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In the book 'Homotopical Topology', exercise 4, chapter 2 asks:

Prove that no closed (that is, without holes) classical surface except $S^2$ is homeomorphic to a suspension over any other space.

It looks like, we have to show any genus g surface (for g > 0) cannot be written as a suspension of some space. All we know at this point in the book is the definition of a suspension. No invariants (fundamental groups, homology groups, ...) have been introduced yet.

I don't know how to approach this question. A hint will be very helpful.

Isomorphism
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  • Normally closed means compact and without boundary. Hint for the problem: compute $\pi_1$. – Tyrone Jun 26 '20 at 15:52
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    If $T$ is a topological space, then small neighborhoods of each endpoint of the suspension of $T$ are homeomorphic to a cone over $T$. Since surfaces are manifolds, small neighborhoods of points are homeomorphic to disks. So if a surface is a suspension, it must be the suspension over a circle. – Neal Jun 26 '20 at 15:52
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    @Tyrone this question appears before $\pi_1$ is introduced – Osama Ghani Jun 26 '20 at 15:53
  • @NoelLundström the question says fundamental groups and homology groups have not been introduced yet – Osama Ghani Jun 26 '20 at 15:54
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    @Neal this assumes that the only think that disks are cones over are circles (which is not true). Disks can also be considered as cones over $[0,1]$ so it could be a suspension over an interval, which yields a disk again. I don't personally know if there are other spaces over which disks are cones. Although I suppose these would be the only spaces to consider since we want a cone over a 1-dimensional compact manifold, of which these are the only two. – Osama Ghani Jun 26 '20 at 15:58
  • Maybe you should include a definition of what a 'classical surface' is. We have no idea what tools are supposed to be available. – Tyrone Jun 26 '20 at 16:07
  • @Tyrone: Classical surfaces are not defined but the polygonal representation of surfaces has been discussed. – Isomorphism Jun 26 '20 at 16:14
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    I highly doubt there is any proof you can reasonably be expected to come up with at this stage of the book. Probably you can make a geometric argument along the lines that Neal is suggesting but the details are very nontrivial--for instance, it is far from obvious that the cone on $T$ actually must be homeomorphic to a disk (a priori, the cones over $T$ could be weird sets inside of disks that form a different neighborhood base). – Eric Wofsey Jun 26 '20 at 17:10

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