MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2

2

Suppose I have a higher category, say a simplicial category $C$, and I want to invert a certain type of morphism $W$. Simplicial localization (for example hammock localization), gives a localized simplicial categetory $W^{-1}C$. Then I truncate this category by passing to $\pi_0$ of hom-sets to obtain the 1-category $\pi_0(W^{-1}C)$.

Conversely, I may pass to the truncated 1-category $\pi_0(C)$, and then localize, to get the 1-category $W^{-1}\pi_0(C)$.

I guess there is a functor $\pi_0(W^{-1}C)\to W^{-1}\pi_0(C)$. Is it an equivalence of categories? I am also interested in 2-category truncations defined by $\Pi_1$ of hom-sets.

If it is not true, are there easy to understand counterexamples?

flag
If you are interested in 2-categories, you might as well consider Pronk's bicategorical localisation from her 1996 paper. I have a potential (counter)example in mind, let me get back to you. – David Roberts Jul 6 2011 at 22:24
5 
Yes, the map $\pi_0(W^{-1}C)\to W^{-1}\pi_0(C)$ is always an equivalence of categories (this follows immediately by comparing the corresponding universal properties). The same remains true (for formal reasons as well) if you look at the truncations of hom-spaces defined by $\pi_1$ and work up to an adequate notion of equivalence of $2$-categories (in fact, you may as well truncate in dimension $n$ and get an equivalence of $(n+1,1)$-categories). – Denis-Charles Cisinski Jul 6 2011 at 22:47
Thanks! Turns out it's so simple. – euklid345 Jul 7 2011 at 9:22

1 Answer

1

This is Denis-Charles Cisinski's answer, given in the comments:

Yes, the map $\pi_0(W^{-1}C)\to W^{-1}\pi_0(C)$ is always an equivalence of categories (this follows immediately by comparing the corresponding universal properties). The same remains true (for formal reasons as well) if you look at the truncations of hom-spaces defined by $\pi_1$ and work up to an adequate notion of equivalence of 2-categories (in fact, you may as well truncate in dimension $n$ and get an equivalence of (n+1,1)-categories).

link|flag
1 
Thanks for posting it as an answer. It's annoying when questions answered in comments keep appearing on the front page and this is a nice way to resolve the issue without anyone gaining unfair reputation. – David White Sep 1 2011 at 17:21

Your Answer

Get an OpenID
or

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