%! TEX root = CT.tex % vim: tw=50 % 04/12/2024 09AM \begin{fclemma}[] \label{lemma:7.14} % Lemma 7.14 Assuming: - $\mathcal{C}$ a \gls{small} \gls{regc} and \gls{ats} \gls{cat} Then: there exists an isomorphism \glsref[reflects]{reflecting} \gls{reg} \gls{func} $\mathcal{C} \to \hat{\mathcal{C}}$ where $\mathcal{C}$ is \gls{cap}. Hence in particular, there is an isomorphism-\glsref[reflects]{reflecting} \gls{reg} \gls{func} $\mathcal{C} \to \Set$. \end{fclemma} \begin{proof} Consider the sequence \[ \mathcal{C} = \mathcal{C}_0 \to \mathcal{C}_1 \to \mathcal{C}_2 \to \cdots ,\] where each $\mathcal{C}_{n + 1}$ is obtained from $\mathcal{C}_n$ by the construction of \cref{lemma:7.13}. We define $\hat{\mathcal{C}}$ to be the pseudo-\gls{colim} of this sequence: objects are pairs $(n, A)$ where $A \in \ob \mathcal{C}_n$, and morphisms $(n, A) \to (m, B)$ are \gls{reped} by pairs $(p, f)$ where $p \ge \max \{m, n\}$ and $F : IA \to I'B$ in $\mathcal{C}_p$, modulo the identification of $(p, f)$ with $(p', f')$ if $p \le p'$ and $f' = If$. The proof that $\mathcal{C}$ is \gls{regc}, and that the embeddings $\mathcal{C}_n \to \hat{\mathcal{C}}$ are isomorphism-\glsref[reflects]{reflecting} \gls{reg} \glspl{func}, is as in \cref{lemma:7.13}. Given any non-invertible \gls{mono} $A' \monic A$ in $\hat{\mathcal{C}}$, it lives in $\mathcal{C}_n$ for some $n$, so there exists $1 \to A$ in $\mathcal{C}_{n + 1}$ not factoring through $A' \monic A$. But if $A \stackrel{f}{\to} B$ isn't \gls{monic} in $\hat{\mathcal{C}}$, the legs $R \twofuncs ab A$ of its kernel-pair aren't equal, so there exists $1 \stackrel{r}{\to} R$ not factoring through their equation, so $1 \twofuncs{ar}{br} A$ are distinct but have the same composite with $f$. So $\hat{\mathcal{C}}(1, \blank)$ \gls{reflects} \glspl{mono} and hence \gls{reflects} isomorphisms. \end{proof} \begin{fcthm}[] \label{thm:7.15} % Theorem 7.15 Assuming: - $\mathcal{C}$ \gls{small} and \gls{regc} Then: there exists a set $I$ and an isomorphism-\glsref[reflects]{reflecting} \gls{reg} \gls{func} $\mathcal{C} \to \Set^I$. \end{fcthm} \begin{proof} Let $I$ be a representative set of \glspl{subobj} of $1$ in $\mathcal{C}$, and for each $U \in I$ consider the composite \[ \mathcal{C} \stackrel{(!_U)^*}{\to} \mathcal{C} / U \to (\mathcal{C} / U)_\tv \to \widehat{(\mathcal{C} / U)_\tv} \to \Set ,\] where the third factor is the \gls{func} of \cref{lemma:7.14} and the fourth is \gls{reped} by $1$. Given any non-invertible morphism $A \stackrel{f}{\to} B$ in $\mathcal{C}$, if $U$ is the \gls{supp} of $B$ then $(!)_H^* f$ remains non-invertible in $\mathcal{C} / U$ and its codomain is \gls{ws} there, so it remains non-invertible in $(\mathcal{C} / U)_\tv$ and hence in $\Set$. So these \glspl{func} collectively \glsref[reflects]{reflect} isomorphisms. \end{proof} \begin{remark} \label{remark:7.16} % Remark 7.16 \phantom{} \begin{enumerate}[(a)] \item Barr's original embedding theorem produces a \gls{full} and \gls{faith} \gls{reg} \gls{func} $\mathcal{C} \to \funccat[\mathcal{D}, \Set]$ for some \gls{small} \gls{cat} $\mathcal{D}$. Moreover if $\mathcal{C}$ is \gls{ats} we can take $\mathcal{D}$ to be a monoid. \item \cref{thm:7.15} yields a `meta theorem' saying that `anything we can prove in $\Set$ is true in all \gls{reg} \glspl{cat}'. For example to prove \cref{prop:7.5} (\gls{cover} implies \gls{reg} \gls{epic}), given a \gls{cover} $A \stackrel{f}{\covers} B$ in a \gls{regc} \gls{cat} $\mathcal{C}$, and a $A \stackrel{g}{\to} C$ having equal composites with the kernel-pair $R \twofuncs{}{} A$ of $f$, we can cut down to a \gls{small} sub\gls{cat} $\mathcal{C}'$ containing $f$ and $g$ and closed under finite \glspl{lim} and \glspl{image}, and then show that the first component of $I \stackrel{(h, k)}{\monic} A \times C$ becomes an isomorphism in $\Set^I$. \item Abelian \glspl{cat} are \gls{regc} \glspl{cat} enriched over $\AbGp$ (i.e. for any two objects $A$ and $B$, $\mathcal{A}(A, B)$ has an abelian group structure and composition distributes over addition). Abelian \glspl{cat} are \gls{ts} since their \glspl{tobj} are \gls{initial}, so for any \gls{small} abelian $\mathcal{A}$ we get an isomorphism-\glsref[reflects]{reflecting} \gls{reg} \gls{func} $\mathcal{A} \to \Set$ and hence an isomorphism-\glsref[reflects]{reflecting} \gls{func} $\mathcal{A} \cong \AbGp(\mathcal{A}) \to \AbGp(\Set) = \AbGp$. \end{enumerate} \end{remark}