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Assume $S$ is a Noetherian scheme and $f:\operatorname{Spec}(R)\longrightarrow S$ is any morphism from an affine scheme to $S$. Assume $\mathcal{Q}$ is a coherent sheaf over $S$ and $\operatorname{Sym}_{\mathcal{O}_{S}}\mathcal{Q}:=\operatorname{Tens}_{\mathcal{O}_{S}}\mathcal{Q}/(a\otimes b-b\otimes a)$ is its associated symmetrical algebra. Nitsure (https://arxiv.org/abs/math/0504590, p.17) makes the claim that \begin{equation*} \operatorname{Mor}_{S}(\operatorname{Spec}(R),\operatorname{Spec}\operatorname{Sym}_{\mathcal{O}_{S}}\mathcal{Q})\cong\operatorname{Hom}_{\mathcal{O}_{S}-\operatorname{mod}}(\mathcal{Q},f_{*}\mathcal{O}_{R})\text{.} \end{equation*} Why is that the case?

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$\renewcommand{\Spec}{\operatorname{Spec}}\renewcommand{\Sym}{\operatorname{Sym}}$ Let's work locally on $S$, say over $\Spec A$ an open subscheme of $S$, and suppose $\mathcal{Q}|_{\Spec A}\cong \widetilde{Q}$ for some $A$-module $Q$. Then $(\Spec\Sym\mathcal{Q})|_{\Spec A}\cong \Spec \Sym Q$, and $S$-maps $f^{-1}(\Spec A)\to \Spec \Sym Q$ are in bijection with $A$-maps $\Sym Q \to \Gamma(\mathcal{O}_{f^{-1}(\Spec A)},f^{-1}(\Spec A))$ by a relative version of 01I1. But such a morphism is determined by where $Q$ goes, and $\Gamma(\mathcal{O}_{f^{-1}(\Spec A)},f^{-1}(\Spec A))=(f_*\mathcal{O}_R)(\Spec A)$, so such a morphism over $\Spec A$ is determined by a map $Q\to (f_*\mathcal{O}_R)(\Spec A)$. Assembling all of this data together over all affine open $\Spec A\subset S$, we have the result.

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