Lemma 62.4.10. Consider a cartesian square

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

of schemes with $f$ locally quasi-finite. There is an isomorphism $g^{-1}f_!\mathcal{F} \to f'_!(g')^{-1}\mathcal{F}$ functorial for $\mathcal{F}$ in $\textit{Ab}(X_{\acute{e}tale})$ which is compatible with the descriptions of stalks given in Lemma 62.4.5 (see proof for the precise statement).

**Proof.**
With conventions as in Remark 62.4.9 we will explicitly construct a map

\[ c : f_{p!}\mathcal{F} \longrightarrow g_*f'_{p!}(g')^{-1}\mathcal{F} \]

of abelian presheaves on $Y_{\acute{e}tale}$. By the discussion in Remark 62.4.9 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 (62.4.0.1) 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 (62.4.0.1) 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

\[ \xymatrix{ (g^{-1}f_!\mathcal{F})_{\overline{y}'} \ar@{=}[r] \ar[d] & (f_!\mathcal{F})_{\overline{y}} \ar[r] \ar[ld] & \bigoplus \nolimits _{f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}} \ar[d] \\ (f'_!(g')^{-1}\mathcal{F})_{\overline{y}'} \ar[rr] & & \bigoplus \nolimits _{f'(\overline{x}') = \overline{y}'} (g')^{-1}\mathcal{F}_{\overline{x}'} } \]

commutes where the horizontal arrows are given in the proof of Lemma 62.4.2 and where the right vertical arrow is an equality by what we just said above. The southwest arrow is described in Remark 62.4.9 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 62.4.2 immediately shows the diagram commutes. This finishes the proof of the lemma.
$\square$

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