Show that $f^{-1}\Omega_{\mathcal{O}_2/\mathcal{O}_1}=\Omega_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$

Question

Let $f:Y\to X$ be a continuous map of topological spaces, and $\varphi:\mathcal{O}_1\to\mathcal{O}_2$ a map of sheaves of rings on $X$, then I claim that there is a natural isomorphism $$f^{-1}\Omega_{\mathcal{O}_2/\mathcal{O}_1}\cong\Omega_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}.$$ This is the content of Stacks Lemma 25.6 in Sheaves of Modules (Tag 08RR). However, in the proof, we define, for any sheaf $\mathcal{F}$ of sets of $X$, the sheaf of $\mathcal{O}_2$-modules $\mathcal{O}_2[\mathcal{F}]$ to be the sheaf associated to the presheaf $$U\mapsto\mathcal{O}_2(U)[\mathcal{F}(U)]$$ with the natural restriction maps. However, the lemma states without proof that $$f^{-1}(\mathcal{O}_2[\mathcal{F}]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{F}]$$ and I'm wondering why that's the case. I'm unfortunately not very good at unpacking repeated sheafifications, so I'm a bit lost here.

Answer 1

Recall the fact that we have a pair of adjoint functors $\mathrm{free}:\mathbf{Sets}\to\mathbf{Ab}$ and $\mathrm{forg}:\mathbf{Ab}\to\mathbf{Sets}$ such that $\mathrm{free}\dashv\mathrm{forg}$. We will replicate this with the category $\mathbf{pMod}(X,\mathcal{O})=\mathbf{pMod}(\mathcal{O})$ of presheaves of $\mathcal{O}$-modules, the category $\mathbf{Mod}(X,\mathcal{O})=\mathbf{Mod}(\mathcal{O})$ of $\mathcal{O}$-modules, the category $\mathbf{pShv}(X)$ of presheaves of sets on $X$, and the category $\mathbf{Shv}(X)$ of sheaves of sets on $X$. So we first define a pair of adjoint functors $$\mathbf{pShv}(X)\xrightarrow{\mathrm{pfree}}\mathbf{pMod}(X,\mathcal{O})\xrightarrow{\mathrm{pforg}}\mathbf{pShv}(X)$$ such that $\mathrm{pfree}\dashv\mathrm{pforg}$ and we use sheafification to define a pair of adjoint functors $$\mathbf{Shv}(X)\xrightarrow{\mathrm{free}}\mathbf{Mod}(X,\mathcal{O})\xrightarrow{\mathrm{forg}}\mathbf{Shv}(X)$$ such that $\mathrm{free}\dashv\mathrm{forg}$. Some details are given below. We define, for $\mathscr{F}\in\mathbf{pShv}(X)$, the presheaf $\mathrm{pfree}(\mathscr{F})\in\mathbf{pMod}(X,\mathcal{O})$ by setting $$\mathrm{pfree}(\mathscr{F})(U) = \mathcal{O}(U)[\mathscr{F}(U)]$$ to be the free $\mathcal{O}(U)$-module generated by $\mathscr{F}(U)$, with the obvious restriction maps, and the obvious induced morphisms $\mathrm{pfree}(\varphi):\mathrm{pfree}(\mathscr{F})\to\mathrm{pfree}(\mathscr{G})$. If we let $$\mathrm{shf}:\mathbf{pMod}(X,\mathcal{O})\to\mathbf{Mod}(X,\mathcal{O})$$ be the sheafification functor, then we let $$\mathrm{free} = \mathrm{shf}\circ\mathrm{pfree}\vert_{\mathbf{Shv}(X)}:\mathbf{Shv}(X)\to\mathbf{Mod}(X,\mathcal{O})$$ and $$\mathrm{forg} = \mathrm{pforg}\vert_{\mathbf{Mod}(X,\mathcal{O})}:\mathbf{Mod}(X,\mathcal{O})\to\mathbf{Shv}(X)$$ be the forgetful functor, so it is straightforward (though tedious) to show that $\mathrm{free}\dashv\mathrm{forg}$.

Now, we will use the Yoneda lemma. Fix $\mathscr{F}\in\mathbf{Shv}(Y)$. Let $\mathscr{G}\in\mathbf{Mod}(X,f^{-1}\mathcal{O})$, so we note that \begin{align*} \mathrm{Hom}_{\mathbf{Mod}(X,f^{-1}\mathcal{O})}(f^{-1}\mathcal{O}[f^{-1}\mathscr{F}],\mathscr{G}) &\cong \mathrm{Hom}_{\mathbf{Shv}(X)}(f^{-1}\mathscr{F},\mathscr{G}) \\ &\cong\mathrm{Hom}_{\mathbf{Shv}(Y)}(\mathscr{F},f_{*}\mathscr{G}) \\ &\cong\mathrm{Hom}_{\mathbf{Mod}(Y,\mathcal{O})}(\mathcal{O}[\mathscr{F}],f_{*}\mathscr{G}) \end{align*} where all isomorphisms are natural. Therefore, by the Yoneda lemma, it suffices to show that for any continuous map $f:X\to Y$ and sheaf of rings $\mathcal{O}$ on $Y$, we have a natural isomorphism $$\mathrm{Hom}_{\mathbf{Mod}(Y,\mathcal{O})}(\mathscr{F},f_{*}\mathscr{G})\cong\mathrm{Hom}_{\mathbf{Mod}(X,f^{-1}\mathcal{O})}(f^{-1}\mathscr{F},\mathscr{G})$$

But if we define the map of ringed spaces $f:(X,f^{-1}\mathcal{O})\to(Y,\mathcal{O})$ via the natural map $\mathcal{O}\to f_{*}f^{-1}\mathcal{O}$ (induced by $\mathrm{id}_{f^{-1}\mathcal{O}}:f^{-1}\mathcal{O}\to f^{-1}\mathcal{O}$), but note that with this map, $f^{-1}\mathscr{F}=f^{*}\mathscr{F}$ so since $f^{*}\dashv f_{*}$, we have the final natural isomorphism, and therefore $$\mathrm{Hom}_{\mathbf{Mod}(X,f^{-1}\mathcal{O})}(f^{-1}\mathcal{O}[f^{-1}\mathscr{F}],\mathscr{G})\cong \mathrm{Hom}_{\mathbf{Mod}(X,f^{-1}\mathcal{O})}(f^{-1}(\mathcal{O}[\mathscr{F}]),\mathscr{G})$$ for all $\mathscr{G}$, and therefore, by the Yoneda lemma, $f^{-1}\mathcal{O}[f^{-1}\mathscr{F}]\cong f^{-1}(\mathcal{O}[\mathscr{F}])$.