(Homotopy theory) When does the 2 of 3 property not imply 2 of 6?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $0$).

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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 $6$.

Answer 4

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.