4
$\begingroup$

Given a diagram $X_1 \rightarrow X_{12} \leftarrow X_2$ of spaces (though I think the question applies more generally), there are two cosimplicial resolutions which I've seen used to compute the homotopy pullback. The first one, which I'll call $\mathcal A$ is given by $$\mathcal A^p = X_1 \times X_{12}^{\times p} \times X_2$$ and the coface maps are like in a bar (cobar?) construction. This is the one you see, for example, in the construction of the Eilenberg-Moore spectral sequence.

For the other one, let's let $\mathscr P$ denote the pullback category, and $F$ our functor. Then the standard cosimplicial replacement, which I'll write $\mathcal B$, is given by $$\mathcal B^p = \prod_{\substack{x_0 \rightarrow \cdots \rightarrow x_p \\ \in \mathscr P}} F(x_p)$$

It seems that I always encounter one or the other, but not both, which leads me to my question: how do I know they give the same answer? The first one computes the homotopy pullback basically by definition, whereas the second one is a bit mysterious to me. Is there a simple way to compare these two cosimplicial spaces? They do not appear to be level-wise equivalent. Can we say anything about the relationship of their $\mathrm{Tot}$ towers?

$\endgroup$

1 Answer 1

5
$\begingroup$

The first one has codegeneracies as well as cofaces, yes? And when you say that it computes the htpy pullback basically by definition I suppose you mean that its Tot is homeomorphic to the space whose points are triples consisting of a point in $X_1$ a point in $X_2$ and a path connecting their images in $X_{12}$? That's how I think of it.

And if you think about Tot of the second one the same way, you should get the space in which a point is a $5$-tuple, one point in each of $X_1$, $X_{12}$, $X_2$ and two paths in $X_{12}$ connecting the middle point with the images of the other two. Which is homeomorphic to the same old thing.

EDIT In reply to Eric's comment: The way I think about it is that holim is the derived functor of lim, therefore well-defined (up to canonical weak eq in some strong sense) when it exists; and then the question is how do you make it? In a simplicial model cat you can always make it by Tot of Bousfield-Kan cosimplicial replacement, I guess, though in general you have to make things fibrant first. For particular indexing categories, there are sometimes even simpler models. For pullbacks, one model is replace the two maps in the diagram by fibrations and then make lim (= pullback). This passes some general test for being a derived functor, so no need for direct comparison with the B-K model. Likewise the cobar construction model should pass that test.

$\endgroup$
2
  • $\begingroup$ Well, right. Except that somehow this explanation is not entirely satisfying to me. Shouldn't you expect a similar result in basically any (simplicial model) category? Spectra, for example? Or is this wrong? If it's true, then it leads me to believe there's some kind of "more general reason" that they are always equivalent. Maybe I am way off base here. $\endgroup$ Oct 6, 2010 at 14:09
  • $\begingroup$ Yes! I think I understand what you are saying better now. The difference between the two can be thought of as just a single subdivision of the path in $X_{12}$. So I wonder if the "canonical equivalence" you mention above can be actually realized as some kind of map induced by this subdivision. I'll give it some thought. Thanks for your help. $\endgroup$ Oct 8, 2010 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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