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First of all, since $M_S$ belongs to a prefactorization system, it is closed under composites, pullbacks, and any intersections which exist. Therefore, if we define $M' \coloneqq M \cap M_S$, then $M'$ satisfies the hypotheses of Theorem \ref{ConstructingOFS}, and so we have an OFS $(E',M')$.

Moreover, it is useful to notice that $E_S=$: this is an easy consequence of the fact that if $S\dashv T$, then $Sa\perp b\iff a\perp Tb$, since $f\perp Tu\iff Sf\perp u$ for each $u\in\hom(C)$, so that $S f$ is an isomorphism.

Now suppose given $f\colon A\to B$; we want to construct an $(E_S,M_S)$-factorization. Let $v$ be the pullback of $T S f$ along the unit $\eta_B \colon B \to T S B$. The naturality square for $\eta$ at $f$ shows that $f$ factors through $v$, say $f = v w$.

$\begin{array}{ccccc}
A & \\
&\overset{w}\searrow\\
&& P &\overset{u}\to& T S A \\
&& {}^v\downarrow && \downarrow^{T S f}\\
&& B &\underset{\eta_B}\to& T S B
\end{array}$

Since $T S f$ is evidently in $M_S=({}^\perp T(\hom(C)))^\perp\supseteq T(\hom (C))$, so is $v$; thus it suffices to find an $(E_S,M_S)$-factorization of $w$.

Let $w = n g$ be the $(E',M')$-factorization of $w$. Since $M' \subseteq M_S$, it suffices to show that $g\in E_S$. Note also that since $w$ is a first factor of the unit $\eta_A$, by passing to adjuncts we find that $S w$ is split monic: in the former diagram we have $u w=\eta_A$, so that the adjunct $\epsilon_{S A} \cdot S u\cdot S n\cdot S g=1$, hence also $S g$ is a split monic. But $T S g$ is then also split monic, hence belongs to $M$ and thus also to $M'$ (since it obviously belong to $M_S=({}^\perp T(\hom(C)))^\perp\supseteq T(\hom (C))$). Therefore, since $g\in E'$, the naturality square for $\eta$ at $g$ contains a lift: there is an $\alpha\colon X\to T S A$ such that in the diagram

$\begin{array}{ccc}
A & \overset{\eta_A}\to & T S A \\
{}^g\downarrow && \downarrow^{T S g} \\
X &\overset{\eta_X}\to& T S X
\end{array}$

$\alpha\cdot g=\eta_A$ and $T S g \cdot \alpha=\eta_X$. Passing to adjuncts again, we find that $S g$ is also split epic, since we can consider the diagram

$\begin{array}{ccccc}
S A & \overset{S\eta_A}\longrightarrow & S T S A &\overset{\epsilon_{SA}}\longrightarrow & S A\\
{}^{S g}\downarrow && \phantom{aaa}\downarrow_{S T S g} &&\downarrow^{S g}\\
S X &\underset{S\eta_X}\longrightarrow & S T S X &\underset{\epsilon_{SX}}\longrightarrow & S X
\end{array}$

and the commutativity

$S g \cdot \epsilon_{S A} \cdot S\alpha = \epsilon_{S X} S T S g \cdot S\alpha = \epsilon_{S X}\cdot S\eta_X = 1$

Hence $S g$ is an isomorphism; thus $g\in E_S$ as desired. $\blacksquare$

?? Grothendieck inequality $<>$ ***

- $[\mathcal{I},\mathcal{A}]$ is really just the underlying category with hom-collections given by $A_0(A,B)=V_0(I,\mathcal{A}(A,B))$.
- $\mathcal{A}(-,-)$ is the fully faithful two-variable hom-functor from $A_0^{op}\times A_0\to V_0$, with $\mathcal{A}(f,g)$ defined as the composite $\mathcal{A}(B,C)\stackrel{l^{-1}r^{-1}}{\to}I\otimes\mathcal{A}(B,C)\otimes I\stackrel{f\otimes id\otimes g}{\to}\mathcal{A}(C,D)\otimes\mathcal{A}(B,C)\otimes\mathcal{A}(A,B)\stackrel{(\circ^{\mathcal{A}})^2}\mathcal{A}(A,D)$ in $V_0$
- $[\mathcal{I},F]$ is the functor from $A_0$ to $B_0$ underlying the enriched functor $F$. This is defined by letting $Ff$ be the composite $I\stackrel{f}{\to}\mathcal{A}(A,B)\stackrel{F_{A,B}}\mathcal{B}(FA,FB)$ where $F_{A,B}$ is the family of morphisms in $V_0$ defining the enriched functor $F$.
- The natural transformation $\bar F\colon\cat A(-,-)\to\cat B(F-,F-)$ has for its components exactly the maps $F_{A,B}$ above: i.e. $\bar F_{A,B}=F_{A,B}$. * * category: meta

Revised on April 20, 2014 09:13:30
by Fosco Loregian
(79.1.178.57)