Let $C$,$D$ be two category, and $F:C \rightarrow D$, $G:D \rightarrow C$ are two functors such that $FG \simeq Id_{D}$ and $GF \simeq Id_{C}$, Show that $(F,G)$ is an adjoint pair.
To show this, we need to construct an isomorphism $Hom(F(A),B)\simeq Hom(A,G(B))$ for any $(A,B)\in Ob(C)\times Ob(D)$
I start from $f\in Hom(F(A),B)$
then got $Gf: GF(A) \rightarrow G(B)$
becasue we have $GF \simeq Id_{C}$, we have an isomorphism $\varphi(A): A \xrightarrow{\sim} GF(A)$
So we have $\varphi(A)\circ Gf: A \rightarrow G(B)$.
Similarly we can get a map from $Hom(A,G(B))$ to $Hom(F(A),B)$
But it's difficult to prove that the composition of these two maps is identity.
The difficulty is to show $F\varphi(A)$=$\psi(F(A))$
Best Answer
This is a well-known result in category theory: that any equivalence can be improved to an adjoint equivalence. Given an equivalence in the form of isomorphisms $\eta: 1_C \stackrel{\sim}\to GF$ and $\xi: FG \stackrel{\sim}\to 1_D$, what we would like is for $\eta$ to be the unit and $\xi$ the counit of an adjunction, and for this we need the famous triangular or zig-zag equations for the pair $(\eta, \xi)$:
$$1_F = \left(F \stackrel{F \eta}{\to} FGF \stackrel{\xi F}{\to} F\right)$$
$$1_G = \left(G \stackrel{\eta G}{\to} GFG \stackrel{G \xi}{\to} G\right)$$
It turns out that if one of the zigzag equations is satisfied, then so is the other. Of course neither may be satisfied for the given pair $(\eta, \xi)$, but, if one replaces $\xi$ with another isomorphism $\epsilon: FG \to 1_C$ defined by the composite
$$FG \stackrel{FG\xi^{-1}}{\to} FGFG \stackrel{F\eta^{-1}G}{\to} FG \stackrel{\xi}{\to} 1_C$$
then one can show one and therefore both of the zig-zag equations are satisfied for the pair $(\eta, \epsilon)$. Full details, using string diagrams, may be found in the nLab here. See also Categories for the Working Mathematician, IV.4.