# How to show that S^1 is not contractible just by definition without going into Fundamental group.

Suppose $$X$$ and $$M$$ be separable metric spaces, these are some of the definitions:

• Two maps $$f_0:X\rightarrow M$$ and $$f_1:X\rightarrow M$$ are said to be homotopic in M if there is a continuous mapping $$F: X×[0,1] \rightarrow M$$ such that $$F(x,0)=f_0 (x) \text{ and } F(x,1)=f_1 (x) \forall x\in X$$.

• A subset $$X$$ of $$M$$ is said to be deformable into a subset $$Y$$ of $$M$$ if there exists a continuous map $$f:X\rightarrow Y$$ which is homotopic to $$Id _X$$, the identity function on X

• A subset $$X$$ of $$M$$ is said to be contractible if it is deformable into a singleton subset of $$M$$ i.e. if a constant map on $$X$$ in $$M$$ is homotopic to identity map on $$X$$

How can I show that the space $$S^1=\{(x,y)\in\mathbb{R}^2|x^2+y^2=1\}$$ under usual topology is not contractible.

I don't want to go into Fundamental Group stuff since I am studying Lusternik Schnirelmann category initial papers which has no mention of Fundamental Group. Here is the link of the paper

This is what I started with. For sure, it can be proved by assuming the contrary. How can I complete this proof?

$$(1,0)\in S^1$$. Without loss of generality, suppose $$S^1$$ is contractible into {(1,0)} i.e there exists a homotopy $$F:S^1×[0,1] \rightarrow S^1$$ such that $$F(x,0)=(1,0)$$ (constant map) and $$F(x,1)=x$$, $$\forall x\in S^1$$(identity map on $$S^1$$),

I have to show some contradiction, how do I continue ?

• You could do it via homology, but that is more work then just computing the fundamental group of $S^1$. Just because a paper does not make explicit mention of a construction does not mean it is not the best way to prove background facts that the articles assumes. – Mees de Vries Nov 5 '18 at 11:26
• True. I suppose everyone agrees with you – Vinay Sipani Nov 5 '18 at 11:34
• But I am not allowed to use it unless I can justify other way of proof completely at this point. – Vinay Sipani Nov 5 '18 at 11:36
• Another approach, the most direct one I know, goes via any number of proofs of Brouwer's fixed-point theorem, which reduce to showing that a disk does not retract onto its boundary. For instance, this can be done with reasonably elementary differential topology. If the circle were contractible, then such a retraction would exist. Another option is to use a groupoid version of van Kampen's theorem, which avoids all the covering space business. If your goal is, however, to just directly manipulate your assumption and get the proof, then I think you will be disappointed. – Kevin Carlson Nov 5 '18 at 16:42
• By the way, I was very skeptical of your claim that the article linked "has no mention of Fundamental Group". In general, any article on homotopy theory and algebraic topology is very highly likely to have implicit connections with fundamental group, if not explicit. So I downloaded the article and did a simple text search. The fundamental group is explicitly mentioned on pages 344, 348, 350, 355, 362, and 363, and there is an entire section regarding relations between category and fundamental group starting on page 353. – Lee Mosher Nov 5 '18 at 18:29