Category theory - Prove that $\operatorname{Hom}$ preserves representations for quasi-inverse functors

Question

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 $F(X)$.

$\DeclareMathOperator\Hom{Hom}$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...

Answer 1

HINT: The map $\phi$ is defined, if $f:G(Y)\to X$, by $\phi(f)=F(f)\circ \varepsilon^{-1}_Y$. Here, $\varepsilon^{-1}_Y:Y\to FG(Y)$ 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 $g:Y\to F(X)$ to $\eta_X^{-1}\circ G(g)$, where $G(g):G(Y)\to GF(X)$ and $\eta_X:X\to GF(X)$ 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.

Answer 2

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

Proof. Let $\bar\eta:\id_{\cat D}\to G\circ F$ be a natural isomorphism.

The functor $F$ is faithful, for if $u,v:A\rightrightarrows B$ and $F(u)=F(v)$ then \begin{align} u\bar\eta_B &=\bar\eta_A(G\circ F)(u)\\ &=\bar\eta_A(G\circ F)(v)\\ &=v\bar\eta_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 $x=\bar\eta_AG(y)\bar\eta_B^{-1}$, then \begin{align} \bar\eta_A(G\circ F)(x) &=x\bar\eta_B\\ &=\bar\eta_AG(y) \end{align} which implies $y=F(x)$ (since $G$ is faithful).

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

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

$\DeclareMathOperator\Hom{Hom} $For all objects $A$ of $\cat C$ we define \begin{align} &\varphi_A:\Hom_{\cat C}(G(A),X)\to\Hom_{\cat D}(A,F(X))& &f\mapsto\varepsilon_A^{-1}F(f) \end{align} and \begin{align} &\psi_A:\Hom_{\cat D}(A,F(X))\to\Hom_{\cat C}(G(A),X)& &g\mapsto G(g)\eta_X^{-1} \end{align}

We have to show that $\varphi_A$ is natural in $A$ and $\varphi_A\circ\psi_A$ and $\psi_A\circ\varphi_A$ are identity functions.

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

For all $g:A\to F(X)$ we have \begin{align} (\varphi_A\circ\psi_A)(g) &=\varepsilon_A^{-1}F(G(g)\eta_X^{-1})\\ &=\varepsilon_A^{-1}(F\circ G)(g)F(\eta_X)^{-1}\\ &=\varepsilon_A^{-1}(F\circ G)(g)\varepsilon_{F(X)}\\ &=\varepsilon_A^{-1}\varepsilon_Ag\\ &=g \end{align}

Let $u:B\to A$ 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}(F\circ G)(u)F(f)\\ &=u\varepsilon_A^{-1}F(f)\\ &=u\varphi_A(f) \end{align}