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I know the prove $S^n$ is not contractible using homology.But I don't know how to prove it from definition of contractibility.Can anyone help me in this direction? Thanks.

Ripan Saha
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    "But I don't know how to prove it from definition of contractibility." It's not something that follows immediately from the definition of contractibility. There are direct algebraic topology proofs without using homology, but a proof for $S^2$ provided here is already maddeningly huge. Alternatively, you can look for proofs of Brouwer fixed point theorem (which implies contractibility of $S^n$) using differential topology... – Balarka Sen Apr 16 '15 at 11:57
  • Okkk...I was trying to it prove this from definition,,That is why I wrote "But I don't know how to prove it from definition of contractibility.",,anyway thank you for suggesting me the link... – Ripan Saha Apr 16 '15 at 12:01
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    Ever since the beginning of time, humanity has yearned to have the answer to the question: "is this space contractible?". Algebraic topology is designed around this :) Why do you want to do this without homology? Do you have other tools at your disposal? (Homotopy groups, something else...?) – Najib Idrissi Apr 16 '15 at 13:30
  • @NajibIdrissi are you sure " Algebraic topology is designed around this"?...your comment is not relevant to my question...if you have some answer then you are welcome to share...otherwise plz don't make such comment...I guess contractibility is a homotopic property...So there should be some answer to this question in homotopic aspect...Isn't it? – Ripan Saha Apr 16 '15 at 15:37
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    Yes, algebraic topology was (in many ways) designed to study homotopy theory; whether a space is contractible or not is part of that. There are certainly ways to show that $S^n$ is not contractible without using homology, but you're not going to be able to do it without some sort of machinery. (Smooth manifold theory is once nice way to do it; you can define the degree of a map without mention of homology; you then need to show that it's a homotopy invariant.) –  Apr 16 '15 at 15:46
  • @BalarkaSen: That homework set is essentially a long joke; spending an afternoon building a minigun so that one can kill a fly. It's not intended as a serious proof that $S^2$ is contractible. –  Apr 16 '15 at 15:50
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    Maybe you noticed that I asked some questions at the end of my comment, to try to clarify your question and understand what was the best help I could give you? Anyway, I wish you the best of luck in your endeavors, as I don't really care about this question anymore. (And yes, I'm sure one of the main goals of algebraic topology is recognizing contractible spaces). – Najib Idrissi Apr 16 '15 at 16:24
  • It follows from Borsuk's degree theorem and Tietze's theorem. – Hulkster Feb 21 '17 at 18:01
  • There's also homotopy: $\pi_n(S^n) \simeq \mathbb{Z}$ so $S^n$ can't be contractible – Maxime Ramzi Jun 21 '18 at 09:51

2 Answers2

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Though not as easy as homological proofs, there is a known purely analytical proof that the sphere is not contractible. It follows from a purely analytical proof of Brouwer's fixed point theorem due to Milnor here.

Once Brouwer's fixed point theorem is proved, it follows that the sphere is not a retract of the closed ball and this means that the sphere can not be contractible.

  • If the sphere were a retract of the ball, then the retract followed by the antipodal map on the sphere would be a fixed point free mapping of the ball into itself - contradicting Brouwer's fixed point theorem.

  • A contraction may be thought of as a retraction mapping of the cone on a space into the space.

$$ S \rightarrow Cone(S) \rightarrow S$$

The closed ball is homeomorphic to the cone on the sphere.

Kelvin Lois
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Joe S
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Following is taken from Fixed point theory - Dugundji, p. 95.

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Hulkster
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