Krdytkyu

Let $F: mathcal C to mathcal D$ and $G: mathcal D to mathcal C$ be quasi-inverse functors, and let $H : mathcal C to Set$ be a representable (contravariant) functor with representative $X in mathcal C$. Prove that $H circ G$ is representable by $F(X)$.

$DeclareMathOperatorHom{Hom}$As ismorphisms are transitive, it suffices to consider the case when $H = Hom( -, X)$. To this end, we wish to find $phi : Hom(-,X) circ G to Hom(-,F(X))$ an ismorphism, from which we quickly deduce that for any $f: B to A$ and $g: GA to X$ it must be that $phi_A(g) circ f = phi_B(g circ Gf)$. I am not sure how to dind such a $phi$ though. It seems like I have to somehow use the fact that $F$ and $G$ are quasi-inverses...

HINT: The map $phi$ is defined, if $f:G(Y)to X$, by $phi(f)=F(f)circ varepsilon^{-1}_Y$. Here, $varepsilon^{-1}_Y:Yto FG(Y)$ is the inverse of the "counit of the adjunction", that is, one of the natural isomorphisms that proves that $F$ and $G$ are quasi-inverses. In the other direction, $phi^{-1}$ sends $g:Yto F(X)$ to $eta_X^{-1}circ G(g)$, where $G(g):G(Y)to GF(X)$ and $eta_X:Xto GF(X)$ is the other natural isomorphism coming with the equivalence. Now you just have to check that $phi$ and $phi^{-1}$ are mutually inverse and (that one, thus the other is) natural.

$newcommandcatmathscrDeclareMathOperatorid{id}$Let $F:cat Crightleftarrowscat D:G$ be quasi-inverse functors.
Then $F,G$ are fully faithful and there exists natural isomorphisms $varepsilon:Fcirc Gtoid_{cat C}$ and $eta:id_{cat D}to Gcirc F$ such that
begin{align}
&eta_GG(varepsilon)=1_G&
&F(eta)varepsilon_F=1_F
end{align}

Proof.
Let $bareta:id_{cat D}to Gcirc F$ be a natural isomorphism.

The functor $F$ is faithful, for if $u,v:Arightrightarrows B$ and $F(u)=F(v)$ then
begin{align}
ubareta_B
&=bareta_A(Gcirc F)(u)\
&=bareta_A(Gcirc F)(v)\
&=vbareta_B
end{align}
which implies $u=v$. Similarly, $G$ is faithful.

The functor $F$ is full, for if $y:F(A)to F(B)$ and $x=bareta_AG(y)bareta_B^{-1}$, then
begin{align}
bareta_A(Gcirc F)(x)
&=xbareta_B\
&=bareta_AG(y)
end{align}
which implies $y=F(x)$ (since $G$ is faithful).

Let $varepsilon:Fcirc Gtoid_{cat C}$ be a natural isomorphism.
Since $F$ is full and faithful, for each object $A$ in $cat C$ there exists one and only one isomorphism $eta_A:Ato (Gcirc F)(A)$ such that $F(eta_A)=varepsilon_{F(A)}^{-1}$.
Then $eta:id_{cat C}to Gcirc F$ is a natural isomorphism (again using faithfulness of $F $) and $F(eta)varepsilon_F=1_F$.

By naturalness of $varepsilon$, we have $varepsilon_{Fcirc G}varepsilon=(Fcirc G)(varepsilon)varepsilon$ from which we get $varepsilon_{Fcirc G}=(Fcirc G)(varepsilon)$.
Consequently,
begin{align}
F(eta_GG(varepsilon))
&=F(eta_G)(Fcirc G)(varepsilon)\
&=F(eta_G)varepsilon_{Fcirc G}\
&=1_{Fcirc G}\
&=F(1_G)
end{align}
from which $eta_GG(varepsilon)=1_G$. $square$

$DeclareMathOperatorHom{Hom} $For all objects $A$ of $cat C$ we define
begin{align}
&varphi_A:Hom_{cat C}(G(A),X)toHom_{cat D}(A,F(X))&
&fmapstovarepsilon_A^{-1}F(f)
end{align}
and
begin{align}
&psi_A:Hom_{cat D}(A,F(X))toHom_{cat C}(G(A),X)&
&gmapsto G(g)eta_X^{-1}
end{align}

We have to show that $varphi_A$ is natural in $A$ and $varphi_Acircpsi_A$ and $psi_Acircvarphi_A$ are identity functions.

For all $f:G(A)to X$ we have
begin{align}
(psi_Acircvarphi_A)(f)
&=G(varepsilon_A^{-1}F(f))eta_X^{-1}\
&=G(varepsilon_A)^{-1}(Gcirc F)(f)eta_X^{-1}\
&=eta_{G(A)}(Gcirc F)(f)eta_X^{-1}\
&=feta_Xeta_X^{-1}\
&=f
end{align}

For all $g:Ato F(X)$ we have
begin{align}
(varphi_Acircpsi_A)(g)
&=varepsilon_A^{-1}F(G(g)eta_X^{-1})\
&=varepsilon_A^{-1}(Fcirc G)(g)F(eta_X)^{-1}\
&=varepsilon_A^{-1}(Fcirc G)(g)varepsilon_{F(X)}\
&=varepsilon_A^{-1}varepsilon_Ag\
&=g
end{align}

Let $u:Bto A$ be a morphism in $cat C$.
Then naturalness of $varphi_A$ means
$$require{AMScd}
begin{CD}
Hom_{cat C}(G(A),X) @>varphi_A>> Hom_{cat D}(A,F(X))\
@VVV @VVV\
Hom_{cat C}(G(B),X) @>>varphi_B> Hom_{cat D}(B,F(X))
end{CD}$$
For all $f:G(A)to X$ we have
begin{align}
varphi_B(G(u)f)
&=varepsilon_B^{-1}F(G(u)f)\
&=varepsilon_B^{-1}(Fcirc G)(u)F(f)\
&=uvarepsilon_A^{-1}F(f)\
&=uvarphi_A(f)
end{align}