Definitions of adjoint functors

Question

Definition. Let $F:\mathcal A\rightleftarrows\mathcal B:G$ be a pair of functors. We say that $F$ is left adjoint to $G$ and write $F\dashv G$ if there are exists natural transformations $\varepsilon:1_\mathcal A\to GF$ and $\eta:FG\to1_\mathcal B$ s.t. $$F\overset{F\varepsilon}{\to}FGF\overset{\eta F}{\to}F$$ and $$G\overset{\varepsilon G}{\to}GFG\overset{G\eta}{\to}G$$ are identical natural transforms (we write $(\eta F)_A$ for $\eta_{FA}$).

I am trying to prove the following

Theorem. Let $F:\mathcal A\rightleftarrows\mathcal B:G$ be a pair of functors. The next three conditions are equivalent.

1. $F\dashv G;$
2. There is a bifunctor isomorphism $\alpha:\mathcal B(F-,*)\overset\sim\longrightarrow\mathcal A(-,G*);$

$1\Rightarrow2.$ We define $\alpha_{X,Y}:\mathcal B(FX,Y)\overset\sim\longrightarrow\mathcal A(X,GY)$ and $\beta_{X,Y}:\mathcal A(X,GY)\overset\sim\longrightarrow\mathcal B(FX,Y)$ as $$\alpha_{X,Y}(h)=G(h)\circ\varepsilon_X,\qquad\beta_{X,Y}(f)=\eta_Y\circ F(f).$$ These are natural transforms. We shall show they are mutually inverse.

Let $h:FX\to Y.$ Then $\beta_{X,Y}\alpha_{X,Y}(h)=\beta_{X,Y}G(h)\varepsilon_X=\eta_YFG(h)\varepsilon_X.$ It's sufficient to show that $G(h)\varepsilon_X=F\varepsilon_X(h)$ but I don't understand how.

$2\Rightarrow1.$ We set $\varepsilon_X=\alpha_{X,FX}(1_{FX})$ and $\eta_Y=\beta_{GY,Y}(1_{GY}).$ Then we need to check these are mutually inverse natural transforms and it's not so clear to me.

Could you please help me?

Answer 1

$\newcommand{\A}{\mathcal{A}}\newcommand{\B}{\mathcal{B}}\newcommand{\op}{^{\mathsf{op}}}\newcommand{\set}{\mathsf{Set}}\newcommand{\hom}{\mathsf{Hom}}$If $\A,\B$ are locally small categories, what is true is that $F\dashv G$ if and only if the following two functors are naturally isomorphic: $$\B(F(-),-):\A\op\times\B\overset{F\op\times1}{\longrightarrow}\B\op\times\B\overset{\hom_\B}{\longrightarrow}\set\\\A(-,G(-)):\A\op\times\B\overset{1\times G}{\longrightarrow}\A\op\times\A\overset{\hom_\A}{\longrightarrow}\set$$

I apologise in advance: you are using $\epsilon,\eta$ entirely contrary to how I have been using them in adjunction notation for the last few months. I will use my convention.

Say $\alpha:\B(F(-),-)\implies\A(-,G(-))$ is a natural transformation with inverse $\beta$. For an object $X\in\B$, set $\epsilon_X=\beta_{GX,X}(1_{GX})$ and for an object $Y\in\A$ set $\eta_Y=\alpha_{Y,FY}(1_{FY})$.

Your task is to show: $$G(\epsilon_X)\circ\eta_{GX}:GX\to GFGX\to GX$$Is equal to the identity for all $X\in\B$, and to show: $$\epsilon_{FY}\circ F(\eta_Y):FY\to FGFY\to FY$$Is equal to the identity for all $Y\in\A$. Moreover you must show both $\epsilon,\eta$ assemble to natural maps $FG\to 1_\B,1_\A\to GF$ respectively.

To show the first, consider that: $$G(\epsilon_X)\eta_{GX}=G(\beta_{GX,X}(1_{GX}))\circ\alpha_{GX,FGX}(1_{FGX})$$The key observation is this: carefully unwinding definitions, the naturality of $\alpha$ means that: $$\alpha_{S,T'}(g\circ\gamma\circ F(f))=G(g)\circ\alpha_{S',T}(\gamma)\circ f:S\to GT'$$For all $S,S'\in\A$, $T,T'\in\B$ and for all $f\in\A(S,S')$, $g\in\B(T,T')$ and $\gamma\in\B(F(S'),T)$.

Fixing an $X\in\B$, let $S=S'=GX$ and $f=1_{GX}$. Let $T'=X$ and $T=FGX$. Let $g=\beta_{GX,X}(1_{GX})$ and let $\gamma:=1_{FGX}$.

Using the naturality formula, we find: $$G(\beta_{GX,X}(1_{GX}))\circ\alpha_{GX,FGX}(1_{FGX})\circ 1_{GX}=\alpha_{GX,X}(\beta_{GX,X}(1_{GX})\circ 1_{FGX}\circ F(1_{GX}))$$

The left hand side is just $G(\epsilon_X)\circ\eta_{GX}$ and the right hand side resolves to $1_{GX}$ ($\alpha,\beta$ are inverse). Hence, we have shown the first triangle identity, as these tend to be called.

I leave it to you to do the rest, feel free to ask for clarifications. Reminder: you must check $\epsilon$ and $\eta$ are natural, check the second triangle identity, and my $\epsilon,\eta$ are different to your $\epsilon,\eta$ (sorry about that).