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Suppose $F:\mathcal{C}\hookrightarrow \mathcal{D}$ is a fully faithful symmetric monoidal functor of symmetric monoidal $\infty$-categories. Suppose $x\in\mathcal{C}$. Is it true that $\mathcal{C}[x^{-1}]\rightarrow\mathcal{D}[F(x)^{-1}]$ is also fully faithful? Here $\mathcal{C}[x^{-1}]$ is the colimit of $\mathcal{C}\xrightarrow{x\otimes-}\mathcal{C}\xrightarrow{x\otimes-}\rightarrow\dots$. $D[F(x)^{-1}]$ is similarly defined.

Intuition says that this should be true. Also, we can reduce it to the case where $\mathcal{C}$ is a full symmetric monoidal subcategory of the symmetric monoidal $\infty$-category $\mathcal{D}$. If I am wrong, is there a counterexample?

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    $\begingroup$ Generally, stabilization is usually defined as a limit, using a right adjoint functor $v$ for the transition arrows in the tower. Then the answer to your question is yes, provided that the inclusion $F$ commutes with $v$, because limits of fully faithful functors are fully faithful. In the presentable case you can compute this limit as a colimit in the category of presentable $\infty$-categories and left adjoints (with the transition arrows in the tower being the left adjoint of $v$), but beware that this is not the same as the colimit in the category of $\infty$-categories. $\endgroup$
    – AAK
    Commented Apr 30, 2017 at 11:53

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