Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
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
8  
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
4  
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
add comment

3 Answers

up vote 10 down vote accepted

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
2  
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
2  
This example comes up in all applications, for any other example receives functors from this one. –  Fernando Muro Feb 6 '13 at 7:55
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.