acknowledge the existence of manifolds without ball decompositions
authorKevin Walker <kevin@canyon23.net>
Fri, 06 May 2011 17:46:07 -0700
changeset 766 823999dd14fd
parent 765 c7c4c0d0e240
child 767 f7265c20425b
acknowledge the existence of manifolds without ball decompositions
text/a_inf_blob.tex
text/ncat.tex
--- a/text/a_inf_blob.tex	Fri May 06 17:24:08 2011 -0700
+++ b/text/a_inf_blob.tex	Fri May 06 17:46:07 2011 -0700
@@ -38,7 +38,8 @@
 
 
 \begin{thm} \label{thm:product}
-Let $Y$ be a $k$-manifold.
+Let $Y$ be a $k$-manifold which admits a ball decomposition
+(e.g.\ any triangulable manifold).
 Then there is a homotopy equivalence between ``old-fashioned" (blob diagrams) 
 and ``new-fangled" (hocolimit) blob complexes
 \[
--- a/text/ncat.tex	Fri May 06 17:24:08 2011 -0700
+++ b/text/ncat.tex	Fri May 06 17:46:07 2011 -0700
@@ -1025,6 +1025,10 @@
 Roughly, a permissible decomposition is like a ball decomposition where we don't care in which order the balls
 are glued up to yield $W$, so long as there is some (non-pathological) way to glue them.
 
+(Every smooth or PL manifold has a ball decomposition, but certain topological manifolds (e.g.\ non-smoothable
+topological 4-manifolds) do nat have ball decompositions.
+For such manifolds we have only the empty colimit.) 
+
 Given permissible decompositions $x = \{X_a\}$ and $y = \{Y_b\}$ of $W$, we say that $x$ is a refinement
 of $y$, or write $x \le y$, if there is a ball decomposition $\du_a X_a = M_0\to\cdots\to M_m = W$
 with $\du_b Y_b = M_i$ for some $i$,