In a model category, when weak equivalences are inverted, nothing else gets inverted. It follows that weak equivalences satsify 2-out-of-6. But the first sentence takes some work to show. Is there a more direct way to demonstrate 2-out-of-6?

For reference, 2-out-of-6 says that if $v \circ u$ and $w\circ v$ are weak equivalences, then $w\circ v \circ u$ is a weak equivalence (in light of 2-out-of-3, the conclusion is equivalent to saying that any of $u,v,w$ is a weak equivalence). This property apparently features in the definition of a homotopical category. It follows from the property above because isomorphisms satisfy 2-out-of-6, so the preimages of isomorphisms under a functor also satsify 2-out-of-6.

I'm interested in having a direct argument because 2-out-of-6 leads to a simple proof that if $f$ is a homotopy equivalence (i.e. there exists $g$ such that $gf$ and $fg$ are both either left or right homotopic to identity maps) then $f$ is a weak equivalence. I would expect this fact to have a simple proof, but in fact Dwyer-Spalinski and Hovey at least only prove this directly with some fibrancy/cofibrancy restrictions on the objects involved. The proof using 2-out-of-6 goes like this: it's easy to see that if $h$ is left or right homotopic to $k$, then if one is a weak equivalence then so is the other; so if $f$ has homotopy inverse, then $fg$ and $gf$ are both weak equivalences, so 2-out-of-6 implies that $f$ and $g$ are each weak equivalences.

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    $\begingroup$ I don't think there's any easy way to do this. If you can show that homotopy equivalences are weak equivalences then you are already halfway to showing the class of weak equivalences is saturated. $\endgroup$ – Zhen Lin Dec 30 '14 at 5:03
  • $\begingroup$ But in actual examples, this is by far the easier half! Showing that homotopy equivalences induce isomorphisms of homotopy groups is easy -- proving Whitehead's theorem is hard... I suppose this is ultimately a weak argument since the axioms of a model category are geared toward proving Whitehead's theorem. But it shocks my conscience that the formalism should make easy things become hard! :-/ $\endgroup$ – tcamps Dec 30 '14 at 5:06
  • $\begingroup$ No, that's not true. In order to construct the model structure on $\mathbf{Top}$ you need to do almost all the hard work needed to prove Whitehead's theorem anyway. At least that's what Moerdijk remarked. There's never any free lunch. $\endgroup$ – Zhen Lin Dec 30 '14 at 6:08
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    $\begingroup$ It's certainly true that in actual examples we often get 2-out-of-6 or even saturation for free. That's why [Dwyer, Hirschhorn, Kan, and Smith] advocate changing the definition. $\endgroup$ – Zhen Lin Dec 30 '14 at 6:09

For some reason, I have the urge to write this out; it's not that hard really. The proof I sketch is only just what you get if you unwind the proof of saturation in Hovey. I refer to a sequence $(w,v,u)$ of composable maps, as in the question: I want to show that if $wv$ and $vu$ are weak equivalences, so are the others.

  1. The special case when either $wv$ or $vu$ is an identity map is immediate using the 2-of-3 and retract axioms. For instance, if $wv=1$, then $u$ is a retract of $vu$.

  2. In particular, we can apply point (1) to $(s,r,s)$ when $rs=1$: if $sr$ is a weak equivalence, so are $r$ and $s$.

  3. The general problem really happens in the model category of factorizations of the map $wvu$. Thus, without loss of generality, we may reduce to the following Claim: In any model category, if $T'$ is weakly equivalent to the terminal object $T$, and $I'$ is weakly equivalent to the initial object $I$, then any map $r\colon T'\to I'$ is a weak equivalence (if such a map exists). [E.g., $T'$ is $(wv,u)$, $I'$ is $(w,vu)$, and $r$ is $v$.]

  4. By a straightforward argument using factorization and 2-of-3, you can show that without loss of generality we may assume that $I'$ and $T'$ are cofibrant and that $r$ is a fibration.

  5. As $I\to I'$ is a trivial cofibration and $r\colon T'\to I'$ is a fibration, by lifting the map $r$ admits a section $s$, so $rs=1_{I'}$.

  6. 2-of-3 implies that any self-map $T'\to T'$ of $T'$ is a weak equivalence, since the unique map $T'\to T$ is a weak equivalence.

  7. Thus $sr\colon T'\to T'$ is a weak equivalence by point (6), whence $r$ must be a weak equivalence by point (2), as desired.


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