I'm trying to show that the join $S^n * S^m$ is $S^{n+m+1}$ via seeing them as CW complexes and checking that they have the same cell structure. I'm aware of other proof of this homeomorphism but I want to use cell structures. In particular, following Hatcher's book a cell structure of $S^n * S^m$ is: $S^n$, $S^m$ are subcomplexes and the rest of cells are the ones of $S^n \times S^m \times (0,1)$. I know that in a natural cell decomposition of the product, the cells of $X\times Y$ are the cells $e_X \times e_Y$, with $e_X$ a cell of $X$ and $e_Y$ of $Y$, but in this case, what is a cell structure of $(0,1)$? I thought that as it is homeomorphic to $\mathbf{R}$ and this one has a cell structure consisting of the integers as 0-cells and then the intervals between them as 1-cells the same holds for $(0,1)$ via the inverse homeomorphism taking it to $\mathbf{R}$, but then, solving the above problem seems very hard.

  • $\begingroup$ When one writes $(0,1)$ in this context it is really thought of as the interior of a segment seen as one cell. In a cartesian product, it adds $1$ to the dimension of the cells of the other factor. See Topology II: Homotopy and Homology. Classical Manifolds, by Viro and Fuchs, Section 7.5.D. $\endgroup$ – Arnaud Mortier Feb 22 '18 at 23:41
  • $\begingroup$ Any CW complex has infinitely many cell structures (unless it is zero dimensional). Sentences which start like "the cell structure of $S^n * S^m$" are therefore nonsensical; the best you can do is to say "a cell structure of $S^n * S^m$". So it is quite possible, in fact highly likely, that you have a cell structure on $S^{n+m+1}$ and a cell structure on $S^n * S^m$ that are quite unlike each other, and they are of no help in determining that $S^{n+m+1}$ and $S^n * S^m$ are homeomorphic. $\endgroup$ – Lee Mosher Feb 23 '18 at 0:20
  • $\begingroup$ @LeeMosher you are completely right, I will edit that part of the question. Hopefully, someone will come up with a cell decomposition of one space that agrees with the other one. $\endgroup$ – inquisitor Feb 23 '18 at 0:32

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