Contents

Definition

An (orthogonal) factorization system in a category $C$ consists of two classes of morphisms $\left(E,M\right)$ such that every morphism in $C$ factors as an $E$-morphism followed by an $M$-morphism, and we have $e$ satisfies the left lifting property –uniquely for the orthogonal case, $e\perp m$, see orthogonality – for any $e\in E$ and $m\in M$ .

To enrich this, we should first consider enriched orthogonality. The statement $e\perp m$ for maps $e:A\to B$ and $m:X\to Y$ can be rephrased by saying that the square

$\begin{array}{ccc}C\left(B,X\right)& \to & C\left(A,X\right)\\ ↓& & ↓\\ C\left(B,Y\right)& \to & C\left(A,Y\right)\end{array}$\array{C(B,X) & \to & C(A,X)\\ \downarrow && \downarrow \\ C(B,Y) & \to & C(A,Y)}

is a pullback in Set.

More generally, the statement that there exists at least one lift is to say that the canonical morphism

$C\left(B,X\right)\to C\left(A,X\right)\prod _{C\left(A,Y\right)}C\left(B,Y\right)$C(B,X) \to C(A,X) \prod_{C(A,Y)} C(B,Y)

into the pullback is a split epimorphism.

It is then clear that if $C$ is enriched over some monoidal category $V$, to say that $e\perp m$ in an enriched sense, we should instead require this square to be a pullback of enriched hom-objects in $V$. Note, though, that $e$ and $m$ are still maps in the underlying ordinary category ${C}_{0}$ of $C$. Likewise, the factorization property still only makes sense for maps in ${C}_{0}$.

Therefore, we define an enriched (orthogonal) factorization system on an enriched category $C$ to consist of two classes of maps $\left(E,M\right)$ in ${C}_{0}$ such that

1. $e\perp m$ in the above enriched sense for every $e\in E$ and $m\in M$, and
2. Every map in ${C}_{0}$ factors as an $E$-map followed by an $M$-map.

By the definition of ${C}_{0}$, enriched orthogonality implies ordinary orthogonality. Therefore, an enriched factorization system on $C$ induces an ordinary factorization system on ${C}_{0}$. Conversely, if $C$ has powers that preserve the maps in $M$, or copowers that preserve the maps in $E$, then unenriched orthogonality in ${C}_{0}$ implies enriched orthogonality by a Yoneda lemma argument.

Functoriality

Moreover, the factorization functor can be made into an enriched functor in the following way. There is a $V$-category ${C}^{2}$ whose objects are morphisms in ${C}_{0}$ and whose hom-objects are defined by, for ${f}_{1}:{X}_{1}\to {Y}_{1}$ and ${f}_{2}:{X}_{2}\to {Y}_{2}$, a pullback

$\begin{array}{ccc}{C}^{2}\left({f}_{1},{f}_{2}\right)& \to & C\left({X}_{1},{X}_{2}\right)\\ ↓& & ↓\\ C\left({Y}_{1},{Y}_{2}\right)& \to & C\left({X}_{1},{Y}_{2}\right).\end{array}$\array{C^{\mathbf{2}}(f_1,f_2) & \to & C(X_1,X_2)\\ \downarrow && \downarrow \\ C(Y_1,Y_2) & \to & C(X_1,Y_2).}

(This is the power of $C$ by $2$ in the 2-category $V-\mathrm{Cat}$, and also the $V$-functor category $\left[{2}_{V},C\right]$, where ${2}_{V}$ denotes the free $V$-category on $2$.)

Likewise, we have ${C}^{3}$ whose objects are composable pairs $X\stackrel{f}{\to }Y\stackrel{g}{\to }Z$ of morphisms in ${C}_{0}$, and whose hom-objects are defined by pullbacks

$\begin{array}{ccc}{C}^{3}\left(\left({f}_{1},{g}_{1}\right),\left({f}_{2},{g}_{2}\right)\right)& \to & {C}^{2}\left({f}_{1},{f}_{2}\right)\\ ↓& & ↓\\ {C}^{2}\left({g}_{1},{g}_{2}\right)& \to & C\left({Y}_{1},{Y}_{2}\right).\end{array}$\array{C^{\mathbf{3}}((f_1,g_1),(f_2,g_2)) & \to & C^{\mathbf{2}}(f_1,f_2)\\ \downarrow && \downarrow \\ C^{\mathbf{2}}(g_1,g_2) & \to & C(Y_1,Y_2).}

By functoriality we then mean that the factorization is given by a functor ${C}^{2}\to {C}^{3}$ which, when composed with the “composition” functor ${C}^{3}\to {C}^{2}$, gives the identity of ${C}^{2}$.

The interesting part of the enrichment of this functor is the following: given ${f}_{1}:{X}_{1}\to {Z}_{1}$ and ${f}_{2}:{X}_{2}\to {Z}_{2}$ in ${C}^{2}$, with factorizations ${X}_{1}\stackrel{{m}_{1}}{\to }{Y}_{1}\stackrel{{e}_{1}}{\to }{Z}_{1}$ and ${X}_{2}\stackrel{{m}_{2}}{\to }{Y}_{2}\stackrel{{e}_{2}}{\to }{Z}_{2}$, by enriched orthogonality we have a pullback

$\begin{array}{ccc}C\left({Y}_{1},{Y}_{2}\right)& \to & C\left({X}_{1},{Y}_{2}\right)\\ ↓& & ↓\\ C\left({Y}_{1},{Z}_{2}\right)& \to & C\left({X}_{1},{Z}_{2}\right)\end{array}$\array{C(Y_1,Y_2) & \to & C(X_1,Y_2)\\ \downarrow & & \downarrow\\ C(Y_1,Z_2) & \to & C(X_1,Z_2)}

and we also have a commutative square

$\begin{array}{ccccc}{C}^{2}\left({f}_{1},{f}_{2}\right)& \to & C\left({X}_{1},{X}_{2}\right)& \to & C\left({X}_{1},{Y}_{2}\right)\\ ↓& & & & \\ C\left({Z}_{1},{Z}_{2}\right)& & & & ↓\\ ↓& & & & \\ C\left({Y}_{1},{Z}_{2}\right)& & ⟶& & C\left({X}_{1},{Z}_{2}\right)\end{array}$\array{C^{\mathbf{2}}(f_1,f_2) & \to & C(X_1,X_2) & \to & C(X_1,Y_2)\\ \downarrow &&&& \\ C(Z_1,Z_2) &&&& \downarrow\\ \downarrow &&&& \\ C(Y_1,Z_2) & & \longrightarrow & & C(X_1,Z_2) }

inducing a map ${C}^{2}\left({f}_{1},{f}_{2}\right)\to C\left({Y}_{1},{Y}_{2}\right)$. It is then straightforward to construct the rest of the functor.

This argument, as it depends crucially on the universality of the pullback and hence the uniqueness part of orthogonality, fails for weak factorization systems. Although they can be made functorial in many cases, rarely can their functoriality be made enriched (as far as is known).

References

Enriched lifting and enriched factorization are discussed around from page 133 on (section “April 3”) in

Revised on April 6, 2012 05:18:55 by Urs Schreiber (82.169.65.155)