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Stream: theory: category theory

Topic: When is the factorization constructed in one step?

Arshak Aivazian (Apr 05 2025 at 08:32):

Given a small set of morphisms $M$ in a locally presentable category $C$, there exists an factorization system $(M^{rl}, M^r)$ generated by $M$. In the general case, the factorization is constructed via a transfinite process as described in Adámek–Rosický.

The construction I describe below works in some cases. For example, this is how ring homomorphisms are factorized into

• ... a composition of a surjection and an injection. For a homomorphism $f: A \to B$, we extract the preimage of zero $f^{-1}(0)$ and freely impose the relations of it vanishing (i.e., take the quotient by it): $A \to A/f^{-1}(0) \to B$

• ... a composition of a localization and a conservative homomorphism. For a homomorphism $f: A \to B$, we extract the preimage of the invertible elements $f^{-1}(B^\times)$ and freely impose the relations of their invertibility (i.e., localize at them) $A \to A_{f^{-1}(B^\times)} \to B$

I’m interested in understanding the range of applicability of this construction.

Let $C$ be a locally presentable category, and let $m$ be a morphism in it (for a set of arrows, the process is completely analogous). The morphism $m \colon U \to V$ defines a natural transformation of representable functors $E \to B$ where $E(c) := \mathrm{Hom}(V, c)$ and $B(c) := \mathrm{Hom}(U, c)$

Given a morphism $f \colon c \to d$, we consider the square induced by this natural transformation and form the pullback $S := B(f)^*(E(d))$.

Now consider the diagram
$\coprod_{S} V \longleftarrow \coprod_{S} U \longrightarrow c$
where the left arrow is the coproduct of $m$ and the right arrow corresponds to the map $S \to B(c)$ (arising from the pullback structure). Taking the pushout of this diagram yields an object $c_S$. By the sturcute and the universal property of the pushout, we obtain a factorization $c \to c_S \to d$. The first morphism is in the left class of the factorization system, since it is a pushout of a coproduct of copies of $m$.

The question is:

1. Under what conditions is the second morphism in the right class of the factorization system? In other words, when does the pullback along it (meaning along $B$(it)) send $E(d)$ to $E(c_S)$? Note that in this case the left class is a pushouts of coproducts of m.

2. Are there any examples in $\mathrm{CRing}$ or other finitary algebraic categories where this does not work?

For example, as mentioned, in the case of commutative rings $\rm{CRing}$, if we take as $m$:

• $\mathrm{ev}_0: \mathbb{Z}[x] \to \mathbb{Z}$
• $\mathrm{in}: \mathbb{Z}[x] \to \mathbb{Z}[x, x^{-1}]$

This construction yields the corresponding factorization.

P.S. I asked this on MO and hope the question might be of interest to many here.

Chaitanya Leena Subramaniam (Apr 16 2025 at 17:35):

Disclaimer: these are only partial answers to your questions :)

As Zhen Lin mentioned on your mathoverflow post, your construction is the iterative step in the standard (a.k.a. "Quillen's") [[small object argument]] to construct the weak factorisation system $({}^\pitchfork (\{m\}^\pitchfork),\{m\}^\pitchfork)$ cofibrantly generated by the single morphism $m$. This is because your pullback $S$ is the homset $\mathrm{Hom}_{C^\to}(m,f)$ of the arrow category, and the iterative step consists of taking the pushout in the outer (counit) square $\mathrm{Hom}_{C^\to}(m,f)\otimes m\to f$. (The outer square is the counit of the density comonad associated to the functor $m : 1\to C^\to$.)

But your question is about the orthogonal factorisation system $({}^\bot (\{m\}^\bot),\{m\}^\bot)$. The following useful facts are true for any pair of morphisms $f,g$ in any 1-category:

• ($f\perp g$) iff ($f\pitchfork g$ and $\nabla_f \pitchfork g$).
• ($\nabla_f\pitchfork g$) iff ($f\pitchfork \Delta_g$)

(N.B.: $\nabla_f : B+_A B \to B$ is the codiagonal of $f: A\to B$, and $\Delta_f : A\to A\times_B A$ is the diagonal of $f$.)

Hence $({}^\bot (\{m\}^\bot),\{m\}^\bot) = ({}^\pitchfork (\{m,\nabla_m\}^\pitchfork),\{m,\nabla_m\}^\pitchfork)$.
Since $m$ is an epi iff $\nabla_m$ is an iso, we have that when $m$ is an epimorphism, the weak and orthogonal factorisation systems generated by $m$ coincide. Both your examples are of this form. So it makes sense that the standard small object argument will construct the orthogonal factorisation you're looking for.

This still doesn't answer why the SOA construction succeeds after one iteration in your two examples. I think this is for two different reasons.

• For the $(\mathrm{RegEpi},\mathrm{Mono})$ o.f.s. on CRing (the one generated by $\mathbb{Z}[x]\to\mathbb{Z}$), the object $\mathbb{Z}[x]$ is regular-projective (projective for surjections). Hence for every lifting problem as below, there is a lift of $\mathbb{Z}[x]\to A/f^{-1}(0)$ to $A$; thus the lifting problem factors through $A\to A/f^{-1}(0)$ thanks to the iterative step of the SOA.
  image.png

  

• For the $(\mathrm{Loc},\mathrm{Cons})$ o.f.s. on CRing, I don't have a good conceptual reason for why the SOA construction succeeds in one step. In one of his papers, Mathieu Anel suggests (in section 4.9.4) that the (RegEpi, Mono) and (Loc, Cons) o.f.s.'s on CRing are dual in a certain sense, analogous to how the (Final, DiscFib) and (Initial, DiscOpFib) o.f.s.'s are $(-)^{\mathrm{op}}$-dual in Cat. If this is the case, then perhaps the 1-step (RegEpi, Mono)-factorisation implies that of the (Loc, Cons)-factorisation.

To answer your Question 2, there are plenty of examples where the standard SOA does not construct the orthogonal factorisation generated by a set of maps --- it suffices to choose generators that are not epimorphisms. For a simple counterexample, consider Set with the map $m:\emptyset\to 1$. The o.f.s. generated by $m$ is (All, Iso), but the factorisation generated by the standard SOA is that of the w.f.s. (Mono, Epi).