The Stacks project

Proof. Let $V$ be an object of $Y_{\acute{e}tale}$. If $Z \subset X_ V$ is locally closed and finite over $V$, then, since $f$ is separated, we see that the morphism $Z \to X_ V$ is a closed immersion. Moreover, if $Z_ i$, $i = 1, \ldots , n$ are closed subschemes of $X_ V$ finite over $V$, then $Z_1 \cup \ldots \cup Z_ n$ (scheme theoretic union) is a closed subscheme finite over $V$. Hence in this case the colimit (63.4.0.2) defining $f_{p!}\mathcal{F}(V)$ is directed and we find that $f_{!p}\mathcal{F}(V)$ is simply equal to the set of sections of $\mathcal{F}(X_ V)$ whose support is finite over $V$. Since any closed subset of $X_ V$ which is proper over $V$ is actually finite over $V$ (as $f$ is locally quasi-finite) we conclude that this is equal to $f_!\mathcal{F}(V)$ by its very definition. $\square$

Proof. Recall that the stalk at $\overline{y}$ of a presheaf is defined by the usual colimit over étale neighbourhoods $(V, \overline{v})$ of $\overline{y}$, see Étale Cohomology, Definition 59.29.6. Accordingly suppose $s = \sum _{i = 1, \ldots , n} (Z_ i, s_ i)$ as in (63.4.0.1) is an element of $f_{p!}\mathcal{F}(V)$ where $(V, \overline{v})$ is an étale neighbourhood of $\overline{y}$. Then since

and since $s_ i$ is a section of $\mathcal{F}$ on an open neighbourhood of $Z_ i$ in $X_ V$ we can send $s$ to

We omit the verification that this is compatible with restriction maps and that the relations (1) $(Z, s) + (Z, s') - (Z, s + s')$ and (2) $(Z, s) - (Z', s)$ if $Z \subset Z'$ are sent to zero. Thus we obtain a map

Let us prove this arrow is surjective. For this it suffices to pick an $\overline{x}$ with $f(\overline{x}) = \overline{y}$ and prove that an element $s$ in the summand $\mathcal{F}_{\overline{x}}$ is in the image. Let $s$ correspond to the element $s \in \mathcal{F}(U)$ where $(U, \overline{u})$ is an étale neighbourhood of $\overline{x}$. Since $f$ is locally quasi-finite, the morphism $U \to Y$ is locally quasi-finite too. By More on Morphisms, Lemma 37.41.3 we can find an étale neighbourhood $(V, \overline{v})$ of $\overline{y}$, an open subscheme

and a geometric point $\overline{w}$ mapping to $\overline{u}$ and $\overline{v}$ such that $W \to V$ is finite and $\overline{w}$ is the only geometric point of $W$ mapping to $\overline{v}$. (We omit the translation between the language of geometric points we are currently using and the language of points and residue field extensions used in the statement of the lemma.) Observe that $W \to X_ V = X \times _ Y V$ is étale. Choose an affine open neighbourhood $W' \subset X_ V$ of the image $\overline{w}'$ of $\overline{w}$. Since $\overline{w}$ is the only point of $W$ over $\overline{v}$ and since $W \to V$ is closed, after replacing $V$ by an open neighbourhood of $\overline{v}$, we may assume $W \to X_ V$ maps into $W'$. Then $W \to W'$ is finite and étale and there is a unique geometric point $\overline{w}$ of $W$ lying over $\overline{w}'$. It follows that $W \to W'$ is an open immersion over an open neighbourhood of $\overline{w}'$ in $W'$, see Étale Morphisms, Lemma 41.14.2. Shrinking $V$ and $W'$ we may assume $W \to W'$ is an isomorphism. Thus $s$ may be viewed as a section $s'$ of $\mathcal{F}$ over the open subscheme $W' \subset X_ V$ which is finite over $V$. Hence by definition $(W', s')$ defines an element of $j_{p!}\mathcal{F}(V)$ which maps to $s$ as desired.

Let us prove the arrow is injective. To do this, let $s = \sum _{i = 1, \ldots , n} (Z_ i, s_ i)$ as in (63.4.0.1) be an element of $f_{p!}\mathcal{F}(V)$ where $(V, \overline{v})$ is an étale neighbourhood of $\overline{y}$. Assume $s$ maps to zero under the map constructed above. First, after replacing $(V, \overline{v})$ by an étale neighbourhood of itself, we may assume there exist decompositions $Z_ i = Z_{i, 1} \amalg \ldots \amalg Z_{i, m_ i}$ into open and closed subschemes such that each $Z_{i, j}$ has exactly one geometric point over $\overline{v}$. Say under the obvious direct sum decomposition

the element $s_ i$ corresponds to $\sum s_{i, j}$. We may use relations (1) and (2) to replace $s$ by $\sum _{i = 1, \ldots , n} \sum _{j = 1, \ldots , m_ i} (Z_{i, j}, s_{i, j})$. In other words, we may assume $Z_ i$ has a unique geometric point lying over $\overline{v}$. Let $\overline{x}_1, \ldots , \overline{x}_ m$ be the geometric points of $X$ over $\overline{y}$ corresponding to the geometric points of our $Z_ i$ over $\overline{v}$; note that for one $j \in \{ 1, \ldots , m\} $ there may be multiple indices $i$ for which $\overline{x}_ j$ corresponds to a point of $Z_ i$. By More on Morphisms, Lemma 37.41.3 applied to both $X_ V \to V$ after replacing $(V, \overline{v})$ by an étale neighbourhood of itself we may assume there exist open subschemes

and a geometric point $\overline{w}_ j$ of $W_ j$ mapping to $\overline{x}_ j$ and $\overline{v}$ such that $W_ j \to V$ is finite and $\overline{w}_ j$ is the only geometric point of $W_ j$ mapping to $\overline{v}$. After shrinking $V$ we may assume $Z_ i \subset W_ j$ for some $j$ and we have the map $H_{Z_ i}(\mathcal{F}) \to H_{W_ j}(\mathcal{F})$. Thus by the relation (2) we see that our element is equivalent to an element of the form

for some $t_ j \in H_{W_ j}(\mathcal{F})$. Clearly, this element is mapped simply to the class of $t_ j$ in the summand $\mathcal{F}_{\overline{x}_ j}$. Since $s$ maps to zero, we find that $t_ j$ maps to zero in $\mathcal{F}_{\overline{x}_ j}$. This implies that $t_ j$ restricts to zero on an open neighbourhood of $\overline{w}_ j$ in $W_ j$, see Étale Cohomology, Lemma 59.31.2. Shrinking $V$ once more we obtain $t_ j = 0$ for all $j$ as desired. $\square$

Proof. Please read the proof of Étale Cohomology, Lemma 59.70.6 before reading the proof of this lemma. Let $V$ be an object of $X_{\acute{e}tale}$. Recall that

Given $\varphi $ we obtain an open subscheme $Z_\varphi \subset U_ V = U \times _ X V$, namely, the image of the graph of $\varphi $. Via $\varphi $ we obtain an isomorphism $V \to Z_\varphi $ over $U$ and we can think of an element

as a section of $\mathcal{F}$ over $Z_{\varphi }$. Since $Z_\varphi \subset U_ V$ is open, we actually have $H_{Z_\varphi }(\mathcal{F}) = \mathcal{F}(Z_\varphi )$ and we can think of $s_\varphi $ as an element of $H_{Z_\varphi }(\mathcal{F})$. Having said this, our map $j_{p!}\mathcal{F} \to f_{p!}\mathcal{F}$ is defined by the rule

with right hand side a sum as in (63.4.0.1). We omit the verification that this is compatible with restriction mappings and functorial in $\mathcal{F}$.

To finish the proof, we claim that given a geometric point $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ there is a commutative diagram

where the top horizontal arrow is constructed in the proof of Étale Cohomology, Proposition 59.70.3, the bottom horizontal arrow is constructed in the proof of Lemma 63.4.2, the right vertical arrow is the obvious equality, and the left veritical arrow is the map defined in the previous paragraph on stalks. The claim follows in a straightforward manner from the explicit description of all of the arrows involved here and in the references given. Since the horizontal arrows are isomorphisms we conclude so is the left vertical arrow. Hence we find that our map induces an isomorphism on sheafifications by Étale Cohomology, Theorem 59.29.10. $\square$

Proof. The formula for the stalks is immediate (and in fact equivalent) to Lemma 63.4.2. The exactness of the functor follows immediately from this and the fact that exactness may be checked on stalks, see Étale Cohomology, Theorem 59.29.10. $\square$

Proof. Here as usual we set $X_{i_0 \ldots i_ p} = X_{i_0} \cap \ldots \cap X_{i_ p}$ and we denote $f_{i_0 \ldots i_ p}$ the restriction of $f$ to $X_{i_0 \ldots i_ p}$. The maps in the complex are the maps constructed in Remark 63.4.6 with sign rules as in the Čech complex. Exactness follows easily from the description of stalks in Lemma 63.4.5. Details omitted. $\square$

Proof. With conventions as in Remark 63.4.9 we will explicitly construct a map

of abelian presheaves on $Y_{\acute{e}tale}$. By the discussion in Remark 63.4.9 this will determine a canonical map $g^{-1}f_!\mathcal{F} \to f'_!(g')^{-1}\mathcal{F}$. Finally, we will show this map induces isomorphisms on stalks and conclude by Étale Cohomology, Theorem 59.29.10.

Construction of the map $c$. Let $V \in Y_{\acute{e}tale}$ and consider a section $s = \sum _{i = 1, \ldots , n} (Z_ i, s_ i)$ as in (63.4.0.1) defining an element of $f_{p!}\mathcal{F}(V)$. The value of $g_*f'_{p!}(g')^{-1}\mathcal{F}$ at $V$ is $f'_{p!}(g')^{-1}\mathcal{F}(V')$ where $V' = V \times _ Y Y'$. Denote $Z'_ i \subset X'_{V'}$ the base change of $Z_ i$ to $V'$. By (2) there is a pullback map $H_{Z_ i}(\mathcal{F}) \to H_{Z'_ i}((g')^{-1}\mathcal{F})$. Denoting $s'_ i \in H_{Z'_ i}((g')^{-1}\mathcal{F})$ the image of $s_ i$ under pullback, we set $c(s) = \sum _{i = 1, \ldots , n} (Z'_ i, s'_ i)$ as in (63.4.0.1) defining an element of $f'_{p!}(g')^{-1}\mathcal{F}(V')$. We omit the verification that this construction is compatible the relations (1) and (2) and compatible with restriction mappings. The construction is clearly functorial in $\mathcal{F}$.

Let $\overline{y}' : \mathop{\mathrm{Spec}}(k) \to Y'$ be a geometric point with image $\overline{y} = g \circ \overline{y}'$ in $Y$. Observe that $X'_{\overline{y}'} = X_{\overline{y}}$ by transitivity of fibre products. Hence $g'$ produces a bijection $\{ f'(\overline{x}') = \overline{y}'\} \to \{ f(\overline{x}) = \overline{y}\} $ and if $\overline{x}'$ maps to $\overline{x}$, then $((g')^{-1}\mathcal{F})_{\overline{x}'} = \mathcal{F}_{\overline{x}}$ by Étale Cohomology, Lemma 59.36.2. Now we claim that the diagram

commutes where the horizontal arrows are given in the proof of Lemma 63.4.2 and where the right vertical arrow is an equality by what we just said above. The southwest arrow is described in Remark 63.4.9 as the pullback map, i.e., simply given by our construction $c$ above. Then the simple description of the image of a sum $\sum (Z_ i, z_ i)$ in the stalk at $\overline{x}$ given in the proof of Lemma 63.4.2 immediately shows the diagram commutes. This finishes the proof of the lemma. $\square$

Proof. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. With conventions as in Remark 63.4.9 we will explicitly construct a map

of abelian presheaves on $Y_{\acute{e}tale}$. By the discussion in Remark 63.4.9 this will determine a canonical map $c^\# : f_!\mathcal{F} \to g_*f'_!\mathcal{F}$. We will show that $c^\# $ has image contained in the subsheaf $g_!f'_!\mathcal{F}$, thereby obtaining a map $c' : f_!\mathcal{F} \to g_!f'_!\mathcal{F}$. Next, we will prove (a), (b), and (c) that. Finally, part (b) will allow us to show that $c'$ is an isomorphism.

Construction of the map $c$. Let $V \in Y_{\acute{e}tale}$ and let $s = \sum (Z_ i, s_ i)$ be a sum as in (63.4.0.1) defining an element of $f_{p!}\mathcal{F}(V)$. Recall that $Z_ i \subset X_ V = X \times _ Y V$ is a locally closed subscheme finite over $V$. Setting $V' = Y' \times _ Y V$ we get $X_{V'} = X \times _{Y'} V' = X_ V$. Hence $Z_ i \subset X_{V'}$ is locally closed and $Z_ i$ is finite over $V'$ because $g$ is separated (Morphisms, Lemma 29.44.14). Hence we may set $c(s) = \sum (Z_ i, s_ i)$ but now viewed as an element of $f'_{p!}\mathcal{F}(V') = (g_*f'_{p!}\mathcal{F})(V)$. The construction is clearly compatible with relations (1) and (2) and compatible with restriction mappings and hence we obtain the map $c$.

Observe that in the discussion above our section $c(s) = \sum (Z_ i, s_ i)$ of $f'_!\mathcal{F}$ over $V'$ restricts to zero on $V' \setminus \mathop{\mathrm{Im}}(\coprod Z_ i \to V')$. Since $\mathop{\mathrm{Im}}(\coprod Z_ i \to V')$ is proper over $V$ (for example by Morphisms, Lemma 29.41.10) we conclude that $c(s)$ defines a section of $g_!f'_!\mathcal{F} \subset g_*f'_!\mathcal{F}$ over $V$. Since every local section of $f_!\mathcal{F}$ locally comes from a local section of $f_{p!}\mathcal{F}$ we conclude that the image of $c^\# $ is contained in $g_!f'_!\mathcal{F}$. Thus we obtain an induced map $c' : f_!\mathcal{F} \to g_!f'_!\mathcal{F}$ factoring $c^\# $ as predicted in the first paragraph of the proof.

Proof of (a). Let $Y'_1 \subset Y'$ be an open subscheme and set $X_1 = (f')^{-1}(W')$. We obtain a diagram

where the horizontal arrows are open immersions. Then our claim is that the diagram

commutes where the left vertical arrow is Remark 63.4.6 and the right vertical arrow is Remark 63.3.5. The equality sign in the diagram comes about because $f'_1$ is the restriction of $f'$ to $Y'_1$ and our construction of $f'_!$ is local on the base. Finally, to prove the commutativity we choose an object $V$ of $Y_{\acute{e}tale}$ and a formal sum $s_1 = \sum (Z_{1, i}, s_{1, i})$ as in (63.4.0.1) defining an element of $f_{1, p!}\mathcal{F}|_{X_1}(V)$. Recall this means $Z_{1, i} \subset X_1 \times _ Y V$ is locally closed finite over $V$ and $s_{1, i} \in H_{Z_{1, i}}(\mathcal{F})$. Then we chase this section across the maps involved, but we only need to show we end up with the same element of $g_*f'_!\mathcal{F}(V) = f'_!\mathcal{F}(Y' \times _ Y V)$. Going around both sides of the diagram the reader immediately sees we end up with the element $\sum (Z_{1, i}, s_{1, i})$ where now $Z_{1, i}$ is viewed as a locally closed subscheme of $X \times _{Y'} (Y' \times _ Y V) = X \times _ Y V$ finite over $Y' \times _ Y V$.

Proof of (b). Let $b : Y_1 \to Y$ be a morphism of schemes. Let us form the commutative diagram

with cartesian squares. We claim that our construction is compatible with the base change maps of Lemmas 63.4.10 and 63.3.12, i.e., that the top rectangle of the diagram

commutes. The verification of this is completely routine and we urge the reader to skip it. Since the arrows going from the middle row down to the bottom row are injective, it suffices to show that the outer diagram commutes. To show this it suffices to take a local section of $b^{-1}f_!\mathcal{F}$ and show we end up with the same local section of $g_{1, *}f'_{1, !}a^{-1}\mathcal{F}$ going around either way. However, in fact it suffices to check this for local sections which are of the the pullback by $b$ of a section $s = \sum (Z_ i, s_ i)$ of $f_{p!}\mathcal{F}(V)$ as above (since such pullbacks generate the abelian sheaf $b^{-1}f_!\mathcal{F}$). Denote $V_1$, $V'_1$, and $Z_{1, i}$ the base change of $V$, $V' = Y' \times _ Y V$, $Z_ i$ by $Y_1 \to Y$. Recall that $Z_ i$ is a locally closed subscheme of $X_ V = X_{V'}$ and hence $Z_{1, i}$ is a locally closed subscheme of $(X_1)_{V_1} = (X_1)_{V'_1}$. Then $b^{-1}c'$ sends the pullback of $s$ to the pullback of the local section $c(s) \sum (Z_ i, s_ i)$ viewed as an element of $f'_{p!}\mathcal{F}(V') = (g_*f'_{p!}\mathcal{F})(V)$. The composition of the bottom two base change maps simply maps this to $\sum (Z_{i, 1}, s_{1, i})$ viewed as an element of $f'_{1, p!}a^{-1}\mathcal{F}(V'_1) = g_{1, *}f'_{1, p!}a^{-1}\mathcal{F}(V_1)$. On the other hand, the base change map at the top of the diagram sends the pullback of $s$ to $\sum (Z_{1, i}, s_{1, i})$ viewed as an element of $f_{1, !}a^{-1}\mathcal{F}(V_1)$. Then finally $c'_1$ by its very construction does indeed map this to $\sum (Z_{i, 1}, s_{1, i})$ viewed as an element of $f'_{1, p!}a^{-1}\mathcal{F}(V'_1) = g_{1, *}f'_{1, p!}a^{-1}\mathcal{F}(V_1)$ and the commutativity has been verified.

Proof of (c). This follows from comparing the definitions for both maps; we omit the details.

To finish the proof it suffices to show that the pullback of $c'$ via any geometric point $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ is an isomorphism. Namely, pulling back by $\overline{y}$ is the same thing as taking stalks and $\overline{y}$ (Étale Cohomology, Remark 59.56.6) and hence we can invoke Étale Cohomology, Theorem 59.29.10. By the compatibility (b) just shown, we conclude that we may assume $Y$ is the spectrum of $k$ and we have to show that $c'$ is an isomorphism. To do this it suffices to show that the induced map

is an isomorphism. The equalities hold by Lemmas 63.4.5 and 63.3.11. Recall that $X$ is a disjoint union of spectra of Artinian local rings with residue field $k$, see Varieties, Lemma 33.20.2. Since the left and right hand side commute with direct sums (details omitted) we may assume that $\mathcal{F}$ is a skyscraper sheaf $x_*A$ supported at some $x \in X$. Then $f'_!\mathcal{F}$ is the skyscraper sheaf at the image $y'$ of $x$ in $Y$ by Lemma 63.4.5. In this case it is obvious that our construction produces the identity map $A \to H^0_ c(Y', y'_*A) = A$ as desired. $\square$

Proof. If $f$ and $g$ are separated, then this is a special case of Lemma 63.3.13. If $g$ is separated, then this is a special case of Lemma 63.4.11 which moreover agrees with the case where $f$ and $g$ are separated.

Construction in the general case. Choose an open covering $Y = \bigcup Y_ i$ such that the restriction $g_ i : Y_ i \to Z$ of $g$ is separated. Set $X_ i = f^{-1}(Y_ i)$ and denote $f_ i : X_ i \to Y_ i$ the restriction of $f$. Also denote $h = g \circ f$ and $h_ i : X_ i \to Z$ the restriction of $h$. Consider the following diagram

By Lemma 63.4.7 the top and bottom row in the diagram are exact. By Lemma 63.4.11 the top left square commutes. The vertical arrows in the lower left square come about because $(f_!\mathcal{F})|_{Y_{i_0i_1}} = f_{i_0i_1, !}\mathcal{F}|_{X_{i_0i_1}}$ and $(f_!\mathcal{F})|_{Y_{i_0}} = f_{i_0, !}\mathcal{F}|_{X_{i_0}}$ as the construction of $f_!$ is local on the base. Moreover, these equalities are (of course) compatible with the identifications $((f_!\mathcal{F})|_{Y_{i_0}})|_{Y_{i_0i_1}} = (f_!\mathcal{F})|_{Y_{i_0i_1}}$ and $(f_{i_0, !}\mathcal{F}|_{X_{i_0}})|_{Y_{i_0i_1}} = f_{i_0i_1, !}\mathcal{F}|_{X_{i_0i_1}}$ which are used (together with the covariance for open embeddings for $Y_{i_0i_1} \subset Y_{i_0}$) to define the horizontal maps of the lower left square. Thus this square commutes as well. In this way we conclude there is a unique dotted arrow as indicated in the diagram and moreover this arrow is an isomorphism.

Proof of properties (1) – (5). Fix the open covering $Y = \bigcup Y_ i$. Observe that if $Y \to Z$ happens to be separated, then we get a dotted arrow fitting into the huge diagram above by using the map of Lemma 63.4.11 (by the very properties of that lemma). This proves (2) and hence also (1) by the compatibility of the maps of Lemma 63.4.11 and Lemma 63.3.13. Next, for any scheme $Z'$ over $Z$, we obtain the compatibility in (5) for the map $(g' \circ f')_! \to g'_! \circ f'_!$ constructed using the open covering $Y' = \bigcup b^{-1}(Y_ i)$. This is clear from the corresponding compatibility of the maps constructed in Lemma 63.4.11. In particular, we can consider a geometric point $\overline{z} : \mathop{\mathrm{Spec}}(k) \to Z$. Since $X_{\overline{z}} \to Y_{\overline{z}} \to \mathop{\mathrm{Spec}}(k)$ are separated maps, we find that the base change of $(g \circ f)_!\mathcal{F} \to g_! f_! \mathcal{F}$ by $\overline{z}$ is equal to the map of Lemma 63.3.13. The reader then immediately sees that we obtain property (3). Of course, property (3) guarantees that our transformation of functors $(g \circ f)_! \to g_! \circ f_!$ constructed using the open covering $Y = \bigcup Y_ i$ doesn't depend on the choice of this open covering. Finally, property (4) follows by looking at what happens on stalks using the already proven property (3). $\square$