# How was the functorial factorization axiom "frequently misstated"?

In modern times, a model category is frequently required to have functorial factorizations of morphisms into (fibration/acyclic cofibration) and (acyclic fibration/cofibration). But the precise meaning of "functorial factorization" has apparently evolved over time.

As far as I know, a correct formulation is found in this $n$Lab entry, or in definition 12.1.1 of Emily Riehl's book Categorical homotopy theory:

Let $\underline{n} = \{ 1 < \dots < n \}$ be the poset seen as a category. Let $\mathsf{C}$ be a category, and let $$c = (d^1)^* : \mathsf{Fun}(\underline{3}, \mathsf{C}) \to \mathsf{Fun}(\underline{2}, \mathsf{C})$$ be the composition (where $d^1 : \underline{2} \to \underline{3}$ is given by $d^1(1) = 1$, $d^1(2) = 3$). Then a functorial factorization is a section of $c$.

In other words it's a functor $F$ that maps $f : X \to Y$ to a sequence $X \xrightarrow{F_L(f)} \bar{F}(f) \xrightarrow{F_R(f)} Y$ which composes to $f$, and if $$\require{AMScd} \begin{CD} X @>f>> Y \\ @VVV @VVV \\ X' @>f'>> Y \end{CD}$$ is a commutative diagram in $\mathsf{C}$, then we have a morphism $\bar{F}(f) \to \bar{F}(f')$ that makes the obvious diagrams commute.

But this formulation wasn't apparently the standard back then, and other ones were used that were incorrect (cf. the $n$Lab entry, I've also been told that orally by different people). "Unfortunately", I've only been taught the correct one, thus I have difficulty imagining what the wrong one is. It doesn't help that a few references only state that the factorization is to be functorial without defining what it means (unless I missed it).

Question. What was the incorrect formulation of the functoriality of factorization? Why was it incorrect?

I'm not sure if this is what people are referring to, but in Hovey's book, the definition of a functorial factorization misses the requirement of the arrow $\bar F(f) \to \bar F(f')$. It also misses something else that There's something you're missing -- the functoriality of these arrows. If we have commutative squares

$$\require{AMScd} \begin{CD} X @>f>> Y \\ @VgVV @VVhV \\ X' @>f'>> Y' \\ @Vg'VV @VVh'V \\ X" @>f">> Y" \end{CD}$$

Then we get maps $\bar F (g,h) : \bar F(f) \to \bar F(f')$ and $\bar F (g',h'): \bar F (f') \to \bar F (f")$, and we need to require that $\bar F(g',h') \circ \bar F(g,h) = \bar F( g' \circ g, h' \circ h)$ (as well as an identity equation $\bar F(1,1) = 1$). These equations come from the requirement that $\bar F$ be a functor.

In the context of a weak factorization system, the existence of $\bar F(g,h)$ can be obtained from a lifting property. But its functoriality can not, although I don't know any examples to illustrate this (probably the experts do, though).

EDIT

I was wrong about what Hovey's definition gets wrong. Hovey actually asks for two functors $F_L$ and $F_R$, $\mathrm{Mor} C \to \mathrm{Mor} C$ which must be functorial in the way I described. On objects, this means from an $X \overset{f}{\to} Y$ we get $F_L^0(f) \overset{F_L(f)}{\to} F_L^1(f)$ and $F_R^1(f) \overset{F_R(f)}{\to} F_R^2(f)$.

Hovey requires that $F_L^0(f) = X$, $F_L^1(f) = F_R^1(f)$ (I'll just call this $F(f)$), and $F_R^2(f) = Y$, and that $F_R(f) \circ F_L(f) = f$. So far so good.

But when it comes to arrows, he forgets to say some obvious things. From a square

$$\require{AMScd} \begin{CD} X @>f>> Y \\ @VgVV @VVhV \\ X' @>f'>> Y' \end{CD}$$

He gets two squares

$$\require{AMScd} \begin{CD} X @>F_L(f)>> F(f) \\ @VF_L^0(g,h)VV @VVF_L^1(g,h)V \\ X' @>F_L(f')>> F(f') \end{CD}$$

$$\require{AMScd} \begin{CD} F(f) @>F_R(f)>> Y \\ @VF_R^1(g,h)VV @VVF_R^2(g,h)V \\ F(f') @>F_R(f')>> Y' \end{CD}$$

but he forgets to require that $F_L^0(g,h) = g$, $F_L^1(g,h) = F_R^1(g,h)$, and $F_R^2(g,h) = h$. This sort of "goes without saying", I suppose, but I guess the advantage of the definition you cite is that all this data is included automatically, and you avoid the possibility of oversights like this.

• Ah, you're right, I was a bit too quick when I tried to reformulate the axiom. Maybe that was actually the problem of earlier formulations...? Anyway I would be very interested in an example for your last paragraph. Mar 7, 2015 at 16:04