Translation functors are exact and biadjoint

Each of $T_{\chi ,E,{\chi }^{\mathrm{\prime }}}$ and $T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }$ is a composite of exact functors by [F1], hence exact.

For $M\in {\mathcal{O}}_{\chi }$ and $N\in {\mathcal{O}}_{{\chi }^{\mathrm{\prime }}}$ the natural isomorphisms of [F3] and [F2] compose to ${\mathrm{Hom}}_{{\mathcal{O}}_{{\chi }^{\mathrm{\prime }}}}(T_{\chi ,E,{\chi }^{\mathrm{\prime }}}M,N)\cong {\mathrm{Hom}}_{\mathcal{O}}(E\otimes {\mathrm{incl}}_{\chi }M,{\mathrm{incl}}_{{\chi }^{\mathrm{\prime }}}N)\cong {\mathrm{Hom}}_{\mathcal{O}}({\mathrm{incl}}_{\chi }M,E^{\ast }\otimes {\mathrm{incl}}_{{\chi }^{\mathrm{\prime }}}N)\cong {\mathrm{Hom}}_{{\mathcal{O}}_{\chi }}(M,T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }N)$, natural in $M$ and $N$, so $T_{\chi ,E,{\chi }^{\mathrm{\prime }}}$ is left adjoint to $T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }$.

Composing the other pair of isomorphisms gives ${\mathrm{Hom}}_{{\mathcal{O}}_{{\chi }^{\mathrm{\prime }}}}(N,T_{\chi ,E,{\chi }^{\mathrm{\prime }}}M)\cong {\mathrm{Hom}}_{\mathcal{O}}({\mathrm{incl}}_{{\chi }^{\mathrm{\prime }}}N,E\otimes {\mathrm{incl}}_{\chi }M)\cong {\mathrm{Hom}}_{\mathcal{O}}(E^{\ast }\otimes {\mathrm{incl}}_{{\chi }^{\mathrm{\prime }}}N,{\mathrm{incl}}_{\chi }M)\cong {\mathrm{Hom}}_{{\mathcal{O}}_{\chi }}(T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }N,M)$, natural in $M$ and $N$, so $T_{\chi ,E,{\chi }^{\mathrm{\prime }}}$ is also right adjoint to $T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }$; equivalently $T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }$ is both a left and a right adjoint of $T_{\chi ,E,{\chi }^{\mathrm{\prime }}}$.

By [F4] a left adjoint of the exact functor $T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }$ carries projectives to projectives, so $T_{\chi ,E,{\chi }^{\mathrm{\prime }}}$ preserves projectives; symmetrically $T_{{\chi }^{\mathrm{\prime }},E^{\ast },\chi }$ preserves projectives as a left adjoint of the exact $T_{\chi ,E,{\chi }^{\mathrm{\prime }}}$. A right adjoint of an exact functor preserves injectives, so each of the two functors preserves injectives.

Specializing $E=L(\nu )$ and $E^{\ast }=L(\nu {)}^{\ast }$ gives that $T_{\mu }^{\lambda }$ is both a left and a right adjoint of $T_{\lambda }^{\mu }$ and that both are exact, which is the stated consequence.