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

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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You mean $X\to Z$, not $X\to A$. – Eric Wofsey Feb 5 '13 at 16:42
Oh, if that's the case then that completely negates my answer. – Simon Rose Feb 5 '13 at 16:46
@Eric, thanks for catching that. – Jesse Wolfson Feb 5 '13 at 17:48
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. – Omar Antolín-Camarena Feb 6 '13 at 3:48
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". – Karol Szumiło Feb 6 '13 at 6:03

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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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. – Jesse Wolfson Feb 6 '13 at 19:52

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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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. – Jesse Wolfson Feb 5 '13 at 17:50
This example comes up in all applications, for any other example receives functors from this one. – Fernando Muro Feb 6 '13 at 7:55

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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