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% 13/11/2024 09AM

\glssymboldefn{Adj}%
Given a \gls{monad} $\Tbb$ on $\mathcal{C}$, we
write $\Adjinternal(\Tbb)$ for the \gls{cat} whose
objects are \glspl{adjunc} $(\mathcal{C} \funcslr
FG \mathcal{D})$ inducing $\Tbb$, and morphisms
$(\mathcal{C} \funcslr FG \mathcal{D}) \to
(\mathcal{C} \funcslr{F'}{G'} \mathcal{D}')$ are
\glspl{func} $\mathcal{D} \stackrel{K}{\to}
\mathcal{D}'$ satisfying $KF = F'$ and $G'K = G$.

\begin{fcthm}[]
  \label{thm:5.7}
  % Theorem 5.7
  The \gls{kls} \gls{adjunc} $(\mathcal{C}
  \funcslr{}{} \mathcal{C}\kct)$ is an
  \gls{iobj} of $\Adj(\Tbb)$, and the \gls{emoo}
  \gls{adjunc} $(\mathcal{C} \funcslr{}{}
  \mathcal{C}\EMaT$ is \gls{terminal}.
\end{fcthm}

\begin{proof}
  Suppose given $(\mathcal{C} \funcslr FG
  \mathcal{D})$ in $\Adj(\Tbb)$. We define $K :
  \mathcal{D} \to \mathcal{C}\EMaT$ by $KB = (GB,
  G\eps_B)$ (an algebra by one of the triangular
  identities for $\eta$ and $\eps$, and \gls{naty}
  of $\eps$), $K(B \stackrel{g}{\to} B') = Gg$ (a
  homomorphism by \gls{naty} of $\eps$). $K$ is
  \gls{funcal} since $G$ is, $G\EMaT K = G$ is
  obvious, and $KFA = (GFA, G\eps_{FA}) = (TA,
  \mu_A) = F\EMaT A$.

  So $K$ is a morphism of $\Adj(\Tbb)$.

  Suppose $K' : \mathcal{D} \to \mathcal{C}\EMaT$
  is another such: then we must have $K'B = (GB,
  \beta_B)$ where $\beta : GFG \to G$ is a
  \gls{natt} since $K'g = Gg$ is a homomorphism
  $K'B \to K'B'$ for all $g : B \to B'$. Also,
  since $K'F = F\EMaT$, we have $\beta_{FA} =
  \mu_A = G \eps_{FA}$ for all $A$.

  For any $B$, we have \glspl{natsq}
  \begin{picmath}
    \begin{tikzcd}
      GFGFGB \ar[r, "GFG\eps_B"]
      \ar[d, shift right=5pt, swap, "G\eps_{FGB}"]
      \ar[d, shift left=5pt, "\beta_{FGB}"]
      & GFGB
      \ar[d, shift right=5pt, swap, "G\eps_B"]
      \ar[d, shift left=5pt, "\beta_B"]
      \\
      GFGB \ar[r, "G\eps_B"]
      & GB
    \end{tikzcd}
  \end{picmath}
  whose left edges are equal, and whose top edge
  is \gls{splitme} \gls{epic}, so we obtain
  $G\eps_B = \beta_B$ for all $B$. So $K' = K$.

  We define $H : \mathcal{C}\kct \to \mathcal{D}$
  by $HA = FA$ and $H(A \kto[f] B) = FA
  \stackrel{Ff}{\to} FGFB
  \stackrel{\eps_{FB}}{\to} FB$. $H$ preserves
  identities and satisfies $HF\kct = F$, by the
  first triangular identity for $\eta$ and $\eps$.

  $H$ preserves the composite $A \kto[f] B \kto[g]
  C$ by
  \begin{picmath}
    \begin{tikzcd}
      FA \ar[r, "Ff"]
      & FGFB \ar[r, "FGFg"] \ar[d, "\eps_{FB}"]
      & FGFGFC \ar[r, "FG\eps_{FC}"] \ar[d,
      "\eps_{FGFC}"]
      & FGFC \ar[d, "\eps_{FC}"] \\
      & FB \ar[r, "Fg"]
      & FGFC \ar[r, "\eps_{FC}"]
      & FC
    \end{tikzcd}
  \end{picmath}
  Also $GHA = GFA = TA = G\kct A$ and
  \[
    GH(A \kto[f] B) = (TA \stackrel{Tf}{\to} TTB
    \stackrel{\mu_B}{\to} TB) = G\kct(A
    \stackrel{f}{\to} B)
  .\]
  So $H$ is a morphism of $\Adj(\Tbb)$. Note that
  $H$ is \gls{full} and \gls{faith}, since it
  sends $A \stackrel{f}{\to} GFB$ to its traspose
  across $(F \adjoint G)$.

  If $H' : \mathcal{C}\kct \to \mathcal{D}$ is any
  morphism of $\Adj(\Tbb)$, we must have $H'A = FA
  = HA$ for all $A$, and since $GH' = G\kct$ and
  the \glspl{adjunc} have the same \gls{unit},
  $H'$ must send the transpose $A \kto[f] B$ of $A
  \kto[f] GFB$ to its transpose across $(F
  \adjoint G)$. So $H' = H$.
\end{proof}

$\mathcal{C}\kct$ has \glspl{coprod} if
$\mathcal{C}$ does, but has few other \glspl{lim}
or \glspl{colim}. In contrast, we have:

\begin{fcprop}[]
  \label{prop:5.8}
  % Proposition 5.8
  Assuming:
  - $\Tbb$ a \gls{monad} on $\mathcal{C}$
  Thens:[(i)]
  - The forgetful \gls{func} $G\EMaT :
  \mathcal{C}\EMaT \to \mathcal{C}$
  \gls{creates} all \glspl{lim} which exist in
  $\mathcal{C}$.
  - If $\mathcal{C}$ has \glspl{colim} of
  shape $J$, then $G\EMaT$ \gls{creates}
  \glspl{colim} of shape $J$ if and only if $T$
  preserves them.
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Suppose given $D : J \to
      \mathcal{C}\EMaT$; write $D(j) = (GD(j),
      \delta_j)$, and let $(L, (\lambda_j : L \to
      GD(j) \mid j \in \ob J))$ be a \gls{lim} for
      $GD$. The composites $TL
      \stackrel{T\lambda_j}{\to} TGD(j)
      \stackrel{\delta_j}{\to} GD(j)$ form a
      \gls{cone} over $GB$. So they induce a
      unique $\lambda : TL \to L$. And $\lambda$
      is a $\Tbb$-algebra structure on $L$, since
      the identities $\lambda \eta_L =
      \identity{L}$ and $\lambda(T\lambda) =
      \lambda\mu_L$ follow from uniqueness of
      factorisations through \glspl{lim} and it's
      the unique lifting of the \gls{lim}
      \gls{cone} in $\mathcal{C}$ to a \gls{cone}
      in $\mathcal{C}\EMaT$. The fact that it's a
      \gls{lim} \gls{cone} is straightforward.
    \item If $G\EMaT$ \gls{creates} \glspl{colim}
      then it \gls{preses} them, but so does
      $F\EMaT$ since it's a \gls{ladj}, so $T$
      preserves them too.

      Conversely, given $D : J \to
      \mathcal{C}\EMaT$ and a \gls{colim}
      \gls{cone} $(GD(j) \stackrel{\lambda_j}{\to}
      L \mid j \in \ob J)$ under $GD$, we need to
      know that $(TGD(j)
      \stackrel{T\lambda_j}{\to} TL \mid j \in \ob
      J)$ is a \gls{colim} \gls{cone} to obtain
      $TL \stackrel{\lambda}{\to} L$ (and that
      $TTL$ is a \gls{colim} to verify that
      $\lambda$ is a $\Tbb$-algebra structure).
      Otherwise, the argument is as before.
      \qedhere
  \end{enumerate}
\end{proof}

Given $(\mathcal{C} \funcslr FG \mathcal{D})$, $(F
\adjoint G)$, how can we tell when $K :
\mathcal{D} \to \mathcal{C}\EMaT$ is part of an
\gls{equivc}?

Note: $H : \mathcal{C}\kct \to \mathcal{D}$ is an
\gls{equivc} if and only if $F$ is \gls{esss}.

\glsadjdefn{monadic}{monadic}{\gls{adjunc}}%
We call $(F \adjoint F)$ (or the \gls{func}
$G$) \emph{monadic} if $K : \mathcal{D} \to
\mathcal{C}\EMaT$ is part of an \gls{equivc}.

% \begin{fclemma}[]
%   \label{lemma:5.9}
%   % Lemma 5.9
%   Assuming:
%   - $(\mathcal{C} \funcslr FG \mathcal{D})$ an
%   \gls{adjunc} inducing $\Tbb$
%   - for every $\Tbb$-algebra $(A, \alpha)$, the
%   pair $FGFA \twofuncs{F\alpha}{\eps_{FA}} FA$ has
%   \gls{coequaler} in $\mathcal{D}$
%   Then:
%   $K$ has a \gls{ladj} $L : C\emat \to
%   \mathcal{D}$.
% \end{fclemma}