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 ?