The Stacks project

Lemma 17.28.6. Let $f : Y \to X$ be a continuous map of topological spaces. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings on $X$. Then there is a canonical identification $f^{-1}\Omega _{\mathcal{O}_2/\mathcal{O}_1} = \Omega _{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$ compatible with universal derivations.

Proof. This holds because the sheaf $\Omega _{\mathcal{O}_2/\mathcal{O}_1}$ is the cokernel of the map ( and a similar statement holds for $\Omega _{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$, because the functor $f^{-1}$ is exact, and because $f^{-1}(\mathcal{O}_2[\mathcal{O}_2]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_2]$, $f^{-1}(\mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_2 \times f^{-1}\mathcal{O}_2]$, and $f^{-1}(\mathcal{O}_2[\mathcal{O}_1]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_1]$. $\square$

Comments (4)

Comment #2361 by Dominic Wynter on

Is the functor on sheaves of rings truly exact, or simply right-exact (as is the case for abelian sheaves)? This doesn't change the content of the proof, however.

Comment #2424 by on

What was meant was that is an exact functor on the category of sheaves. But I also do not understand your comment as is an exact functor on the category of abelian sheaves.

Comment #8563 by on

In case anyone needs it, here's a very detailed proof of the last three equalities:

Along the proof we will use the adjunction pointed out in the second paragraph of this comment.

Lemma 1. Let , be sheaves over , respectively, of rings and of sets, and let be a point. There is an isomorphism of -modules , that is natural on .

Proof. Consider the unit of the adjunction and take map on stalks . This induces an -linear map (natural on , by naturality of the unit). Namely, it maps to . Conversely, for each open neighborhood of , there is an obvious map . Since is an -module, this induces a map . Since the latter is compatible with restrictions to open neighborhoods of , we obtain a canonical map . But it maps to , so the two maps are mutually inverse.

Lemma 2. Same notation as in Lemma 1, and suppose that is a sheaf of -modules. Then the following square commutes: where the left arrow is obtained from the induced map on stalks plus the isomorphism of Lemma 1.

We leave the proof to the reader. In the diagram, the horizontal maps are \eqref{1} and the universal property of the free module over a set.

Lemma 3. Notations as in Lemma 1. Let be a continuous map of topological spaces. Then .

Proof. Pulling back the unit of the adjunction gives a map . Note that is a -module. By the adjunction \eqref{1}, we get a map . On stalks, this map equals where we have used Lemma 1. By Lemma 2, \eqref{2} is adjoint to , i.e., to . In other words, the map \eqref{2} equals the canonical morphism of Lemma 1, which is an isomorphism.

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