-1}\mathcal{F} \to \mathcal{F}$ having the prescribed behaviour on stalks.

Proof. By Lemma 62.5.1 we have an identification $f_!f^{-1}\mathcal{F} = f_!\underline{\mathbf{Z}} \otimes \mathcal{F}$ compatible with the description of stalks of these sheaves at geometric points. Hence it suffices to produce the map

$\text{Tr}_{f, w, \underline{\mathbf{Z}}} : f_!\underline{\mathbf{Z}} \longrightarrow \underline{\mathbf{Z}}$

having the prescribed behaviour on stalks. By Definition 62.4.4 we have $f_!\underline{\mathbf{Z}} = (f_{p!}\underline{\mathbf{Z}})^\#$ where $f_{p!}\underline{\mathbf{Z}}$ is the presheaf constructed in Section 62.4. Thus it suffices to construct a map

$f_{p!}\underline{\mathbf{Z}} \longrightarrow \underline{\mathbf{Z}}$

of presheaves on $Y_{\acute{e}tale}$. Let $V$ be an object of $Y_{\acute{e}tale}$. Recall from Section 62.4 that

$f_{p!}\underline{\mathbf{Z}}(V) = \mathop{\mathrm{colim}}\nolimits _ Z H_ Z(\underline{\mathbf{Z}})$

Here the colimit is over the (partially ordered) collection of locally closed subschemes $Z \subset X_ V$ which are finite over $V$. For each such $Z$ we will define a map

$H_ Z(\underline{\mathbf{Z}}) \longrightarrow \underline{\mathbf{Z}}(V)$

compatible with the maps defining the colimit.

Let $Z \subset X_ V$ be locally closed and finite over $V$. Choose an open $U \subset X_ V$ containing $Z$ as a closed subset. An element $s$ of $H_ Z(\underline{\mathbf{Z}})$ is a section $s \in \underline{\mathbf{Z}}(U)$ whose support is contained in $Z$. Let $U_ n \subset U$ be the open and closed subset where the value of $s$ is $n \in \mathbf{Z}$. By the support condition we see that $Z \cap U_ n = U_ n$ for $n \not= 0$. Hence for $n \not= 0$, the open $U_ n$ is also closed in $Z$ (as the complement of all the others) and we conclude that $U_ n \to V$ is finite as $Z$ is finite over $V$. By the very definition of a weighting this means the function $\int _{U_ n \to V} w|_{U_ n}$ is locally constant on $V$ and we may view it as an element of $\underline{\mathbf{Z}}(V)$. Our construction sends $(Z, s)$ to the element

$\sum \nolimits _{n \in \mathbf{Z},\ n \not= 0} n \left(\int _{U_ n \to V} w|_{U_ n}\right) \quad \in \quad \underline{\mathbf{Z}}(V)$

The sum is locally finite on $V$ and hence makes sense; details omitted (in the whole discussion the reader may first choose affine opens and make sure all the schemes occuring in the argument are quasi-compact so the sum is finite). We omit the verification that this construction is compatible with the maps in the colimit and with the restriction mappings defining $f_{p!}\underline{\mathbf{Z}}$.

Let $\overline{y}$ be a geometric point of $Y$ lying over the point $y \in Y$. Taking stalks at $\overline{y}$ the construction above determines a map

$(f_!\underline{\mathbf{Z}})_{\overline{y}} = \bigoplus \nolimits _{f(\overline{x}) = \overline{y}} \mathbf{Z} \longrightarrow \mathbf{Z} = \underline{\mathbf{Z}}_{\overline{y}}$

To finish the proof we will show this map is given by multiplication by $w(\overline{x})$ on the summand corresponding to $\overline{x}$. Namely, pick $\overline{x}$ lying over $\overline{y}$. We can find an étale neighbourhood $(V, \overline{v}) \to (Y, \overline{y})$ such that $X_ V$ contains an open $U$ finite over $V$ such that only the geometric point $\overline{x}$ is in $U$ and not the other geometric points of $X$ lifting $\overline{y}$. This follows from More on Morphisms, Lemma 37.41.3; some details omitted. Then $(U, 1)$ defines a section of $f_!\underline{\mathbf{Z}}$ over $V$ which maps to $1$ in the summand corresponding to $\overline{x}$ and zero in the other summands (see proof of Lemma 62.4.2) and our construction above sends $(U, 1)$ to $\int _{U \to V} w|_ U$ which is constant with value $w(\overline{x})$ in a neighbourhood of $\overline{v}$ as desired. $\square$

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