Let $B^2 = \{x\in \Bbb R^2\mid \|x\|\leq 1\}$ be the closed unit disk. Let $r $ be an interior point. I need to show that $B^2\setminus\{r\} $ and $S^1$ have the same homotopy type.
First I defined maps $i:S^1\to B^2\setminus\{r\}, i (x)=x $ and $f:B^2\setminus\{r\}\to S^1, f (x)=\dfrac {x-r}{\| x-r\|} $.
Then I showed that $f\circ i $(which is same as $f $ restricted to $S^1$) is homotopic to the identity map of $S^1$ by defining the homotopy $H (x,t):S^1×[0,1]\to S^1$ by $$H (x,t)=\dfrac {(1-t)x+t f (x)}{\|(1-t)x+t f (x)\|}.$$
To show $i\circ f=f$ is homotopic to the identity map of $B^2\setminus\{r\} $, I defined $G (x,t): B^2\setminus\{r\}×[0,1]\to B^2\setminus\{r\}$ by $$G (x,t)=(1-t)x+t \left (\dfrac {x-r}{\| x-r\|}\right) .$$
Is $G (x,t) $ well-defined?
Since $$\|(1-t)x+tf (x)\|\leq \|(1-t)x\|+\|tf (x)\|\leq (1-t)×1+t=1,$$ $G (x,t) $ is in $B^2$.
Now I need to show that $G (x,t) $ never takes the value $r $. I tried to use a proof by contradiction, but it was taking me nowhere. Vector diagrams give me the impression that my claim is true, but I'm unable to prove this. Can someone help me ?
Any kind of hints or other solutions are appreciated.
Thank you.
If $r=(0, 0)$, then $G(x, t) =(1-t)x + t\frac x{||x||}$ always lies on the line segment between $x$ and $\frac x{||x||} $ and so $r\neq G(x, t)$.
Now assume $r\neq(0,0)$. Suppose on the contrary that $G(x, t)=r$ at some point (x, t), then $x, r, \frac{x-r}{||x-r||} $ are colinear. And so the vector $x-r$ is a multiple of the vector $\frac{x-r}{||x-r||} - r$. But $\frac{x-r}{||x-r||}$ is a multiple of $x-r$, this implies $x=kr$ for some constant $k$, putting this in $G(x, t)$, we get $G(kr, t)=(1-t) kr+\operatorname{sgn}(k-1)t\frac r{||r||}=r $. So if $k>1$, $(1-t)k +\frac t{||r||} >(1-t)+t=1$, a contradiction. And if $k<1$, then $(1-t)k-\frac{t}{||r||}< 1-2t\leq 1$, is also a contradiction.