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    \item If $A \twofuncs fg B$ is \gls{pprefl},
      then in any commutative square
      \begin{picmath}
        \begin{tikzcd}
          A \ar[r, "f"] \ar[d, "g"]
          & B \ar[d, "h"] \\
          B \ar[r, "k"]
          & C
        \end{tikzcd}
      \end{picmath}
      we have $h = hfr = kgr = k$. So a
      \gls{pushb} for
      \begin{picmath}
        \begin{tikzcd}
          A \ar[r, "f"] \ar[d, "g"]
          & B \\
          B
        \end{tikzcd}
      \end{picmath}
      is a \gls{coequaler} for $A \twofuncs fg B$.
    \item In $\Set$, or more generally in a
      \gls{cc} \gls{cat}, if $A_i
      \twofuncs{f_i}{g_i} B_i \stackrel{h_i}{\to}
      C_i$ ($i = 1, 2$) are \gls{pprefl}
      \glspl{coequaler}, then $A_1 \times A_2
      \twofuncs{f_1 \times f_2}{g_1 \times g_2}
      B_1 \times B_2 \stackrel{h_1 \times
      h_2}{\to} C_1 \times C_2$ is also a
      \gls{coequaler}. To see this, consider
      \begin{picmath}
        \begin{tikzcd}
          A_1 \times A_2
          \ar[r, shift right=3pt]
          \ar[r, shift left=3pt]
          \ar[d, shift right=3pt]
          \ar[d, shift left=3pt]
          & A_1 \times B_2
          \ar[r]
          \ar[d, shift right=3pt]
          \ar[d, shift left=3pt]
          & A_1 \times C_2
          \ar[d, shift right=3pt]
          \ar[d, shift left=3pt] \\
          B_1 \times A_2
          \ar[r, shift right=3pt]
          \ar[r, shift left=3pt]
          \ar[d]
          & B_1 \times B_2
          \ar[r]
          \ar[d]
          & B_1 \times C_2
          \ar[d] \\
          C_1 \times A_2
          \ar[r, shift right=3pt]
          \ar[r, shift left=3pt]
          & C_1 \times B_2
          \ar[r]
          & C_1 \times C_2
        \end{tikzcd}
      \end{picmath}
      in which all rows and columns are
      \glspl{coequaler}. Then the lower right
      square is a \gls{pushb}; but if $B_1 \times
      B_2 \stackrel{k}{\to} D$
      \glsref[coequaler]{coequalises} $A_1 \times
      A_2 \twofuncs{f_1 \times f_2}{g_1 \times
      g_2} B_1 \times B_2$, then is also
      \glsref[coequaler]{coequalises} $A_1 \times
      B_2 \twofuncs{}{} B_1 \times B_2$ and $B_1
      \times A_2 \twofuncs{}{} B_1 \times B_2$, so
      if factors through the top and left edges of
      the lower right square, and hence through
      $B_1 \times B_2 \stackrel{h_1 \times
      h_2}{\to} C_1 \times C_2$.
  \end{enumerate}
\end{remark}

\begin{example}
  \label{eg:5.14}
  % Example 5.14
  \phantom{}
  \begin{enumerate}[(a)]
    \item The forgetful \gls{func} $\Gp \to \Set$
      is \gls{monadic}, and satisfies the
      hypotheses of \cref{thm:5.12}. If $G
      \twofuncs fg H$ is a \gls{pprefl} pair in
      $\Gp$, with \gls{coequaler} $H
      \stackrel{h}{\to} K$ in $\Set$, then $G
      \times G \twofuncs{}{} H \times H \to K
      \times K$ is a \gls{coequaler}, so the
      multiplication $H \times H \to H$ induces a
      binary operation $K \times K \to K$, which
      is the unique group multiplication on $K$
      making $h$ a homomorphism, and it makes $h$
      into a \gls{coequaler} in $\Gp$.

      The same argument works for $\AbGp$, $\Rng$,
      $\Lat$, $\DLat$, \ldots.

      It doesn't work for \glspl{cat} like
      $\CSLat$ or $\CLat$, but here we can use
      \cref{thm:5.11} \emph{provided} the
      forgetful \gls{func} has a \gls{ladj}.
    \item Any \gls{reflon} is \gls{monadic}: this
      can be proved using \cref{thm:5.11}. If
      $\mathcal{D} \subseteq \mathcal{C}$ is a
      \gls{reflsub}, and $A \twofuncs fg B$ is a
      pair in $\mathcal{D}$ for which there exists
      \begin{picmath}
        \begin{tikzcd}
         A
         \ar[r, shift left=3pt, "f"]
         \ar[r, shift right=3pt, "g", swap]
         & B
         \ar[r, "h"]
         \ar[l, "t", bend left=45]
         & C
         \ar[l, "s", bend left=30]
        \end{tikzcd}
      \end{picmath}
      in $\mathcal{C}$ satisfying the equaitions
      of \cref{defn:5.10}(b), then $t \in \mor
      \mathcal{D}$ since $\mathcal{D}$ is
      \gls{full}, so $ft = sh$ is in
      $\mathcal{D}$, but $\mathcal{D}$ is closed
      under splittings of idempotents by
      \cref{eg:4.7}(d), so $h$ belongs to it.
    \item Consider the composite \gls{adjunc}
      \begin{picmath}
        \begin{tikzcd}
          \Set
          \ar[r, shift left=3pt, "F"]
          & \AbGp
          \ar[r, shift left=3pt, "L"]
          \ar[l, shift left=3pt, "G"]
          & \tfAbGp
          \ar[l, shift left=3pt, "I"]
        \end{tikzcd}
      \end{picmath}
      where $(L \adjoint I)$ is the \gls{adjunc}
      of \cref{eg:3.11}(b). The two factors are
      \gls{monadic}, but the composite isn't since
      free abelian groups are torsion free, so
      $GILF \simeq GF$ and its \gls{cat} of
      algebras is $\cong \AbGp$.
    \item The contravariant power-set \gls{func}
      $P^* : \Set^\op \to \Set$ is \gls{monadic},
      and satisfies the hypotheses of
      \cref{thm:5.12}. Its \gls{ladj} is $P^* :
      \Set \to \Set^\op$ by \cref{eg:3.2}(i), and
      it \gls{reflects} isomorphisms by
      \cref{eg:2.9}(a).

      Let $E \stackrel{e}{\to} A \twofuncs fg B$
      be a co\gls{pprefl} \gls{equaler} \gls{diag}
      in $\Set$. Then
      \begin{picmath}
        \begin{tikzcd}
          E \ar[r, "e"] \ar[d, "e"]
          & A \ar[d, "f"] \\
          A \ar[r, "g"]
          & B
        \end{tikzcd}
      \end{picmath}
      is a \gls{pullb} by \cref{remark:5.13}(c),
      so
      \begin{picmath}
        \begin{tikzcd}
          PE \ar[d, "Pe"]
          & PA \ar[l, "P^*e"] \ar[d, "Pf"] \\
          PA
          & PB \ar[l, "P^*g"]
        \end{tikzcd}
      \end{picmath}
      commutes. But we also have $(P^*e)(Pe) =
      \identity{PE}$ and $(P^*f)(Pf) =
      \identity{PB}$ since $e$ and $f$ are
      injective, so
      \begin{picmath}
        \begin{tikzcd}
          PA \ar[r, shift left=3pt, "P^* g"]
          \ar[r, shift right=3pt, swap, "P^* f"]
          & P^* B \ar[r, "P^* e"]
          \ar[l, "Pf", bend left=45]
          & PE \ar[l, "Pe" bend left=30]
        \end{tikzcd}
      \end{picmath}
      is a \gls{splitcoeq} \gls{diag}.
    \item The fogetful \gls{func} $\Top
      \stackrel{U}{\to} \Set$ is not
      \gls{monadic}; the \gls{monad} on $\Set$
      induced by $(D \adjoint U)$ is
      $(\identity{\Set},
      \identity{\identity{\Set}},
      \identity{\identity{\Set}})$ so its
      \gls{cat} of algebras is $\cong \Set$.
    \item The composite \gls{adjunc}
      \begin{picmath}
        \begin{tikzcd}
          \Set
          \ar[r, shift left=3pt, "D"]
          & \Top
          \ar[r, shift left=3pt, "B"]
          \ar[l, shift left=3pt, "U"]
          & \KHaus
          \ar[l, shift left=3pt, "I"]
        \end{tikzcd}
      \end{picmath}
      is \gls{monadic}. We'll prove this using
      \cref{thm:5.11}: suppose given $X \twofuncs
      fg Y$ in $\KHaus$ and a \gls{splitcoeq}
      \begin{picmath}
        \begin{tikzcd}
          UX
          \ar[r, shift left=3pt, "Uf"]
          \ar[r, shift right=3pt, swap, "Ug"]
          & UY
          \ar[r, "h"]
          \ar[l, "t", bend left=45]
          & Z
          \ar[l, "s" bend left=30]
        \end{tikzcd}
      \end{picmath}
      in $\Set$. The quotient topology on $Z$ is
      the unique topology making $h$ into a
      \gls{coequaler} in $\Top$, and it's compact,
      so $h$ will be a \gls{coequaler} in $\KHaus$
      provided $Z$ is Hausdorff. It is also the
      unique topology that could make $h$ into a
      morphism of $\KHaus$.

      But, given an equivalence relation $S$ on a
      compact Hausdorff space $Y$, $Y / S$ is
      Hausdorff if and only if $S$ is closed in $Y
      \times Y$.

      In our case, if $(y_1, y_2) \in S$ (i.e.
      $h(y_1) = h(y_2)$) then $x_1 = t(y_1)$ and
      $x = t(y_2)$ satisfy $g(t_1) = y_1$, $g(x_2)
      = y_2$ and $f(x_1) = f(x_2)$.

      Conversely, if we have $x_1$ and $x_2$ as
      above, then $h(y_1) = h(y_2)$, so $S = g
      \times g(R)$ where $R \subseteq X \times X$
      is $\{(x_1, x_2) \st f(x_1) = f(x_2)\}$. But
      $R$ is closed in $X \times X$ since it's the
      \gls{equaler} of $X \times X
      \twofuncs{f\pi_1}{f\pi_2} Y$. So $R$ is
      compact, so $S$ is compact, so $S$ is closed
      in $Y \times Y$.
  \end{enumerate}
\end{example}

\begin{fcdefn}[Monadic tower]
  \label{defn:5.15}
  % Definition 5.15
  Let $\mathcal{C} \funcslr FG \mathcal{D}$ be an
  \gls{adjunc} where $\mathcal{D}$ has
  \gls{pprefl} \glspl{coequaler}. The
  \emph{monadic tower} of $(F \adjoint G)$ is the
  \gls{diag}
  \begin{picmath}
    \begin{tikzcd}
      & \vdots \\
      & (\mathcal{C}\emat)^{\mathbb{S}}
      \ar[ld, shift left=2pt] \\
      \mathcal{D}
      \ar[rd, "G", shift left=2pt]
      \ar[r, "K", shift left=2pt]
      \ar[ru, shift left=2pt]
      & \mathcal{C}\emat
      \ar[l, "L", shift left=2pt] \\
      & \mathcal{C}
      \ar[ul, "F", shift left=2pt]
    \end{tikzcd}
  \end{picmath}
  where $\Tbb$ is the \gls{monad} induced by $(F
  \adjoint G)$, and $K$ and $L$ are as in
  \cref{thm:5.7} and \cref{lemma:5.9}, and
  $\mathbb{S}$ is the \gls{monad} induced by $(L
  \adjoint K)$ and so on.

  We say $(F \adjoint G)$ has \emph{monadic
  length} $n$ if we reach an equivalence after $n$
  steps.
\end{fcdefn}
