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% 28/10/2024 09AM

\begin{proof}
  First suppose $(F \adjoint G)$. For each $A \in
  \ob \mathcal{C}$, let $\eta_A : A \to GFA$ be
  the morphism corresponding to $\identity{FA} :
  FA \to FA$. Then $(FA, \eta_A)$ is an \gls{iobj}
  of $(A \darr G)$: given any $f : A \to GB$, the
  diagram
  \begin{picmath}
    \begin{tikzcd}
      A \ar[r, "\eta_A"] \ar[rd, "f", swap]
      & GFA \ar[d, "Gg"] \\
      & GB
    \end{tikzcd}
  \end{picmath}
  commutes if and only if $g$ corresponds to $f$
  under the \gls{adjunc}, by \gls{naty} of the
  \gls{adjunc} bijection.

  So there's a unique morphism $(FA, \eta_A) \to
  (B, f)$ in $(A \darr G)$.

  Conversely, suppose given in \gls{iobj} $(FA,
  \eta_A)$ in $(A \darr G)$ for each $A$. We make
  $F$ into a function $\mathcal{C} \to
  \mathcal{D}$: given $A \stackrel{f}{\to} B$,
  $Ff$ is the unique morphism $(FA, \eta_A) \to
  (FB, \eta_B f)$ in $(A \darr G)$.
  \glsref[funcal]{Functoriality} comes from
  uniqueness: given $B \stackrel{g}{\to} C$,
  $(Fg)(Ff)$ and $F(gf)$ are both morphisms $(FA,
  \eta_A) \to (FC, \eta_C gf)$ in $(A \darr G)$.
  The \gls{adjunc} bijection sends $A
  \stackrel{f}{\to} GB$ to the unique morphism
  $(FA, \eta_A) \to (B, f)$ in $(A \darr G)$, with
  inverse sending $FA \stackrel{g}{\to} B$ to
  $(Gg) \eta_A : A \to GB$.
  This is natural in $A$ since $\eta$ is a
  \gls{natt} $\identity{\mathcal{C}} \to GF$ and
  natural in $B$ since $G$ is \gls{funcal}.
\end{proof}

\begin{fccoro}[]
  \label{coro:3.4}
  % Corollary 3.4
  Assuming:
  - $F$ and $F'$ are both \gls{ladj} to $G :
  \mathcal{D} \to \mathcal{C}$
  Then:
  there is a canonical natural isomorphism $\alpha
  : F \to F'$.
\end{fccoro}

\begin{proof}
  $(FA, \eta_A)$ and $(F'A, \eta_A')$ are both
  \gls{initial} in $(A \darr G)$, so there's a
  unique isomorphism $\alpha_A$ between them.
  $\alpha$ is natural: given $A \stackrel{f}{\to}
  B$, $(F'f) \alpha_A$ and $\alpha_B (Ff)$ are
  both morphisms $(FA, \eta_A) \to (F'B, \eta_B'
  f)$ in $(A \darr G)$, so they're equal.
\end{proof}

As a result of this, we will often talk about
``the'' \gls{ladj} of a \gls{func} (when it
exists), because we don't usually care about which
one in the isomorphism class we use.

\begin{fclemma}[]
  \label{lemma:3.5}
  % Lemma 3.5
  Assuming:
  - $\mathcal{C} \funcslr FG \mathcal{D} \funcslr
  HK \mathcal{E}$
  - $(F \adjoint G)$ and $(H \adjoint K)$
  Then:
  $(HF \adjoint GK)$.
\end{fclemma}

\begin{proof}
  Given $A \in \ob \mathcal{C}$, $C \in \ob
  \mathcal{E}$, we have bijections between
  morphisms $HFA \to C$, morphisms $FA \to KC$,
  and morphisms $A \to GKC$ which are both natural
  in $A$ and $C$, $D$.
\end{proof}

\begin{fccoro}[]
  \label{coro:3.6}
  % Corollary 3.6
  Assuming:
  - the diagram
  \begin{picmath}
    \begin{tikzcd}
      \mathcal{C} \ar[r, "F"] \ar[d, "G"]
      & \mathcal{D} \ar[d, "H"] \\
      \mathcal{E} \ar[r, "K"]
      & \mathcal{F}
    \end{tikzcd}
  \end{picmath}
  is a commutative square of \glspl{cat} and
  \glspl{func}
  - all the \glspl{func} have \glspl{ladj}
  Then:
  the square of
  \glspl{ladj} commutes up to natural isomorphism.
\end{fccoro}

\begin{proof}
  By \cref{lemma:3.5}, both ways round are
  \gls{ladj} to $HF = KG$, so by \cref{coro:3.4}
  they're isomorphic.
\end{proof}

\glsnoundefn{unit}{unit}{units}%
\glsnoundefn{counit}{counit}{counits}%
We saw in \cref{thm:3.3} that an \gls{adj} $(F
\adjoint G)$ gives rise to a natural
transformation $\eta : \identity{\mathcal{C}} \to
GF$, called the \emph{unit} of the \gls{adjunc}.
Dually, we have $\eps : FG \to
\identity{\mathcal{D}}$, the \emph{counit} of $(F
\adjoint G)$.

\begin{fcthm}[]
  \label{thm:3.7}
  % Theorem 3.7
  Assuming:
  - $F : \mathcal{C} \to \mathcal{D}$ and $G :
  \mathcal{D} \to \mathcal{C}$ are \glspl{func}
  Then:
  specifying an \gls{adjunc} $(F \adjoint G)$ is
  equivalent to specifying a \gls{natt} $\eta :
  \identity{\mathcal{C}} \to GF$ and $\eps : FG
  \to \identity{\mathcal{D}}$ satisfying the two
  commutative diagrams:
  \begin{picmath}
    \begin{tikzcd}
      F \ar[r, "F\eta"] \ar[rd, "\identity{F}",
      swap]
      & FGF \ar[d, "\eps_F"] \\
      & F
    \end{tikzcd}
    \qquad \text{and} \qquad
    \begin{tikzcd}
      G \ar[r, "\eta_G"] \ar[rd, "\identity{G}",
      swap]
      & GFG \ar[d, "G\eps"] \\
      & G
    \end{tikzcd}
  \end{picmath}
\end{fcthm}

\begin{proof}
  Suppose $(F \adjoint G)$. We defined $\eta$ in
  the proof of \cref{thm:3.3}, and $\eps$ is
  defined dually. Since $\eps_{FA}$ corresponds to
  $\identity{GFA}$, the composite
  $\eps_{FA}(F\eta_A)$ corresponds to
  $\identity{GFA} \eta_A = \eta_A$. But by
  definition $\identity{FA}$ corresponds to
  $\eta_A$. The other identity is dual.

  Conversely, suppose given $\eta$ and $\eps$
  satisfying the triangular identities. Given $FA
  \stackrel{f}{\to} B$, we define $\Phi(f) =
  (Gf)\eta_A : A \to GFA \to GB$. Dually, given $A
  \stackrel{g}{\to} GB$, we define $\Psi(g) =
  \eps_B(Fg)$. Then $\Psi \Phi(f) =
  \Psi((Gf)\eta_A) = \eps_B (FGf) F\eta_A =
  f(\eps_{FA})(F\eta_A) = f$, and dually
  $\Phi\Psi(g) = g$. \glsref[naty]{Naturality} of $\Phi$ and
  $\Psi$ follows from \gls{naty} of $\eta$ and
  $\eps$.
\end{proof}

In \cref{defn:1.9}, we had \glspl{nati} $\alpha :
\identity{\mathcal{C}} \to GF$ and $\beta : FG \to
\identity{\mathcal{D}}$. These look like the
\gls{unit} and \gls{counit} of an \gls{adjunc} $(F
\adjoint G)$: do they satisfy the triangular
identities? No, but we can always change them:

\begin{fcprop}[]
  % Proposition 3.8
  Assuming:
  - $F : \mathcal{C} \to \mathcal{D}$, $G :
  \mathcal{D} \to \mathcal{C}$, $\alpha :
  \identity{\mathcal{C}} \to GF$ and $\beta : FG
  \to \identity{\mathcal{D}}$ be an \gls{equivc}
  of \glspl{cat} as defined in \cref{defn:1.9}
  Then:
  there exist isomorphisms $\alpha' :
  \identity{\mathcal{C}} \to GF$ and $\beta' : FG
  \to \identity{\mathcal{D}}$ satisfying the
  triangular identities. In particular, $(F
  \adjoint G \adjoint F)$.
\end{fcprop}

\begin{proof}
  We define $\alpha' = \alpha$ and take $\beta'$
  to be the composite
  \[
    FG \stackrel{(FG\beta)^{-1}}{\to} FGFG
    \stackrel{(F\alpha_G)^{-1}}{\to} FG
    \stackrel{\beta}{\to} \identity{\mathcal{D}}
  .\]
  Note that $FG\beta = \beta_{FG}$, since
  \begin{picmath}
    \begin{tikzcd}
      FGFG \ar[r, "FG\beta"] \ar[d, "\beta_{FG}"]
      & FG \ar[d, "\beta"] \\
      FG \ar[r, "\beta"]
      & \identity{\mathcal{D}}
    \end{tikzcd}
  \end{picmath}
  commutes by \gls{naty} of $\beta$, and $\beta$
  is \gls{monic}.
  Similarly, $GF\alpha = \alpha_{GF}$.

  To verify the triangular identities, consider
  \begin{picmath}
    \begin{tikzcd}
      F \ar[r, "F\alpha"]
      \ar[rd, "\identity{F}", swap]
      & FGF \ar[r, "(\beta_{FGF})^{-1}"]
      \ar[d, "F_\alpha^{-1}"]
      & FGFGF \ar[d, "(F\alpha_{GF})^{-1} =
      (FGF\alpha)^{-1}"] \\
      & F \ar[r, "(\beta_F)^{-1}"]
      \ar[rd, "\identity{F}", swap]
      & FGF \ar[d, "\beta_F"] \\
      &
      & F
    \end{tikzcd}
  \end{picmath}
  which commutes by \gls{naty} of $\beta^{-1}$.