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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.

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    $\begingroup$ "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... $\endgroup$ Apr 16, 2015 at 11:57
  • $\begingroup$ 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... $\endgroup$
    – Ripan Saha
    Apr 16, 2015 at 12:01
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    $\begingroup$ 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...?) $\endgroup$ Apr 16, 2015 at 13:30
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    $\begingroup$ 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.) $\endgroup$
    – user98602
    Apr 16, 2015 at 15:46
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    $\begingroup$ 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). $\endgroup$ Apr 16, 2015 at 16:24

2 Answers 2

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Following is taken from Fixed point theory - Dugundji, p. 95.

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  • $\begingroup$ Brouwer's theorem can be proved with Sperner's lemma. $\endgroup$
    – Hulkster
    Feb 22, 2017 at 11:16
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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.

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