Let $\mathscr{T}$ be a triangulated category, and $\mathscr{A}$ be a right admissible subcategory, which means that $i\_{\mathscr{A}} : \mathscr{A} \rightarrow \mathscr{T}$ has a right adjoint $i\_{\mathscr{A}}^R$. Let $\mathscr{B}$ another subcategory of $\mathscr{T}$ (not necessarily right/left admissible). Consider the subcategory $\mathscr{A} \cap \mathscr{B} \subset \mathscr{B}$. It's natural to ask whether this subcategory is still right admissible. The first thing one would try is to consider the right adjoint $i\_{\mathscr{A}}^R$ and check whether it sends $\mathscr{A} \cap \mathscr{B}$ to $\mathscr{B}$. If this is the case, we get right admissibility. Let's assume however that this is not the case, *i.e.* there exists an object $E \in \mathscr{A} \cap \mathscr{B}$ such that $i\_{\mathscr{A}}^R(E) \notin \mathscr{A} \cap \mathscr{B}$. Is it possible the there exists another functor which is right adjoint to $i\_{\mathscr{A}} \vert\_{\mathscr{A} \cap \mathscr{B}} : \mathscr{A} \cap \mathscr{B} \rightarrow \mathscr{B}$ and which is not the restriction of $i\_{\mathscr{A}}^R$? I tried to apply the Yoneda lemma to prove that this is not possible, but the only case in which this works is the degenerate case in which $\mathscr{A} \cap \mathscr{B}$ weakly generates $\mathscr{B}$, *i.e.* if $\left( \mathscr{A} \cap \mathscr{B} \right)^{\perp} = 0$ in $\mathscr{B}$, which makes me think that maybe it is possible. Thank you in advance.



Admissibility of intersection of subcategories
It is possible. Given $\mathscr{T}$, $\mathscr{A}$ and $i\_\mathscr{A}^R$, it is often possible to find a subcategory $\mathscr{B}$ of $\mathscr{T}$ so that $\mathscr{A}\cap\mathscr{B}=0$ (and so certainly $\mathscr{A}\cap\mathscr{B}$ is a right admissible subcategory of $\mathscr{B}$), but $i\_\mathscr{A}^R(\mathscr{B})\neq0$.


For example, let $\mathscr{T}=K^-(\text{Mod-}\mathbb{Z})$ be the homotopy category of bounded above complexes of abelian groups, and let $\mathscr{A}=K^-(\text{Proj-}\mathbb{Z})$ be the homotopy category of bounded above complexes of free abelian groups, so $i\_\mathscr{A}^R$ takes a complex to its projective resolution. And finally let $\mathscr{B}$ be the homotopy category of bounded above complexes of torsion abelian groups.


Then $\mathscr{A}\cap\mathscr{B}=0$ (there are no nonzero maps from $\mathscr{B}$ to $\mathscr{A}$), but if $B\in\mathscr{B}$ is not acyclic, then $i\_\mathscr{A}^R(B)\neq0$.


