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Usually when $F: C \rightarrow D$ is a localization functor, the categories $C$ and $D$ are not equivalent. My question is when is it possible for $C, D$ to be abstractly equivalent but $F$ is not an equivalence. Alternatively, when is there a localization functor $F: C \rightarrow C$ that is an endofunctor but not an equivalence?

I think such examples can exist when $C$ is of ``infinite type". For example, if $C \cong \oplus_{\mathbb{N}} D$, then there is a non-trivial localization $C \rightarrow C$ obtained by making the first copy of $D$ trivial. I would guess that if $C$ has some compactness condition, then any localization $F: C \rightarrow C$ must be an equivalence.

More precisely, I am interested in the examples when $C$ is a presentable stable $\infty$-category so that $F$ is equivalent to making some objects $A$ of $C$ trivial, i.e. $C \cong C/A$, and $C$ is smooth (which is a compactness condition).

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    $\begingroup$ What's the definition of "smooth" in this context? $\endgroup$
    – Jeremy Rickard
    Commented Feb 22 at 16:54
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    $\begingroup$ I mean smooth in the sense of Kontsevich, that the identity functor is a compact object in Fun(C, C). In the dg setting, smoothness is the condition that the diagonal bimodule is a perfect complex of product bimodules. $\endgroup$
    – user1077
    Commented Feb 23 at 2:38

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