Functor with two successive adjoints is a localization

Question

Assume that we have a fully faithful functor $f_{!} \colon \mathcal{C} \to T^{-1}\mathcal{P}(\mathcal{G})$ where $\mathcal{G}$ is some small $\infty$-category and $T$ is a strongly saturated class of small generation. Assume further that this functor has a right adjoint which again has another right adjoint, i.e. there are $f^{\ast}$ and $f_{\ast}$ such that $f_{!} \dashv f^{\ast} \dashv f_{\ast}$. Is then $\mathcal{C}$ a localization of $\mathcal{P}(\mathcal{G})$?

Answer 1

We want to show, that $f^{\ast} \dashv f_{\ast}$ is a localization adjunction, meaning $f_{\ast}$ is fully faithful. We prove this by showing that the counit $f^{\ast} f_{\ast} \to 1$ is an isomorphism. Note that since $f_{!}$ is fully faithful, the unit $1 \to f^{\ast} f_{!}$ is an isomorphism. We obtain that $$\mathrm{Mor}(X, f^{\ast} f_{\ast} Y) \cong \mathrm{Mor}(f^{\ast} f_{!} X, Y) \cong \mathrm{Mor}(X, Y) $$ by the adjunction and the assumption. Hence, the counit is an equivalence and $\mathcal{C}$ is a reflective localization of $T^{-1}\mathcal{P}(\mathcal{G})$. Now a localization of a localization is of course again a localization.