That is terrific. I looked up the Whitney result, here: jstor.org/stable/1968197 I see what you mean about recovering the triangulation, but if one hemisphere has a cone point, the chromatic polynomial has the form $x(x-1)(x-2)(x-2)^p(x-3)^q$, not too many of them. My summer computer experiment said this was the most common kind of chromatic polynomial when you chose the two hemispheres randomly.

Hi David, I want (b) or something even stronger, but getting to (a) is already hard for me.

Will that is a very appealing approach.

Piotr if I fix a map $\mathbf{A}^1 \to \mathbf{A}^1$ in the category $C$ (in other words I fix a Laurent series), does its inverse image in $\overline{C}$ have any concrete description?

Thanks Piotr and Will. What are the prospects for making this functorial for C((t))-algebraic maps? It probably doesn't make it any easier, but getting it to work for open smooth varieties is more important to me than getting it to work for singular varieties.

If you draw a picture of building a genus g surface with one boundary component, by attaching handles to a disk "in the standard way", you can just see an immersion to R^2. I looked a bit online for a picture of what I mean, like "pointed matched circles" in this document math.columbia.edu/~lipshitz/CambridgeSlides.pdf , an overhead picture of a highway interchange makes the point too.

I love that article. Nelson is not defining a notion of counting number inside of the peano system, he is extending the peano system by introducing a new predicate C(x), calling it "x is a counting number," and subjecting it to a couple of axioms. But he does not subject C to anything like an induction axiom. Your proposition alpha is also true in this extension of the peano system (call it PA+C?), but you cannot use it to prove that for all x, C(x).

Thanks Will. How is the mass of a spinor genus defined? Should I weight each lattice by the order of a finite subgroup of Spin(24), and if so which subgroup?

Does M_{24} contain a copy of a medium-sized symmetric group, like S_6? Maybe it is less intimidating to compute the composite map H^4(S_{24}) --> H^4(M_{24}) --> H^4(S_6), than either map separately.

The domain is ZxZ/2, with first factor generated by p1 and second factor by the Bockstein of w1w2. If you think the same question for H^{1,2,3} is easy, we can focus on p1. I bet the codomain is unknown, I don't know why. But you can start by restricting the rep. to easy subgroups. In general there's a split injection of H4(G;Z_p) into H4(p-Sylow;Z_p), which is Z/p for p=11,13,23. For p=7 the Sylow is Z/pxZ/p, likely easy. The 2-Sylow is complicated, but it's the same as the 2-Sylow of the normalizer of a Z/2^12-subgroup -- with known H4? I don't know how far this train goes.