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(...) if two topological objects have different homotopy groups, they can't have the same topological structure—a fact that may be difficult to prove using only topological means. For example, the torus is different from the sphere: the torus has a "hole"; the sphere doesn't.

From the wikipedia article. The homotopy group proof is indeed pretty simple. The article claims it's "difficult" to prove using only topological means. How can this be proved using topological means (without homotopy theory etc.)?

Carla is my name
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  • One could use Euler characteristic-like arguments. – Randall May 18 '21 at 13:00
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    The torus $\mathbb{T}^2$ is not simply connected, whereas the sphere $\mathbb{S}^2$ is. – TheSilverDoe May 18 '21 at 13:01
  • You can refer to this link : https://math.stackexchange.com/questions/847929/why-sphere-and-torus-are-not-homeomorphic – p_square May 18 '21 at 13:02
  • See here for a proof https://math.stackexchange.com/questions/3431989/homeomorphism-between-s1%c3%97s1-and-s2/3432013#3432013 – Sumanta May 18 '21 at 13:06
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    @TheSilverDoe: "simply connected" is a homotopy concept. The OP is wondering about a proof that doesn't use homotopy. – Troposphere May 18 '21 at 13:09
  • @Troposphere Sorry, I read "homology" for a reason :) – TheSilverDoe May 18 '21 at 13:22
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    If you want the best proof that uses topological means, it might help to understand that "homotopy theory" is a topological theory. If you took the standard proof that uses "homotopy theory", and replaced all occurrences of "homotopy" concepts with their topological definitions, then you would have a very good topological proof (except it would be expressed in more words due to avoiding the "h" word). – Lee Mosher May 18 '21 at 13:27
  • A cleaner statement would be "without using tools of algebraic topology" rather than "only topological means." – Moishe Kohan May 18 '21 at 13:28
  • @Randall: You are leaving aside a proof of topological invariance of the Euler characteristic. The usual proof goes via homology theory. There is an alternative argument, but it is quite hard, requires a proof of uniqueness of PL structure on surfaces. – Moishe Kohan May 18 '21 at 13:31

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Use the Jordan curve theorem, if there was a homeomorphism $\phi: S^1 \times S^1 \rightarrow S^2$ the curve $\gamma:S^1 \rightarrow S^1 \times S^1$ given by $\gamma(s) = (1,s)$ would get taken to a Jordan curve in $S^2$. The Jordan curve theorem says that the complement of the image of $\phi \circ \gamma$ is not connected however the complement of the image of $\gamma$ is clearly connected.

Thus $S^2$ and $S^1 \times S^1$ are not homeomorphic.

Noel Lundström
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    It's all correct but Jordan curve theorem is a hard result: Much harder result than proving nontriviality of the fundamental group of the torus and triviality of the fundamental group of the sphere. – Moishe Kohan May 18 '21 at 13:26
  • @MoisheKohan Right, but that's not what he asked for... so I fail to see what your point is – Noel Lundström May 18 '21 at 13:49
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    Admittedly, the question in OP is very sloppy, but it is asking for a "topological proof," which presumably means a proof which does not use any tools of algebraic topology. For completeness of the answer, you should either provide a proof of Jordan Curve Theorem which avoids such tools or give a reference to such a proof. This is the point of my comment. – Moishe Kohan May 18 '21 at 15:37