**Proof.**
By Lemma 103.5.3 part (2)(c) the object $Lg_!K$ is in $D_{\textit{LQCoh}^{fbc}}(\mathcal{O}_\mathcal {X})$. It follows readily from this that the map displayed in the lemma is an isomorphism if $\mathcal{O}(x') \to \mathcal{O}(x)$ is a flat ring map; we omit the details.

In this paragraph we argue that the question is local for the étale topology. Let $x \to x'$ be a general morphism of $\mathcal{X}_{affine}$. Let $\{ x'_ i \to x'\} $ be a covering in $\mathcal{X}_{affine, {\acute{e}tale}}$. Set $x_ i = x \times _{x'} x'_ i$ so that $\{ x_ i \to x\} $ is a covering of $\mathcal{X}_{affine, {\acute{e}tale}}$ too. Then $\mathcal{O}(x') \to \prod \mathcal{O}(x'_ i)$ is a faithfully flat étale ring map and

\[ \prod \mathcal{O}(x_ i) = \mathcal{O}(x) \otimes _{\mathcal{O}(x')} \left(\prod \mathcal{O}(x'_ i)\right) \]

Thus a simple algebra argument we omit shows that it suffices to prove the result in the statement of the lemma holds for each of the morphisms $x_ i \to x'_ i$ in $\mathcal{X}_{affine}$. In other words, the problem is local in the étale topology.

Choose a scheme $X$ and a surjective smooth morphism $f : X \to \mathcal{X}$. We may view $f$ as an object of $\mathcal{X}$ (by our abuse of notation) and then $(\mathit{Sch}/X)_{fppf} = \mathcal{X}/f$, see Sheaves on Stacks, Section 95.9. By Sheaves on Stacks, Lemma 95.19.10 for example, there exist an étale covering $\{ x'_ i \to x'\} $ such that $x'_ i : U'_ i = p(x'_ i) \to \mathcal{X}$ factors through $f$. By the result of the previous paragraph, we may assume that $x \to x'$ is a morphism which is the image of a morphism $U \to U'$ of $(\textit{Aff}/X)_{fppf}$ by the functor $(\mathit{Sch}/X)_{fppf} \to \mathcal{X}$. At this point we see use that the restriction to $(\mathit{Sch}/X)_{fppf}$ of $Lg_!K$ is equal to $f^*Lg_!K = L(g')_!(f')^*K$ by Lemma 103.3.2. This reduces us to the case discussed in the next paragraph.

Assume $\mathcal{X} = (\mathit{Sch}/X)_{fppf}$ and $x \to x'$ corresponds to the morphism of affine schemes $U \to U'$. We may still work étale (or Zariski) locally on $U'$ and hence we may assume $U' \to X$ factors through some affine open of $X$. This reduces us to the case discussed in the next paragraph.

Assume $\mathcal{X} = (\mathit{Sch}/X)_{fppf}$ where $X = \mathop{\mathrm{Spec}}(R)$ is an affine scheme and $x \to x'$ corresponds to the morphism of affine schemes $U \to U'$. Let $M^\bullet $ be a complex of $R$-modules representing $R\Gamma (X, K)$. By the construction in More on Algebra, Lemma 15.59.10 we may assume $M^\bullet = \mathop{\mathrm{colim}}\nolimits P_ n^\bullet $ where each $P_ n^\bullet $ is a bounded above complex of free $R$-modules. Details omitted; see also More on Algebra, Remark 15.59.11. Consider the complex of modules $M^\bullet _{flat, fppf}$ on $X_{flat, fppf} = (\mathit{Sch}/X)_{flat, fppf}$ given by the rule

\[ U \longmapsto \Gamma (U, M^\bullet \otimes _ R \mathcal{O}_ U) \]

This is a complex of sheaves by the discussion in Descent, Section 35.8. There is a canonical map $M^\bullet _{flat, fppf} \to K$ which by our initial remarks of the proof produces an isomorphism on sections over the affine objects of $X_{flat, fppf}$. Since every object of $X_{flat, fppf}$ has a covering by affine objects we see that $M^\bullet _{flat, fppf}$ agrees with $K$.

Let $M^\bullet _{fppf}$ be the complex of modules on $X_{fppf}$ given by the same formula as displayed above. Recall that $Lg_!\mathcal{O} = g_!\mathcal{O} = \mathcal{O}$. Since $Lg_!$ is the left derived functor of $g_!$ we conclude that $Lg_!P_{n, flat, fppf}^\bullet = P_{n, fppf}^\bullet $. Since the functor $Lg_!$ commutes with homotopy colimits (or by its construction in Cohomology on Sites, Lemma 21.37.2) and since $M^\bullet = \mathop{\mathrm{colim}}\nolimits P_ n^\bullet $ we conclude that $Lg_!M^\bullet _{flat, fppf} = M^\bullet _{fppf}$. Say $U = \mathop{\mathrm{Spec}}(A)$, $U' = \mathop{\mathrm{Spec}}(A')$ and $U \to U'$ corresponds to the ring map $A' \to A$. From the above we see that

\[ R\Gamma (U, Lg_!K) = M^\bullet \otimes _ R A \quad \text{and}\quad R\Gamma (U', Lg_!K) = M^\bullet \otimes _ R A' \]

Since $M^\bullet $ is a K-flat complex of $R$-modules, by transitivity of tensor product it follows that

\[ R\Gamma (U', Lg_!K) \otimes _{A'}^\mathbf {L} A \longrightarrow R\Gamma (U, Lg_!K) \]

is a quasi-isomorphism as desired.
$\square$

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