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?


1 Answer 1


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).


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.

  • $\begingroup$ 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. $\endgroup$ Mar 7, 2015 at 16:04

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