${\underline{C}}{{^{op}}}$ satisfy the Brown representability property.

2\. Assume that ${\underline{C}}$ satisfies the Brown representability property. Then it has products and any exact functor $F$ from ${\underline{C}}$ (into a triangulated category ${\underline{D}}$) that respects coproducts possesses an exact right adjoint $G$.

I.1. The first part of the assertion is essentially a part of [@neebook Proposition 8.4.1] (note that the simple argument used for the proof of this implication does not require ${\mathcal{P}}$ to be a set); it is also a particular case of assertion I.2.

Now assume that the embedding ${\underline{C}}'\to {\underline{C}}$ possesses a right adjoint. Then Proposition \[pbouloc\](III.\[ibou1\]) implies that ${\underline{C}}'={\underline{C}}$ whenever ${\operatorname{Obj}}{\underline{C}}'{{}^{\perp}}={\{0\}}$ (note here that ${\underline{C}}'$ is a strict subcategory of ${\underline{C}}$). Lastly, ${\operatorname{Obj}}{\underline{C}}'{{}^{\perp}}$ is certainly zero if ${\mathcal{P}}$ Hom-generates ${\underline{C}}$ (since ${\mathcal{P}}\subset {\operatorname{Obj}}{\underline{C}}'$).

2\. Similarly to the proof of loc. cit., if $N$ belongs to ${\mathcal{P}}^{\perp}\cap C$ then the class $C\cap {}^{\perp}N$ contains ${\mathcal{P}}$, coproductive and extension-closed. Hence $N\perp N$ and we obtain $N=0$.

II.1. The Brown representability property for ${\underline{C}}$ is given by Proposition 8.4.2 of [@neebook]. The Brown representability for ${\underline{C}}{{^{op}}}$ is immediate from the combination of Theorem 8.6.1 with Remark 6.4.5 of ibid.

2\. The first part of the assertion is given by Proposition 8.4.6 