Hatcher 0.15

Hatcher, Algebraic Topology, Chapter 0

15. Enumerate all the subcomplexes of S^\infty, with the cell structure on S^\infty that has S^n as its n-skeleton.

Proof. Recall that by definition, S^\infty = \cup_{n=0}^\infty S^n (see page 7 in Chapter 0). Moreover, recall that for n finite, S^n can be given a cell structure in which each of the spheres S^k, k\leq n, is a subcomplex: This is done by way of attaching two k-cells to the S^{k-1} considered as the equator as S^k for all k\leq n. More precisely:

Begin with S^0 as two points e^0_1=x_1 and e^0_2=x_2; obtain S^1 by attaching two “arcs” (i.e., closed segments homeomorphic to the 1-dimensional “disk”) e^1_1 and e^1_2 to create a circle; next, obtain S^2 by attaching two 2-cells (i.e., closed regions homeomorphic to the 2-dimensional disk) e^2_1 and e^2_2. Continue inductively by attaching two n-cells e^n_1 and e^n_2, each homeomorphic to the n-disk D^n, to the resulting S^{n-1}. This yields a CW-complex structure on each S^n which has two cells in each dimension d=0,1,\ldots,n and which has S^k as a subcomplex for each k\leq n. The construction for S^k, k=0,1,2, is shown in this solution to a previous exercise (see Figure 3 in that solution).

Therefore, we have that S^n = \cup_{k=0}^n(e^k_1\cup e^k_2) and that S^\infty is the union of all the S^n. Also, for any n, a subcomplex X of S^n is automatically closed (by definition of subcomplex) and therefore must contain the boundary \partial e^k_i \cong S^{k-1} of any k-cell e^k_i it contains, i=1,2. What does this mean? It means that any (closed) subcomplex X of S^\infty which contains a single k-cell must also include every single j-cell for j<k. Thus, it follows that the only k-dimensional subcomplexes of S^\infty are the hyper-hemispheres1 of the form e^k_i\cup e^{k-1}_{1,2}\cup\cdots\cup e^1_{1,2}\cup e^0_{1,2} and so enumerating these for all k=0,1,\ldots yields the final result.    \square


    1. Thanks to Tarun Chitra for coming up with a good prose way to describe these objects geometrically. His work can be found here.

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