Lemma 20.49.4. Let $f : X \to Y$ be a morphism of ringed spaces such that $f$ is a homeomorphism onto a closed subset. Then (20.49.2.1) 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 20.26.7). Thus the right hand side of (20.49.2.1) is represented by

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

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

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

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

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