Equivalent data for an adjunction

Proof sketch. First assume (1), so that there is a natural equivalence \(\Hom _D(FX,Y) \simeq \Hom _C(X,GY)\). If we fix \(X \in C\), then by the Yoneda lemma the natural transformation \(\Hom _D(FX,-) \iso \Hom _D(X,G(-))\) corresponds to a map \(\eta _X \colon X \to GFX\): we obtain this map by taking \(Y = FX\) and letting \(\eta _X\) be the map that under this equivalence corresponds to the identity map \(\id _{FX}\colon FX \to FX\). Conversely, the map \(\eta _X\) determines the transformation as the composite \[ \Hom _D(FX,-) \xrightarrow {G} \Hom _C(GFX,G(-)) \xrightarrow {- \circ \eta _X} \Hom _C(X,-). \] The naturality of the equivalence in \(X\) corresponds to the naturality of \(\eta \colon \id _C \to GF\), showing the equivalence between (1) and (2).

The equivalence between (1) and (3) is proved similarly, where now the counit map \(\epsilon _Y\colon FGY \to Y\) corresponds to the identity map \(\id _{GY}\colon GY \to GY\).

To show that these conditions are also equivalent to (4), assume first that (1)-(3) are satisfied. For the first triangle identity, observe that by definition the map \(\eta _X\colon X \to GFX\) is a preimage of \(\id _{FX}\colon FX \to FX\) under the equivalence \[ \Hom _C(X,GFX) \xrightarrow {F} \Hom _D(FX,FGFX) \xrightarrow {\epsilon _{FX} \circ -} \Hom _D(FX,FX). \] This precisely says that the composite \(\epsilon _{FX} \circ F\eta _X\) is equivalent to \(\id _{FX}\), naturally in \(X\), which is the first triangle identity. The second triangle identity similarly follows from the fact that the morphism \(\epsilon _Y\colon FGY \to Y\) is a preimage of \(\id _{GY}\colon GY \to GY\) under the equivalence \[ \Hom _D(FGY,Y) \xrightarrow {G} \Hom _C(GFGY,GY) \xrightarrow {- \circ \eta _{GY}} \Hom _C(GY,GY). \] Finally, assume that (4) is satisfied. An easy check shows that in this case the map defined in (2) is inverse to the map defined in (3), producing the desired natural equivalence \(\Hom _D(FX,Y) \simeq \Hom _C(X,GY)\). □