Let's suppose I have a category $\mathcal C$ with two weak factorisation systems $(C,F_W)$ and $(C_W,F)$, where $C_W\subset C$ and $F_W\subset F$.

I would like to have a model structure on $\mathcal C$ such that $C$ (resp. $C_W$) are the cofibrations (resp. the acyclic cofibrations) and $F$ (resp. $F_W$) are the fibrations (resp. the acyclic fibrations).

1. Is there always at most one possible choice for the class of weak equivalences? (I've heard that in a model structure, having two classes of arrows determines uniquely the third one, but I don't know how to do this when it's the weak equivalences that I want)

2. Under what conditions does a model structure indeed exists.

  • $\begingroup$ Thanks for all the answers, I'm accepting Jonathan's answer because I was looking for the simplest characterization, but I'll keep in mind the other answers for when I'll need to use this kind of result. $\endgroup$ Dec 23, 2011 at 17:33
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    $\begingroup$ For the benefit of anyone else wandering in here: (1) is addressed succinctly at mathoverflow.net/questions/29635/… (where it is also labeled question (1)!). Cf. Mark Hovey's response. $\endgroup$
    – Tim Campion
    May 19, 2012 at 8:17

4 Answers 4


When you have a model structure, the cofibrations and fibrations give you the acyclic cofibrations and acyclic fibrations because of the lifting axioms. Then, the weak equivalences are those arrows which can be factored as an acyclic cofibration followed by an acyclic fibration. (Use the factorization and $2$ out of $3$ axioms.) That is how the class of cofibrations and the class of fibrations give you the whole model structure.

EDIT: While I am at it, it may be worth mentioning that you do not need all the fibrations to recover the model structure once you know the cofibrations. The fibrations whose codomain is the terminal object give you a sufficient data. In other words: a model structure is determined by the cofibrations and fibrant objects. I think this observation is due to Joyal. This is Proposition E.1.10. of his text The Theory of Quasi-Categories and its Applications.

  • $\begingroup$ Just to clarify to Jonathan's answer, if there exists such a model structure, then the weak equivalences are exactly those arrows which are composites of an arrow in $C_W$ followed by an arrow in $F_W$. So there is at most one such model structure. Moreover, let this define the class W. Then your factorizations systems give you a model structure precisely if $F_W = W \cap F$ and similarly for $C_W$. $\endgroup$ Dec 22, 2011 at 14:08
  • $\begingroup$ Thanks Jonathan! Chris, is it automatic that this $W$ satisfies 2 out of 3? $\endgroup$ Dec 22, 2011 at 14:15
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    $\begingroup$ "is it automatic W satisfies 2 out of 3?", ahh no. You caught me! You need that too. It looks like Rezk's answer has another characterization. $\endgroup$ Dec 22, 2011 at 14:28
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    $\begingroup$ Je t'en prie ! (Excuse my French.) I should have made clear that I addressed only the first point (the answer to which is standard), sorry about that. The second one is indeed trickier (see Charles Rezk's answer). $\endgroup$ Dec 22, 2011 at 14:59

Starting with your setup of two related weak factorization systems, let $A$ be the class of morphisms $f$ with the following property: there exists a factorization $f=qsi$ such that:

  • $i\in C$ and $qs\in F_W$, and
  • $q\in F$ and $si\in C_W$.

This is better pictured as: a commutative square whose two sides are $(C,F_W)$ and $(C_W,F)$ factorizations of $f$, together with a lift $s$ which goes the "wrong way".

The result is that your setup is a model category if and only if the class $A$ satisfies the "2 out of 6" property (in which case, $A$ is precisely the class of weak equivalences).

Proof: if you are starting with a model category, then it is not too hard to show that $A=W$. Conversely, if you start with your weaker setup, you can show directly that $C_W=C\cap A$ and $F_W=F\cap A$; once you also know that $A$ satisfies 2 out of 6, standard results tell you that $(A,C,F)$ is a model category structure.

Added. You can sharpen this a bit: it turns out that if you know that the class $A$ defined in the first paragraph is closed under 2-out-of-3, then it is also closed under retracts. Thus, you have a model category if and only if $A$ is closed under 2-out-of-3.


Charles and Jonathan have given good answers to (1), but here's another way to recover the weak equivalences from the cofibrations and fibrations, which I think is sometimes convenient.

First of all, the wfss give us notions of "fibrant replacement" (a $(C_W,F)$-factorization of $X\to 1$) and "cofibrant replacement" (a $(C,F_W)$-factorization of $0\to X$). If $F_W\subseteq F$ (equivalently $C_W\subseteq C$), then we have two notions of fibrant-cofibrant replacement.

Now the wfs $(C_W,F)$ defines a notion of "path object", namely a $(C_W,F)$ factorization of the diagonal $X\to P X \to X\times X$. From this we can obtain a notion of "right homotopy": two maps $A\to X$ are right homotopic if the induced map into $X\times X$ factors through some path object for $X$. Dually, $(C,F_W)$ defines a notion of "cylinder object" and thereby "left homotopy".

If the two given wfss underlie a model structure, then the following all characterize the class of weak equivalences:

  • $f\in W$ iff some (hence any) fibrant-cofibrant replacement of $f$ is a left (or right) homotopy equivalence (i.e. becomes an isomorphism upon quotienting by homotopy)
  • $f\in W$ iff some (hence any) cofibrant replacement $f'$ of $f$ has the property that it induces an isomorphism on right homotopy classes of maps into any fibrant object.
  • Dually, $f\in W$ iff some (hence any) fibrant replacement $f'$ of $f$ has the property that it induces an isomorphism on left homotopy classes of maps out of any cofibrant object.

In contrast to Charles' and Jonathan's answers, with these definitions it's almost automatic to get 2-out-of-3 (there's probably some fiddlyness with some vs any, but if you have a given functorial realization of your wfss you could just use that factorization to define the replacements). Moreover, $C_W$ automatically has the second property, while $F_W$ automatically has the third. Thus, if you can show that the three definitions agree, then you get $C_W \subseteq C\cap W$ and $F_W \subseteq F\cap W$. The tricky part would now be showing the reverse inclusions (although a standard argument implies that you only need to show one of them).


I was unaware of the result Charles Rezk gave. When I saw the question I immediately thought of cotorsion pairs, which is a way to get from two $Ext$-orthogonal systems $(C,F_W)$ and $(C_W,F)$ on a (nice) abelian category $\mathcal{M}$ to a model category structure on $\mathcal{M}$. The connection was established by my advisor, Mark Hovey, in this paper and also this one. The connection was used to find certain new model category structures by his students Jim Gillespie and Daniel Bravo.

Anyway, the issues about whether or not $W$ satisfies the 2 out of 3 axiom, etc were a driving reason for these papers. If you're in an abelian setting then it's possible Theorem 2.5 from the second linked paper might be an easier way to check if the given pairs define a model category structure. Also, it's more general than Charles's answer because you actually don't need the two pairs to satisfy lifting. They only need to be $Ext$-orthogonal. You can find all the relevant definitions in the second linked paper above. I put Theorem 2.5 here for completeness.

If $(C,F\cap W)$ and $(C\cap W, F)$ are complete cotorsion pairs on an abelian category $\mathcal{M}$, and if $W$ is thick, then $C,F,W$ define a model category structure on $\mathcal{M}$.


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