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Note that the characterisation of
$\mathcal{T}$-\glspl{model} in any \gls{cat} with
finite products.
Note also that the equations of \cref{lemma:6.10}
allow us to reduce any compound operation
$\alpha(\beta_1(x\cdots), \beta_2(x\cdots),
\ldots, \beta_n(x\cdots))$ to a single operation
$\gamma$.

\begin{fcthm}[]
  \label{thm:6.11}
  % Theorem 6.11
  Assuming:
  - $\mathcal{C}$ a \gls{cat}
  Then:
  the following are equivalent:
  \begin{cenum}[(i)]
    \item $\mathcal{C}$ is \gls{equivt} to a
      finitary algebraic \gls{cat} in the sense of
      \cref{eg:5.14}(a). (Recall that these
      \glspl{cat} are those whose objects are sets
      with finitary operations satisfying some
      equations, and morphisms are homomorphisms
      between them.)
    \item $\mathcal{C}$ is \gls{equivt} to the
      \gls{cat} of $\Set$-\glspl{model} of a
      \gls{Lt}.
    \item $\mathcal{C} \simeq \Set\emat$ for a
      finitary \gls{monad} $\Tbb$ on $\Set$.
  \end{cenum}
\end{fcthm}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i) $\Rightarrow$ (ii)]
    \item[(ii) $\Rightarrow$ (i)] Let
      $\mathcal{T}$ be the \gls{full} sub\gls{cat} of
      $\mathcal{C}$ on the free algebras $F[n]$,
      for $n \in \Nbb$. Then $\mathcal{T}$ is a
      \gls{Lt}, and for every object $A$ of
      $\mathcal{C}$, the \gls{func}
      $\mathcal{C}(\blank, A)$ restricted to
      $\mathcal{T}$ \gls{preses} finite
      \glspl{prod}, so it's a \gls{model} of
      $\mathcal{T}$. This defines a \gls{func}
      $\mathcal{T}-\Mod(\Set)
      \stackrel{Y}{\leftarrow} \Set\emat$; but
      $\mathcal{T}-\Mod(\Set) \simeq \Set^{\Tbb'}$
      for some finitary \gls{monad} $\Tbb'$ on
      $\Set$, so we get a \gls{func} $\Set\emat
      \stackrel{Y}{\to} \Set^{\Tbb'}$ which is the
      identity on underlying sets.

      In this
      situation, $Y$ is induced by a
      \emph{morphism of monads} $\Tbb' \to \Tbb$,
      i.e. a \gls{natt} $\theta : T' \to T$
      commuting with the \glspl{unit} and
      multiplications. (Clearly, such a $\theta$
      induces a \gls{func} $\Set\emat \to
      \Set^{\Tbb'}$ sending $(A, \alpha)$ to $(A,
      \alpha\theta_A)$).
      
      But we know $\theta_{[n]}$ is bijective for
      all $n$, since elements of the free algebras
      on $[n]$ are just morphisms $[1] \to [n]$ in
      $\mathcal{T}$. But both \glspl{func} are
      finitary, so $\theta_A$ is bijective for all
      $A$, i.e. it's an isomorphism of
      \glspl{monad}. \qedhere
  \end{enumerate}
\end{proof}

For a general \gls{monad} $\Tbb$ on $\Set$, this
construction produces a finitary \gls{monad}
$\Tbb'$ which is the co\gls{reflon} of $\Tbb$ in
the \gls{cat} of finitary \glspl{monad}.

For example:
\begin{itemize}
  \item For $\Tbb = (\text{double
    power-set})$, we obtain $\Tbb' = \{\text{Boolean
    algebras}\}$.
  \item For $\Tbb = \text{Stone-\v{C}ech}$, we
    obtain the trivial \gls{monad}
    $(\identity{\Set}, \identity{\identity{\Set}},
    \identity{\identity{\Set}})$.
\end{itemize}

\newpage
\section{Regular Categories}

\begin{fcdefn}[Image, cover]
  \glsnoundefn{image}{image}{images}%
  \glsnoundefn{cover}{cover}{covers}%
  \glssymboldefn{cover}%
  \label{defn:7.1}
  % Definition 7.1
  We say a \gls{cat} $\mathcal{C}$ \emph{has
  images} if, for every $A \stackrel{f}{\to} B$ in
  $\mathcal{C}$, there exists a least $m : B'
  \monic B$ in $\Sub(B)$ through which $f$
  factors. We call $m$ the \emph{image} of $f$,
  and we say $f$ is a \emph{cover} if its image is
  $\identity{B}$.

  We write $A \stackrel{f}{\rightarrowtriangle} B$
  to indicate that $f$ is a cover.
\end{fcdefn}

\begin{fclemma}[]
  \label{lemma:7.2}
  % Lemma 7.2
  Any strong \gls{epi} is a \gls{cover}. The
  converse holds if $\mathcal{C}$ has
  \glspl{equaler} and \glspl{pullb}.
\end{fclemma}

\begin{proof}
  If $f$ is strong \gls{epic}, applying the
  definition to commutative squares of the form
  \begin{picmath}
    \begin{tikzcd}
      A \ar[r, "g"] \ar[d, "f"]
      & B' \ar[d, "m", tail] \\
      B \ar[r, "\identity{B}"]
      \ar[ru, dashed]
      & B
    \end{tikzcd}
  \end{picmath}
  shows that $f$ is a \gls{cover}.

  For the converse, a \gls{cover} $A
  \stackrel{f}{\to} B$ is \gls{epic} since it
  can't factor through the \gls{equaler} of any $B
  \twofuncs gh C$ with $g \neq h$. To verify the
  other condition, suppose given
  \begin{picmath}
    \begin{tikzcd}[ampersand replacement=\&]
      A \ar[r, "g"] \ar[d, "f", -{Triangle[open]}]
      \& C \ar[d, "m", tail] \\
      B \ar[r, "h"]
      \ar[ru, dashed]
      \& D
    \end{tikzcd}
  \end{picmath}
  then the \gls{pullb} of $m$ along $h$ is
  \gls{monic} by \cref{lemma:4.15}, and $f$
  factors through it, so it's an isomorphism. So
  we get $B \to C$ by composing with the top edge
  of the \gls{pullb} square.
\end{proof}

Here, if $\mathcal{C}$ has \glspl{image}, image
facorisation defines a \gls{func} $\funccat[2,
\mathcal{C}] \to \funccat[3, \mathcal{C}]$: given
\begin{picmath}
  \begin{tikzcd}
    A \ar[r, "f"] \ar[d, "g"]
    & B \ar[d, "h"] \\
    C \ar[r, "k"]
    & D
  \end{tikzcd}
\end{picmath}
if we form the image factorisations
\begin{picmath}
  \begin{tikzcd}
    A \ar[d] \ar[r,-{Triangle[open]}]
    & I \ar[d, dashed] \ar[r, tail]
    & B \ar[d] \\
    C \ar[r, -{Triangle[open]}]
    & J \ar[r, tail]
    & D
  \end{tikzcd}
\end{picmath}
we get a unique $I \to J$ making both squares
commute.

\begin{fcdefn}[Regular category]
  \glsadjdefn{regc}{regular}{\gls{cat}}%
  \label{defn:7.3}
  % Definition 7.3
  We say $\mathcal{C}$ is \emph{regular} if it has
  finite \glspl{lim} and \glspl{image}, and image
  factorisations are stable under \gls{pullb},
  i.e. if the left hand square above is a
  \gls{pullb} then so are both right hand squares.
  (This is equivalent to saying that \glspl{cover}
  are stable under \gls{pullb}).
\end{fcdefn}

\begin{example}
  \label{eg:7.4}
  % Example 7.4
  \phantom{}
  \begin{enumerate}[(a)]
    \item $\Set$ is \gls{regc} and co\gls{regc}: all
      \glspl{mono} and \glspl{epi} are strong, and
      so the two factorisations coincide and
      \glspl{epi} (respectively \glspl{mono}) are
      stable under \gls{pullb} (respectively
      \gls{pushb}).
    \item If $\mathcal{C}$ is \gls{regc}, so is any
      $\funccat[\mathcal{D}, \mathcal{C}]$ with
      \glspl{image} constructed pointwise (they're
      stable under \gls{pushb} since \glspl{pullb}
      are also constructed pointwise).
    \item If $\mathcal{C}$ is \gls{regc}, then so
      $\mathcal{C}\emat$ for any \gls{monad}
      $\Tbb$ whose underlying \gls{func} $T$
      \gls{preses} \glspl{cover}. If $f : (A,
      \alpha) \to (B, \beta)$ is a morphism of
      $\mathcal{C}\emat$ and $A \covers I \monic
      B$ is the image factorisation of $f$ in
      $\mathcal{C}$, then in
      \begin{picmath}
        \begin{tikzcd}
          TA \ar[r, -{Triangle[open]}] \ar[d,
          "\alpha"]
          & TI \ar[r] \ar[d, "\iota", dashed]
          & TB \ar[d, "\beta"] \\
          A \ar[r, -{Triangle[open]}]
          & I \ar[r, tail]
          & B
        \end{tikzcd}
      \end{picmath}
      we get a unique $\iota$ making both squares
      commute, making $(I, \iota)$ into a
      $\Tbb$-algebra, and it's the image of $f$ in
      $\mathcal{C}\emat$.

      In particular, any \gls{cat} \gls{monadic}
      over $\Set$ is \gls{regc}.