Skip to main content
6 events
when toggle format what by license comment
May 23 at 11:09 vote accept Michael Albanese
May 22 at 21:16 comment added Tom Goodwillie Here's a simplicial construction. Let $S^n$ be a simplicial model for a based $n$-sphere, such as $\Delta[n]/\partial\Delta^n$. Let $S^n[G]$ be the simplicial abelian group obtained from $S^n$ by applying the functor from based sets to abelian groups (left adjoint to the forgetful functor). By Dold-Kan, the homotopy groups of (the realization of) $S^n[G]$ are the reduced homology groups of $S^n$ with coefficients in $G$; this CW complex is a $K(G,n)$. If $G$ is countable, then in each simplicial degree this simplicial abelian group is countable, so that the CW complex has countably many cells.
May 22 at 20:39 comment added Michael Albanese Thanks. Is it clear that there is a countable CW complex which is a model for $K(G, n)$? I guess it follows from here, but I was hoping for something more direct. My idea was to take a presentation for $G$, build the analogue of a presentation complex using $n$ and $(n+1)$-cells, then kill higher homotopy groups by attaching more cells. However, its not clear to me if this will only require countably many cells. It does if $G$ is finitely presented (the first CW complex will be finite and hence have finitely generated homotopy groups by Serre).
May 22 at 16:11 comment added Tom Goodwillie @MichaelAlbanese $M(k+1)=M(k+1)\times 0$ is contained in the boundary of $M(k+1)\times [0,1]$, so replace $M(k+1)$ by $M(k+1)\times [0,1]$.
May 22 at 13:16 comment added Michael Albanese Given a finite CW complex, I know you can embed it into Euclidean space of large enough dimension, and then take a thickening to obtain a manifold with boundary with the same homotopy type. From this perspective, I can see how we can arrange for $M(k)$ to be embedded in $M(k+1)$, but I don't see how we can arrange for $M(k)$ to be in the boundary of $M(k+1)$. Can this be seen from this thickening idea or should I be constructing these manifolds differently?
May 22 at 11:54 history answered Tom Goodwillie CC BY-SA 4.0