Lemma 21.39.5. Notation and assumptions as in Example 21.39.1. If $\mathcal{C}$ has either an initial or a final object, then $L\pi _! \circ \pi ^{-1} = \text{id}$ on $D(\textit{Ab})$, resp. $D(B)$.

Proof. If $\mathcal{C}$ has an initial object, then $\pi _!$ is computed by evaluating on this object and the statement is clear. If $\mathcal{C}$ has a final object, then $R\pi _*$ is computed by evaluating on this object, hence $R\pi _* \circ \pi ^{-1} \cong \text{id}$ on $D(\textit{Ab})$, resp. $D(B)$. This implies that $\pi ^{-1} : D(\textit{Ab}) \to D(\mathcal{C})$, resp. $\pi ^{-1} : D(B) \to D(\underline{B})$ is fully faithful, see Categories, Lemma 4.24.4. Then the same lemma implies that $L\pi _! \circ \pi ^{-1} = \text{id}$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).