Mapping cone and derived tensor product

Question

This question is in some sense a continuation to this question: Derived Nakayama for complete modules

For the setting: Let $A$ be a ring and let $I$ be some finitely generated ideal in $A$. Let $f\colon \mathcal C\rightarrow \mathcal D$ be a map of chain complexes of derived $I$-complete $A$-modules. I am trying to apply the "derived Nakayama" to the mapping cone of $f$ to produce the following result:

Suppose that $f'\colon \mathcal C\otimes^{\mathbf L} A/I\rightarrow \mathcal D\otimes^{\mathbf L} A/I$ is a quasi-isomorphism. Then $f$ is a quasi-isomorphism.

To do this, I want to relate the mapping cone $cone(f')$ to the mapping cone of $cone(f)\otimes^{\mathbf L}A/I$ but I am at a loss on how to proceed here. Any tips or solutions are welcome.

Answer 1

The comment by skd basically answers your question. I am writing to flesh it out with references, so your question doesn't stay open forever. The derived category of $I$-complete $A$-modules has recently featured in work of Barthel, Heard, and Valenzuela, section 3.4, expanding on earlier work of Greenlees-May. This establishes a closed symmetric monoidal $(\infty,1)$-category of $I$-complete $A$-modules, in which the mapping cone is a homotopy pushout, as explained on nLab. Because the category is closed symmetric monoidal, the functor $-\otimes^L A/I$ commutes with homotopy pushout, so $cone(f') \simeq cone(f)\otimes^L A/I$. A map is a quasi-isomorphism iff its cone is acyclic, so an affirmative answer to your question follows from your previous question, which you linked to.