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Given an adjunction $F \dashv G$ with for all $B$, (5) $GF \eta_{GB} = \eta_{GFGB}$ I proved (directly from the triangle laws) that $G\varepsilon_{FGB} \eta_{GFGB} = 1$ and $\eta_{GFGB} G\varepsilon_{FGB} = 1$ so $G\varepsilon_{FGB}$ is an isomorphism, but I want to show (6) $G\varepsilon_{B}$ is an isomorphism.

How can I finish the proof? Is there a way to deduce that $FG$ is surjective on objects or should I take a different approach?


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up vote 3 down vote accepted

You know that $G\varepsilon \circ \eta_G = 1$ by one of the triangular identities.

To show that $\eta_G \circ G\varepsilon = 1$ you need to look at the naturality square for $\eta : 1 \to GF$, where the arrow we consider is $G\varepsilon_B : GFGB \to GB$ for an arbitrary $\mathcal{D}$-object $B$. Drawing this gives

$$GFG\varepsilon_{B} \circ \eta_{GFGB} = \eta_{GB} \circ G\varepsilon_{B}$$

By your assumption that $\eta_{GFG}=GF\eta_G$, this is the same as

$$GF(G\varepsilon_B \circ \eta_{GB}) = \eta_{GB} \circ G\varepsilon_B$$

Now apply the triangular identity to the left-hand side and use functoriality.

Finally apply the result that a natural transformation is a natural isomorphism if and only if all its components are natural isomorphisms.

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