Lemma 20.54.4. Let f : X \to Y be a morphism of ringed spaces such that f is a homeomorphism onto a closed subset. Then (20.54.2.1) 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 (20.54.2.1) 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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