I found this question in one of the exams given for a topology course and I couldn't get anything out of it; it just seemed overwhelming as a question, but maybe I'm missing something.

Let us define the cone on any topological space $Y$ as $c(Y) = Y \times [0,1]$ where $Y \times \{0\}$ is shrunk to one point. Prove that for any compact, connected surface without boundary with Euler characteristic less than $1$ its cone is not homeomorphic to $c(S^2)$.

The only thing I could do with it was to get it down to this (which is the same thing only with the characteristic requirement made explicit): $\forall n > 1$ $c(U_n)$ is not homeomorphic to $c(S^2)$ and $\forall n \geq 1$ $c(V_n)$ is not homeomorphic to $c(S^2)$. Where $U_n$ and $V_n$ are respectively standard non-orientable and orientable surfaces.

Actually, something I just came up with is that they can't be homeomorphic because $S^2$ is not homeomorphic to any of those surfaces, thus every fiber of those cones isn't homeomorphic to any fiber of the cone on $S^2$; but I don't think that's enough, am I right?

EDIT: the bounty is both for an answer which is different from mine, since I think there might be a "easier" approach to the question, and for an answer that might explain what (if there is) is wrong with my own approach.

  • $\begingroup$ I took the liberty of adding the algebraic topology tag as the cone construction has its uses mostly there. Also: wouldn't "cones on surfaces" be a much more descriptive title? $\endgroup$ – t.b. Jun 20 '11 at 22:51
  • $\begingroup$ I agree and I'll change it. I gave that title because the definition given in the problem was general :) $\endgroup$ – Andy Jun 20 '11 at 22:55
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    $\begingroup$ Did you study homology or the fundamental group in your course? Notice that $C(Y)$ as a space has at least point such that if you remove it, the resulting space is homeomorphic to $Y \times (0,1]$. Perhaps try to prove the "at least one" can be replaced by "at most one". Notice this "at most one" statement does not hold when $Y$ is a sphere. $\endgroup$ – Ryan Budney Jun 20 '11 at 23:00
  • $\begingroup$ @Ryan Budney: would you care to take a look at the answer I posted regarding the proof of the statement? I got the idea from your comment. $\endgroup$ – Andy Jul 14 '11 at 15:35

Thanks to Ryan's comment I was able to do the following, I would appreciate it if anyone pointed out any mistakes or gaps in this proof:

let $\pi(Y): (Y\times [0,1]) \to c(Y)$ the quotient projection; if we remove the point $Y\times \{0\}$ (which we'll call the vertex $v$ of the cone), we have that $c(Y)\setminus \{v\}$ is homeomorphic to $Y \times (0,1]$, since $\pi$ is an open bijection.

$Y \times (0,1]$ retracts on $Y \times \{1\}$ (this is actually a deformation retract), since $F((y,t),s) = (y,1-s+ts)$ is a deformation. Moreover, we have that $Y \times \{1\}$ is homeomorphic to $Y$. Thus we have that $c(Y) \setminus v$ is homotopy equivalent to $Y$, so also the Euler characteristic of the cone minus the vertex is at most 1.

Let's suppose that $c(Y)$ and $c(S^2)$ are homeomorphic and let $g$ be that homeomophism; we also have that $c(Y) \setminus v$ is homeomorphic to $c(S^2)\setminus f(v)$. But $c(S^2) \cong D^3$ ($\cong$ means "is homeomorphic to"), because $g: c(S^2) \to D^2$ such that $g(x,t) = tx$ with $t \in [0,1]$ and $x \in S^2$ is constant on the fibers of the quotient projection, which means that there exists $h: c(S^2) \to D^3$ which is continuous. This is also a bijection and sinche $c(S^2)$ is compact and $D^3$ is T2, $h$ is an homoeomorphism.

At this point we have that $c(Y) \setminus v \cong D^3 \setminus h(g(v))$. If we remove a point from $D^3$, on the other hand, we get that it retracts on its border sphere, which has characteristic $2$, which is absurd.

The case that $h(g(v))$ is on the border of $D^3$ can't present, since there exists a retraction $r: c(Y) \setminus v \to Y \times 1$, which implies the existence of the retraction $hgrg^{-1}h^{-1}: D^3 \setminus h(g(v)) \to h(g(Y \times 1))$, but $h(g(Y \times 1)) \cong Y$, which has non-trivial fundamental group; this is absurd, since the inclusion would induce an injective homomorphism between a trivial group and a non-trivial one ($D^3$ minus a point is contractible).

This concludes the proof.

  • $\begingroup$ This is actually the first time I answer my own question, so I don't really know the policy regarding accepting it as the answer. I would really appreciate it if anyone could tell me if there are any errors here. :) $\endgroup$ – Andy Jul 20 '11 at 9:46
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    $\begingroup$ I think the proof is correct, you can make it more readable by dropping maps whenever possible. You've come to the conclusion that $Y\simeq Y\times (0,1]\approx D^3\setminus\lbrace*\rbrace$ where $*$ is a point in the disk, $\simeq$ means homotopy equivalent, and $\approx$ means homeomorphic, and $D^3\setminus\lbrace*\rbrace\simeq$ either $\lbrace \mathrm{pt}\rbrace$ if $*\in\mathbb{S}^2$, or $\simeq\mathbb{S}^2$ if $*\in D^3\setminus\mathbb{S}^2.$ $\endgroup$ – Olivier Bégassat Jul 25 '11 at 3:54
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    $\begingroup$ Thus your surface $Y$ is either homotopy equivalent to the $2$ sphere $\mathbb{S}^2$ and has Euler characteristic $2$, which goes against the hypothesis you've made on $Y$, or it is contractible, which no borderless compact surface is. $\endgroup$ – Olivier Bégassat Jul 25 '11 at 3:54

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