How is that $S^1$ is not contractible?

Question

It is stated in Wikipedia (and other pages too) that the spheres $S^n$ are all not contractible.

Take $n=1$. Would anyone explain to me why $$S^1\times [0,1]\to S^1$$$$(e^{2\pi i t},s)\mapsto e^{2\pi i ts}$$is not an homotopy between the identity and a point?

Hint: what happens in a neighborhood of $1 \in S^1$? – Fimpellizzeri Aug 11 '16 at 00:40 — Fimpellizzeri, Aug 11 '16 at 00:40

score 6 · Accepted Answer · edited Aug 11 '16 at 01:08

6

$$e^{2\pi i}=1,$$ so that $$(e^{2\pi i},s)=(1,s)$$ but $$(e^{2\pi i},s)$$ is mapped to $e^{2\pi is},$ while $$(1,s)$$ (which corresponds to $t=0$) is mapped to $1.$

For $0<s<1$ you have $$ e^{2\pi is}\not=1$$ which shows that your map is not well-defined on the circle. (Not to talk about continuity.)

edited Aug 11 '16 at 01:08

Aweygan

23,232

answered Aug 11 '16 at 00:46

user39082

2,124

Oh. I see. Thank you. – Talexius Aug 11 '16 at 00:56

score 0 · Answer 2 · answered Oct 19 '21 at 14:05

We can easily prove that S¹ in not contractible. Let's recall some definations and results for the proof.

★Simply Connected Space: A path Connected space X is called as simply connected, if every closed curve in X is a null homotopy. After studying Fundamental groups, one can define the term as: "A path Connected space X is called Simply Connected, if it's fundamental group is trivial.

★RESULT: A contractible space is Simply Connected.

Now we are all set to prove that, S¹ is not contractible.

As we stated above,A contractible space X is Simply Connected, i.e. fundamental group defined on contractible space X at any point x is trivial. But we know that fundamental group π1(S¹,z), for any z in S¹, is Homeomorphic to Z(the set of all integers) which can't be trivial. Hence S¹ in not Simply Connected, implies S¹ is not contractible.

How is that $S^1$ is not contractible?

2 Answers2