Lemma 20.51.4. Let $f : X \to Y$ be a morphism of ringed spaces such that $f$ is a homeomorphism onto a closed subset. Then (20.51.2.1) is an isomorphism always.

Proof. Since $f$ is a homeomorphism onto a closed subset, the functor $f_*$ is exact (Modules, Lemma 17.6.1). Hence $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a K-flat complex $\mathcal{K}^\bullet$ of $\mathcal{O}_ Y$-modules representing $K$ and choose any complex $\mathcal{E}^\bullet$ of $\mathcal{O}_ X$-modules representing $E$. Then $Lf^*K$ is represented by $f^*\mathcal{K}^\bullet$ which is a K-flat complex of $\mathcal{O}_ X$-modules (Lemma 20.26.8). Thus the right hand side of (20.51.2.1) is represented by

$f_*\text{Tot}(\mathcal{E}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{K}^\bullet )$

By the same reasoning we see that the left hand side is represented by

$\text{Tot}(f_*\mathcal{E}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{K}^\bullet )$

Since $f_*$ commutes with direct sums (Modules, Lemma 17.6.3) it suffices to show that

$f_*(\mathcal{E} \otimes _{\mathcal{O}_ X} f^*\mathcal{K}) = f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathcal{K}$

for any $\mathcal{O}_ X$-module $\mathcal{E}$ and $\mathcal{O}_ Y$-module $\mathcal{K}$. We will check this by checking on stalks. Let $y \in Y$. If $y \not\in f(X)$, then the stalks of both sides are zero. If $y = f(x)$, then we see that we have to show

$\mathcal{E}_ x \otimes _{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y) = \mathcal{E}_ x \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y$

(using Sheaves, Lemma 6.32.1 and Lemma 6.26.4). This equality holds and therefore the lemma has been proved. $\square$

Comment #1484 by jojo on

In the statement I think it should say that $f$ is a homeomorphism onto a closed subset (not a homomorphism).

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