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

Let $P$ be a small category, and let $F:P\to M$ be a diagram in a left-proper combinatorial model category $M$. We say that $F$ is a diagram of cofibrations if for every object $p\in P$, $F(p)$ is cofibrant, and for any $f:p\to p'$ in $\mathrm{Arr}(P)$, $F(f)$ is a cofibration.

Recall that a diagram $F: P\to M$ is called projectively cofibrant if the map $\emptyset \to F$ in $M^P$ has the left lifting property with respect to the class of morphisms that are objectwise trivial fibrations.

An easy example of when this holds is a for spans. Such diagrams are projectively cofibrant if and only if they are diagrams of cofibrations. Another example of such a diagram shape is the category of natural numbers (viewed as an ordered set).

Are there any known rules of thumb for when we can say that a diagram of cofibrations is projectively cofibrant?

If no general result exists, I'm specifically interested in the case where $M$ is the category of simplicial sets and $P$ is a poset.

Edit: Since the original question was answered in the comments, I'd like to add the conditions that the poset $P$ is finite and Thomason-contractible (that is, its nerve is a weakly contractible simplicial set).

share|improve this question
3  
It's very rarely true. Note that if this is true for $P$ then it is also true for any subposet $Q\subset P$ such that no element of $Q$ is less than any element of $P-Q$. It also appears to me that, in order for it to hold for $P$, the realization of the nerve of $P$ must be homotopy equivalent to a discrete space. –  Tom Goodwillie Jul 18 '11 at 14:03
4  
If $P$ is a direct category, then projectively cofibrant diagrams coincide with Reedy cofibrant diagrams. It is easily seen that only very uncomplicated direct categories will satisfy your property. For example a constant diagram is always a "diagram of cofibrations", but constant diagrams are Reedy cofibrant only when $P$ is a coproduct of categories with initial objects. The situation is probably only worse for non-direct categories. –  Karol Szumiło Jul 18 '11 at 14:30
1  
Does "Thomason-contractible" mean "with a contractible nerve"? If so this is not sufficient and my previous comment explains why. There are contractible direct categories with no initial object, for example the direct part of $\Delta$ (i.e. the category of finite nonempty totally ordered sets and injective order-preserving maps). There are also examples among posets. –  Karol Szumiło Jul 18 '11 at 18:19
2  
This will fail when $P$ is the product of two copies of $\{0\to 1\}$, so that to be a "finite Thomason-contractible poset" won't be sufficient. I think the case of spans is quite representative: you will need $P$ to be a directed category which is free (i.e. isomorphic to the free category on a graph). Note that, for a directed category, to be free is a very big constraint; this implies, for instance, that for any model category $M$, the functor $Ho(M^P)\to Ho(M)^P$ is full and essentially surjective. –  Denis-Charles Cisinski Jul 18 '11 at 22:40
2  
This condition on a poset $P$ (every "diagram of cofibrations" is a cofibrant diagram) implies that for any element of $P$ the set of all lower bounds is totally ordered. –  Tom Goodwillie Jul 19 '11 at 2:40
show 2 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.