Adjunctions compose with the composite unit and counit formulas

Whiskering gives $G{\eta }^{\mathrm{\prime }}F:GF\Rightarrow G{G}^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F$ and $F^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }}:F^{\mathrm{\prime }}FG{G}^{\mathrm{\prime }}\Rightarrow F^{\mathrm{\prime }}{G}^{\mathrm{\prime }}$, so the displayed composites have the required types; they are natural because whiskering and vertical composition preserve naturality.

Expanding the first triangle composite for $F^{\mathrm{\prime }}F$ gives $(\bar{\epsilon }{F}^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}F\bar{\eta })=({\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}FG{\eta }^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}F\eta )$. Interchange rewrites the two middle factors as $F^{\mathrm{\prime }}[(\epsilon G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F)\circ (FG{\eta }^{\mathrm{\prime }}F)]$, and naturality of $\epsilon$ at the component ${\eta }_{Fc}^{\mathrm{\prime }}$ turns that bracket into $({\eta }^{\mathrm{\prime }}F)\circ (\epsilon F)$.

Expanding the second triangle composite for $G{G}^{\mathrm{\prime }}$ gives $(G{G}^{\mathrm{\prime }}\bar{\epsilon })\circ (\bar{\eta }G{G}^{\mathrm{\prime }})=(G{G}^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }})\circ (G{G}^{\mathrm{\prime }}{F}^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }})\circ (G{\eta }^{\mathrm{\prime }}FG{G}^{\mathrm{\prime }})\circ (\eta G{G}^{\mathrm{\prime }})$. Interchange rewrites the two middle factors as $G[(G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }})\circ ({\eta }^{\mathrm{\prime }}FG{G}^{\mathrm{\prime }})]$, and naturality of ${\eta }^{\mathrm{\prime }}$ at the component ${\epsilon }_{G^{\mathrm{\prime }}d}$ turns that bracket into $({\eta }^{\mathrm{\prime }}{G}^{\mathrm{\prime }})\circ (\epsilon G^{\mathrm{\prime }})$. The composite becomes $G[(G^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }})\circ ({\eta }^{\mathrm{\prime }}{G}^{\mathrm{\prime }})]\circ [(G\epsilon )\circ (\eta G)]G^{\mathrm{\prime }}$, which by the two second triangle identities $G\epsilon \circ \eta G=1_G$ and $G^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }}\circ {\eta }^{\mathrm{\prime }}{G}^{\mathrm{\prime }}=1_{G^{\mathrm{\prime }}}$ is $1_{G{G}^{\mathrm{\prime }}}$.

By step 2.1 the first composite becomes $[({\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }})\circ (F^{\mathrm{\prime }}{\eta }^{\mathrm{\prime }})]F\circ F^{\mathrm{\prime }}[(\epsilon F)\circ (F\eta )]$, which by the two first triangle identities $\epsilon F\circ F\eta =1_F$ and ${\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }}\circ F^{\mathrm{\prime }}{\eta }^{\mathrm{\prime }}=1_{F^{\mathrm{\prime }}}$ is $1_{F^{\mathrm{\prime }}F}$.

Thus $\bar{\eta }$ and $\bar{\epsilon }$ satisfy both triangle identities, and [L1] gives $F^{\mathrm{\prime }}F⊣G{G}^{\mathrm{\prime }}$.