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A relative category is a category $C$ with a subcategory $W$ containing all the objects of $C$.

Given a relative category $(C,W)$, $W$ is said to satisfy the ``2 implies 6'' property if, for any collection of three composable maps,

$$X\rightarrow Y\rightarrow Z\rightarrow A$$

the presence of the composites $X\rightarrow Z$ and $Y\rightarrow A$ in $W$ implies that each individual map is in $W$ (and so also the triple composition).

The property I'm more familiar with from thinking about weak equivalences is the ``2 implies 3" property, which says that, given a pair of composable maps

$$X\rightarrow Y\rightarrow Z$$

the presence of any two of the maps

$$X\rightarrow Y$$ $$Y\rightarrow Z$$ $$X\rightarrow Z$$

in $W$ implies that the third is as well.

The "2 implies 6" property implies the "2 implies 3" property, and I've been told that "2 implies 6" is a strictly stronger property.

QUESTION: What is the basic example of a relative category $(C,W)$ where $W$ satisfies "2 implies 3", but not "2 implies 6"?

Edit: By "basic", I mean what is an example which comes up in applications, or better yet, what is the example to keep in mind?

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    $\begingroup$ You mean $X\to Z$, not $X\to A$. $\endgroup$ Feb 5, 2013 at 16:42
  • $\begingroup$ Oh, if that's the case then that completely negates my answer. $\endgroup$
    – Simon Rose
    Feb 5, 2013 at 16:46
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    $\begingroup$ As Emily Riehl pointed out to me, as long as there are non-identity isomorphisms in your category, the class of identities satisfies 2 out of 3 but not 2 out of 6. $\endgroup$ Feb 6, 2013 at 3:48
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    $\begingroup$ It is a result of Cisinski that a cofibration category satisfies 2 out of 6 if and only if it is saturated (see Theorem 7.2.7 of arxiv.org/abs/math/0610009v4 ). So I would like to raise the bar by asking for an example of non-saturated cofibration category. An answer to this question would also have a better chance of fulfilling the criterion of "coming up in applications". $\endgroup$ Feb 6, 2013 at 6:03
  • $\begingroup$ There is an interesting discussion about the two out of six property in "Homotopy Limit Functors on Model Categories and Homotopical Categories" by Dwyer, Hischhorn, Kan and Smith (Mathematical Surveys and Monographs 113 American Mathematical Society). $\endgroup$ Sep 27, 2016 at 10:14

4 Answers 4

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I'm not sure that any examples naturally come up, of cases where you have the 2 out of 3 condition but not the 2 out of 6. Of course, if membership in W is defined by requiring certain functors to take a morphism to isomorphisms (as is so often the case in applications), then you always have 2 out of 6 (because a morphism that has both a left inverse and a right inverse always has an inverse).

In Quillen's model category axioms 2 out of 3 is an axiom and 2 out of 6 follows from this and the other axioms.

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  • $\begingroup$ Thanks Tom. I was wondering if that was the case. It seems to imply that "2 implies 6" is the right property for doing homotopy theory, and "2 implies 3" should be de-emphasized. $\endgroup$ Feb 6, 2013 at 19:52
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Here's a rather tautological example. Consider the category $$X\rightarrow Y\rightarrow Z\rightarrow A.$$ That is, $X$, $Y$, $Z$, and $A$ are the only objects, and the only morphisms are those appearing in the diagram (and their composites). Then let $W$ consist of the identity maps, the map $X\to Z$, and the map $Y\to A$. Then this satisfies 2 out of 3 but not 2 out of 6.

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    $\begingroup$ Thanks for this answer, and I should have been clearer in what I'm asking for. By "basic", I meant something like most important example which comes up in applications. $\endgroup$ Feb 5, 2013 at 17:50
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    $\begingroup$ This example comes up in all applications, for any other example receives functors from this one. $\endgroup$ Feb 6, 2013 at 7:55
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Here is an example, which I got from a discussion with Karol Szumilo (and maybe he got it from Cisinski?).

Consider the notion of a cofibration category, which means essentially that you have weak equivalences (with 2 out of 3), cofibrations (closed under pushouts and the same is true for trivial cofibrations) and you can factor everything into a cofibration and a weak equivalence. According to the Radulescu-Banu article https://arxiv.org/pdf/math/0610009v4.pdf (Theorem 7.2.7) we have the following theorem:

A cofibration category is saturated iff weak equivalences fulfill 2 out of 6 iff they are closed under retracts.

Thus, we have just to find a cofibration category where weak equivalences are not closed under retracts. Let $R$ be a ring such that its reduced $K_0$ is non-trivial (i.e. there are projective modules that are not stably free). Consider first the cofibration category $Ch_R$ of bounded chain complexes, where weak equivalences are quasi-isomorphisms and the cofibrations $\mathcal{C}$ are injections with levelwise projective cokernel. Its homotopy category $\mathrm{Ho}(Ch_R)$ is the usual bounded derived category of $R$. This is a triangulated category and its $K_0$ agrees with the $K_0$ of the ring (where we define $K_0$ of a triangulated category as the free abelian group on all isomorphism classes of objects modulo the relation that $[X]+[Z] = [Y]$ if there is a triangle $$X \to Y \to Z \to X[1].)$$ We define its reduced K-theory $\tilde{K}_0$ by taking the quotient by the subgroup generated by $R$ itself concentrated in degree $0$.

We want to localize $Ch_R$ at the class of morphisms $f \in \mathcal{W}$ such that the cone $C(f)$ is zero in $\tilde{K}_0$. First we show that $\mathcal{W}$ satisfies $2$ out of $3$: Let $f$ and $g$ be composable morphisms. Then we get by the octahedral axiom a triangle in $\mathrm{Ho}(Ch_R)$ of the form $$C(f) \to C(gf) \to C(g) \to C(f)[1].$$ By the defining relation of K-theory, the third of these cones is zero in $\tilde{K}_0$ if the other two are.

To show now that $(\mathcal{W},\mathcal{C})$ is a cofibration category, we just have to show $\mathcal{W}$ is closed under homotopy pushouts in $Ch_R$. This is clear as the homotopy pushout of a map $f$ has an equivalent cone to $C(f)$.

Last we have to show that $\mathcal{W}$ is not closed under retracts. But this is clear: Just take a projective $R$-module $P$ that is nonzero in $\tilde{K}_0$ and write it as a retract of a free $R$-module $R^n$. Then $0 \to R^n$ is in $\mathcal{W}$, but $0\to P$ is not in $\mathcal{W}$ (where we view this maps as maps of chain complexes concentrated in degree $0$).


Upshot: Most of the time $2$ out of $6$ is fine, but there are some natural examples where one only has $2$ out of $3$. This should be reason enough to set up the theory with $2$ out of $3$ and only later introduced $2$ out of $6$.

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Edit: This answer only makes sense if the actual question was about morphism $X \to A$ and not $X \to Z$.

Couldn't you just pick a category with two objects $x, y$ with only one invertible morphism $f$ between the two of them? Then consider $$ x \to x \to y \to x $$ with the morphism $x \to x$ being the identity. If we let $W$ be the subcategory consisting of $x, y$ but not the morphism between them, then it satisfies your condition that $X \to Y$ and $Y \to A$ (both of which are just the identity $x \to x$), but not that $x \to y$ or $y \to x$ are in $W$.

In this case, $W$ satisfies the 2-implies-3, since if the diagrams $$ X \to Y \to Z $$ have 2 morphisms that are in $W$, then they must be of the form $x \to x \to x$ or $y \to y \to y$, and so it satisfies the 2-implies-3 condition.

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