Lemma 17.22.5 (0C6I)—The Stacks project

Lemma 17.22.5. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$ , $\mathcal{G}$ be $\mathcal{O}_ Y$ -modules. If $\mathcal{F}$ is finitely presented and $f$ is flat, then the canonical map

$f^*\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{F}, \mathcal{G}) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f^*\mathcal{F}, f^*\mathcal{G})$

is an isomorphism.

Proof. Note that $f^*\mathcal{F}$ is also finitely presented (Lemma 17.11.4). Let $x \in X$ map to $y \in Y$ . Looking at the stalks at $x$ we get an isomorphism by Lemma 17.22.4 and More on Algebra, Lemma 15.65.4 to see that in this case $\mathop{\mathrm{Hom}}\nolimits$ commutes with base change by $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ . Second proof: use the exact same argument as given in the proof of Lemma 17.22.4. $\square$

Comments (2)

Comment #2181 by JuanPablo on August 27, 2016 at 18:19

In the statement, the morphism of ringed spaces

$f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ should be flat as in Definition 17.18.3 (tag 08KT).

Comment #2208 by Johan on August 29, 2016 at 19:50

Thank you very much indeed! See here.

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