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Let $f:\operatorname{Spec}B \rightarrow \operatorname{Spec}A$ be a morphism of spectra, how do I show, for any $A$-module $M$, we have $f^*(\tilde{M})$ is isomorphic to $\widetilde{M\otimes_{A}B}$?

This question was asked here Inverse image of the sheaf associated to a module. In Martin's answer, he used the adjointness property. However he also mentioned that one can use the complicated ad hoc definition to compute this inverse image.

How can one prove this without using adjointness? Can I compute directly, for example, the module $f^*(\tilde{M})(D(g))$ for any $g\in B$ and then compare it with that of $(M\otimes_{A}B)_g$?

Bernard
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Xin
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    By "complicated ad hoc definition" do you mean $f^{-1}(\widetilde{M})\otimes_{f^{-1}\mathcal{O}{\operatorname{Spec} A}} \mathcal{O}{\operatorname{Spec B}}$? – KReiser Feb 12 '21 at 22:37
  • @KReiser Yes, that is the definition of inverse image. – Xin Feb 12 '21 at 22:39
  • Just checking - whether that's complicated or ad hoc is a matter of opinion, and I wanted to make sure I knew exactly what definition you wanted before potentially writing an answer. – KReiser Feb 12 '21 at 22:40
  • I wrote an answer about this last year here which might be helpful. – KReiser Feb 13 '21 at 03:28

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