%! TEX root = CT.tex
% vim: tw=50
% 02/12/2024 09AM

\begin{fcdefn}[Support, totally supported, capital]
  \glsnoundefn{supp}{support}{supports}%
  \glsnoundefn{siobj}{strict initial
  object}{strict initial objects}%
  \glsadjdefn{ws}{well-supported}{object}%
  \glsadjdefn{ts}{totally-supported}{\gls{cat}}%
  \glsadjdefn{ats}{almost totally-supported}{\gls{cat}}%
  \glsadjdefn{cap}{capital}{\gls{cat}}%
  \phantom{}
  \begin{cenum}[(a)]
    \item By the \emph{support} of an object $A$
      in a \gls{regc} \gls{cat}, we mean the
      \gls{image} of $A \to 1$. We say $A$ is
      \emph{well-supported} if $A \to 1$ is a
      \gls{cover}.
    \item We say a \gls{regc} \gls{cat}
      $\mathcal{C}$ is \emph{totally supported} if
      every object is well-supported. We say
      $\mathcal{C}$ is \emph{almost totally
      supported} if every object is either
      well-supported or a strict \gls{iobj},
      where we cann an object $0$ \emph{strict} if
      every $A \to 0$ is an isomorphism. (Given
      finite \glspl{lim}, a strict object is
      \gls{initial} since for any $A$ there exists $0
      \stackrel{\pi^{-1}}{\to} 0 \times A
      \stackrel{\pi_2}{\to} A$, and the
      \gls{equaler} of any pair $0 \twofuncs{}{}
      A$ is $a$).
    \item We say a \gls{regc} \gls{cat}
      $\mathcal{C}$ is \emph{capital} if its
      \gls{tobj} $1$ is a \gls{det}, i.e.
      $\mathcal{C}(1, \blank)$ \gls{reflects}
      isomorphisms.
  \end{cenum}
\end{fcdefn}

\begin{example*}
  $\Gp$ and $\AbGp$ are \gls{ts} since their
  \glspl{tobj} are \gls{initial}. $\Set$ is
  \gls{ats} and \gls{cap}. Note that \gls{cap}
  implies \gls{ats} since if $A$ isn't \gls{ws}
  there are no morphisms $1 \to A$.
\end{example*}

\glsadjdefn{cp}{cover-projective}{object}%
A \gls{reple} \gls{func} $\mathcal{C}(A, \blank)$
always \gls{preses} \glspl{lim}, so it's a
\gls{reg} \gls{func} if and only if $A$ is
cover-projective (c.f. \cref{defn:2.10}).

\begin{fclemma}[]
  \label{lemma:7.12}
  % Lemma 7.12
  Assuming:
  - $\mathcal{C}$ a \gls{locsm} \gls{cap} \gls{regc}
  \gls{cat}
  Then:
  $1$ is \gls{cp}.
\end{fclemma}

\begin{proof}
  Since \glspl{cover} are stable under
  \gls{pullb}, we need to show that every $A
  \covers 1$ is \gls{splitme} \gls{epic}. If $A
  \cong 1$, nothing to prove. If not, the
  projections $A \times A \twofuncs{}{} A$ aren't
  equal (since their \gls{coequaler} is $A \covers
  1$, by \cref{prop:7.5}). So there exists $1 \to
  A \times A$ not factoring through their
  \gls{equaler}, so there exists $1 \to A \times A
  \to A$.
\end{proof}

\glssymboldefn{ws}%
If $\mathcal{C}$ is \gls{regc}, the \gls{full}
sub\gls{cat} $\mathcal{C}_{\text{ws}}$ of \gls{ws}
objects is closed under finite \glspl{prod} since
\begin{picmath}
  \begin{tikzcd}
    A \times B
    \ar[r] \ar[d]
    & A \ar[d, {-Triangle[open]}] \\
    B \ar[r, {-Triangle[open]}]
    & 1
  \end{tikzcd}
\end{picmath}
is a \gls{pullb}, and under \glspl{pullb} of
\glspl{cover} since if $A \covers B$ then $A$ and
$B$ have the same \gls{supp}.

\glssymboldefn{tv}%
We write $\mathcal{C}_{\text{tv}}$ for the
\gls{cat} obtained from $\mathcal{C}_\ws$ by
adjoining a \gls{siobj} $0$: this is \gls{regc}
and \gls{ats} and the \gls{func} $\mathcal{C} \to
\mathcal{C}_{\text{tv}}$ sending all non-\gls{ws}
objects to $0$ is \gls{reg} (c.f. Exercise 5.19).

\begin{fclemma}[]
  \label{lemma:7.13}
  % Lemma 7.13
  Assuming:
  - $\mathcal{C}$ a \gls{small} \gls{ats}
  \gls{regc} \gls{cat}
  Then:
  there exists an
  isomorphism-\glsref[reflects]{reflecting}
  \gls{reg} \gls{func} $I : \mathcal{C} \to
  \mathcal{C}'$, where $\mathcal{C}'$ is also
  \gls{small} and \gls{ats}, such that for every
  \gls{ws} $A \in \ob \mathcal{C}$ there exists a
  morphism $1 \to IA$ in $\mathcal{C}'$ not
  factoring through $I(m)$ for any proper
  \gls{subobj} $m : A' \monic A$ in $\mathcal{C}$.
\end{fclemma}

\begin{proof}
  Recall from Exercise 7.17: $\mathcal{C}$
  \gls{regc} implies $\mathcal{C} / A$ \gls{regc}
  for any $A$, and for any $f : A \to B$ in
  $\mathcal{C}$ \gls{pullb} along $f$ defines a
  \gls{reg} \gls{func} $f^* : \mathcal{C} / B \to
  \mathcal{C} / A$, which has a \gls{ladj}
  $\Sigma_f : \mathcal{C} / A \to \mathcal{C} / B$
  sending $g : C \to A$ to $fg$. And $f^*$
  \gls{reflects} isomorphisms if and only if $f$
  is a \gls{cover}.

  We'll define $\mathcal{C}'$ as
  $(\hat{\mathcal{C}})_\tv$ where
  $\hat{\mathcal{C}}$ is easier to describe.

  To satisfy the desired conclusion for a single
  \gls{ws} object $A$, enough to take $(!)_A^* :
  \mathcal{C} \cong \mathcal{C} / 1 \to
  \mathcal{C} / A$, since $(!_A)^* A = (A \times A
  \stackrel{\pi_2}{\to} A)$ acquires a point
  $\Delta : (A \stackrel{1}{\to} A) \to (A \times
  A \to A)$ not factoring through $(A' \times A
  \to A)$ for any proper $A \monic A$.

  More generally, for any finite list $A_1,
  \ldots, A_n$ of \gls{ws} objects, we can take
  $\mathcal{C} / \prod_{i = 1}^{n} A_i$.

  We define a \emph{base} to be a finite list
  $\vec{A} = (A_1, \ldots, A_n)$ of distinct
  \gls{ws} objecs of $\mathcal{C}$. We preorder
  the set $\mathcal{B}$ of bases by $\vec{A} \le \vec{B}$ if
  $\vec{B}$ contains all the members of $\vec{A}$.
  We write $\prod \vec{A}$ for the product
  $\prod_{i = 1}^{n} A_i$ and if $\vec{A} \le
  \vec{B}$ we write $\pi_{\vec{B}, \vec{A}}$ for
  the \gls{prod} projection $\prod \vec{B} \to
  \prod \vec{A}$. This makes $\vec{A} \mapsto
  \prod \vec{A}$ into a \gls{func}
  $\mathcal{B}^\op \to \mathcal{C}$.

  Hence the assignment $\vec{A} \to \mathcal{C} /
  \prod \vec{A}$, $\pi_{\vec{B}, \vec{A}} \mapsto
  \pi_{\vec{B}, \vec{A}}^*$ is `almost' a
  \gls{func} $\mathcal{B} \to \Cat$.

  We now define $\hat{\mathcal{C}}$: its objects
  are pairs $(\vec{B}, f)$ where $\vec{B}$ is a
  base and $f : A \to \prod \vec{B}$ is an object
  of $\mathcal{C} / \prod \vec{B}$. Morphisms
  $(\vec{B}, f) \to (\vec{B}', f')$ are
  \gls{reped} by pairs $(\vec{C}, g)$ where
  $\vec{C}$ is a base containing $\vec{B}$ and
  $\vec{B}'$ and $g : \pi^* f \to \pi'^* f'$ in
  $\mathcal{C} / \prod \vec{C}$, subject to the
  relation which identifies $(\vec{C}, g)$ with
  $(\vec{C}', g')$ if $\vec{\mathcal{C}} \le
  \vec{\mathcal{C}}'$ and the \gls{pullb} of $g$
  to $\mathcal{C} / \prod \vec{\mathcal{C}}$ is
  isomorphic to $g'$.

  Clearly, each $\mathcal{C} / \prod \vec{B}$ sits
  inside $\hat{\mathcal{C}}$ as a non-\gls{full}
  sub\gls{cat}; so in particular $\mathcal{C}
  \cong \mathcal{C} / \prod []$ is a sub\gls{cat}
  of $\hat{\mathcal{C}}$, $\hat{\mathcal{C}}$ is
  \gls{regc}, and the inclusions $\mathcal{C} /
  \prod \vec{B} \to \hat{\mathcal{C}}$ are
  isomorphism-\glsref[reflects]{reflecting}
  \gls{reg} \glspl{func}.

  Given a finite \gls{diag} in
  $\hat{\mathcal{C}}$, we can choose $\vec{B}$
  such that all edges of the \gls{diag} appear as
  morphisms in $\mathcal{C} / \prod \vec{B}$, and
  take the \gls{lim} there, and this is a
  \gls{lim} in $\hat{\mathcal{C}}$. Similarly for
  \glspl{image}.

  Also, if a morphism $f$ becomes an isomorphism
  in $\hat{\mathcal{C}}$, its inverse must live
  $\mathcal{C} / \prod \vec{B}$ for some
  $\vec{B}$, hence $f$ is an isomorphism
  $\mathcal{C} / \prod \vec{B}$.

  We define $\mathcal{C}' =
  (\hat{\mathcal{C}})_\tv$: the induced \gls{func}
  $\mathcal{C} \to \hat{\mathcal{C}} \to
  \mathcal{C}'$ is still isomorphism
  \glsref[reflects]{reflecting} since
  $\mathcal{C}$ is \gls{ats}.
\end{proof}
