Find weak equivalences from fibrations and cofibrations

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.

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.

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).