The Stacks project

Lemma 63.4.11. Let $f' : X \to Y'$ and $g : Y' \to Y$ be composable morphisms of schemes with $f'$ and $f = g \circ f'$ locally quasi-finite and $g$ separated and locally of finite type. Then there is a canonical isomorphism of functors $g_! \circ f'_! = f_!$. This isomorphism is compatible with

  1. covariance with respect to open embeddings as in Remarks 63.3.5 and 63.4.6,

  2. the base change isomorphisms of Lemmas 63.4.10 and 63.3.12, and

  3. equal to the isomorphism of Lemma 63.3.13 via the identifications of Lemma 63.4.1 in case $f'$ is separated.

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

\[ c : f_{p!}\mathcal{F} \longrightarrow g_*f'_{p!}\mathcal{F} \]

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 ( 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

\[ \xymatrix{ X_1 \ar[d]_{f'_1} \ar[r]_ a \ar@/_2em/[dd]_{f_1} & X \ar[d]^{f'} \ar@/^2em/[dd]^ f \\ Y'_1 \ar[d]_{g_1} \ar[r]_{b'} & Y' \ar[d]^ g \\ Y \ar@{=}[r] & Y } \]

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

\[ \xymatrix{ f_{1, !}\mathcal{F}|_{X_1} \ar[r]_{c'_1} \ar[dd] & g_{1, !}f'_{1, !}\mathcal{F}|_{X_1} \ar@{=}[d] \\ & g_{1, !}(f'_!\mathcal{F})|_{Y'_1} \ar[d] \\ f_!\mathcal{F} \ar[r]^{c'} & g_!f'_!\mathcal{F} \ar[r] & g_*f'_!\mathcal{F} } \]

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 ( 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

\[ \xymatrix{ X_1 \ar[d]_{f'_1} \ar[r]_ a \ar@/_2em/[dd]_{f_1} & X \ar[d]^{f'} \ar@/^2em/[dd]^ f \\ Y'_1 \ar[d]_{g_1} \ar[r]_{b'} & Y' \ar[d]^ g \\ Y_1 \ar[r]^ b & Y } \]

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

\[ \xymatrix{ b^{-1}f_!\mathcal{F} \ar[rr] \ar[d]_{b^{-1}c'} & & f_{1, !}a^{-1}\mathcal{F} \ar[d]^{c_1'} \\ b^{-1}g_!f'_!\mathcal{F} \ar[r] \ar[d] & g_{1, !}(b')^{-1}f'_!\mathcal{F} \ar[r] \ar[d] & g_{1, !}f'_{1, !}a^{-1}\mathcal{F} \ar[d] \\ b^{-1}g_*f'_!\mathcal{F} \ar[r] & g_{1, *}(b')^{-1}f'_!\mathcal{F} \ar[r] & g_{1, *}f'_{1, !}a^{-1}\mathcal{F} } \]

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

\[ \bigoplus \nolimits _{x \in X} \mathcal{F}_ x = H^0(Y, f_!\mathcal{F}) \longrightarrow H^0(Y, g_!f'_!\mathcal{F}) = H^0_ c(Y', f'_!\mathcal{F}) \]

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. $\square$

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