The Stacks project

Proof. Recall that $f_!\mathcal{F} = (f_{p!}\mathcal{F})^\# $ by Definition 63.4.4 where $f_{p!}\mathcal{F}$ is the presheaf constructed in Section 63.4. Thus in order to construct the arrow it suffices to construct a map

of presheaves on $Y_{\acute{e}tale}$. Here the symbol $\otimes _{p, \Lambda }$ denotes the presheaf tensor product, see Modules on Sites, Section 18.26. Let $V$ be an object of $Y_{\acute{e}tale}$. Recall that

See Section 63.4. Our map will be defined on pure tensors by the rule

(for notation see below) and extended by linearity to all of $(f_{p!}\mathcal{F} \otimes _{p, \Lambda } \mathcal{G})(V) = f_{p!}\mathcal{F}(V) \otimes _\Lambda \mathcal{G}(V)$. Here the notation used is as follows

1. $Z \subset X_ V$ is a locally closed subscheme finite over $V$,

2. $s \in H_ Z(\mathcal{F})$ which means that $s \in \mathcal{F}(U)$ with $\text{Supp}(s) \subset Z$ for some $U \subset X_ V$ open such that $Z \subset U$ is closed, and

3. $t \in \mathcal{G}(V)$ with image $f^{-1}t \in f^{-1}\mathcal{G}(U)$.

Since the support of $s \in \mathcal{F}(U)$ is contained in $Z$ it is clear that the support of $s \otimes f^{-1}t$ is contained in $Z$ as well. Thus considering the pair $(Z, s \otimes f^{-1}t)$ makes sense. It is immediate that the construction commutes with the transition maps in the colimit $\mathop{\mathrm{colim}}\nolimits _ Z H_ Z(\mathcal{F})$ and that it is compatible with restriction mappings. Finally, it is equally clear that the construction is compatible with the identifications of stalks of $f_!$ in Lemma 63.4.5. In other words, the map $can$ we've produced on stalks at a geometric point $\overline{y}$ fits into a commutative diagram

where the direct sums are over the geometric points $\overline{x}$ lying over $\overline{y}$, where the vertical arrows are the identifications of Lemma 63.4.5, and where the lower horizontal arrow is the obvious isomorphism. We conclude that $can$ is an isomorphism as desired. $\square$

Proof. By Lemma 63.5.1 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

having the prescribed behaviour on stalks. By Definition 63.4.4 we have $f_!\underline{\mathbf{Z}} = (f_{p!}\underline{\mathbf{Z}})^\# $ where $f_{p!}\underline{\mathbf{Z}}$ is the presheaf constructed in Section 63.4. Thus it suffices to construct a map

of presheaves on $Y_{\acute{e}tale}$. Let $V$ be an object of $Y_{\acute{e}tale}$. Recall from Section 63.4 that

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

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

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 occurring 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

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 63.4.2) 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. $\square$

Proof. Part (1) either follows from the construction of the trace map in the proof of Lemma 63.5.2 or more simply because the characterization of the map forces it to be true on all stalks. Let

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 37.75.3. Statement (2) means that the diagram

is commutative where the left vertical equality is given by

with first equality sign given by Lemma 63.4.10 (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 63.4.10 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. $\square$

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. $\square$