Lemma 17.27.6. 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 (17.27.2.1) 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 $f^{-1}(\mathcal{O}_2[\mathcal{O}_1]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_1]$. $\square$

Comment #2361 by Dominic Wynter on

Is the functor $f^{-1}$ 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 $f^{-1}$ is an exact functor on the category of sheaves. But I also do not understand your comment as $f^{-1}$ is an exact functor on the category of abelian sheaves.

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