Prove that it is NOT $F_{0} \simeq_{S_{1} \cup 2 S_{1}} F_{1}$ for $F(x, t)= e^{2\pi t i (|x|-1)} x$

Question

Let $X= \{ x\in R^{2} : 1\leq |x| \leq 2 \}$ be a topological space on $R^{2}$

$A= \partial X$ or $A= S^{1} \cup 2S^{1}$

and Let function $F(x, t) $ be:

I=[0,1]

$F: X\times I \rightarrow X$

$F(x, t)= e^{2\pi t i (|x|-1)} x$

and for each $t \in I$ define $F_{t}(x)$ as follows

$F_{t}: X\rightarrow X$

$F_{t}(x)= F (x, t)$

It is easy to show that F(x,t) is an isomorphism between $F_{0}(x)$ and $F_{1}(x)$

show It is NOt $F_{0}(x) \simeq_{S^{1}\cup 2S^{1}} F_{1}(x)$

Which means $F_{0}$ and $F_{1}$ are not homotopic on A:

$F_{0}(x)=x$ $F_{1}(x)=e^{2\pi i (|x|-1)}x$

I need to show that there is no continous function H(x, t): $X\times I \rightarrow X$ that

$\forall x \in X$

$H(x, 0)=F_{0}(x)$

$H(x, 1)=F_{1}(x)$

$\forall t \in I; \ \forall a \in A$

$H(a, t)=F_{0}(a)=F_{1}(a)$

I tried to show that but I find no contradiction if those two function be homotopic.

I know that $F_{0}(x) \simeq_{S^{1}} F_{1}(x)$

using the H(x, t):=F(x, t) as the homotopy function.

H(x, t): $X\times I \rightarrow X$

$H(x, 0)=F_{0}(x)= x$

$H(x, 1)=F_{1}(x)= e^{2\pi i(|x|-1)}x$

H is continuous and It is suffices to show for every a $\in S^{1}$ and every $t\in I$ : H(x, t)= $1_{X}$ but this is not true for all $t \in I$

x $\in S^{1} \Rightarrow\ $|x|=1 $\Rightarrow$ $H(x, t)=e^{2\pi it(1-1)}x= x$

Any suggestion will be helpful Thanks in advance...

Answer 1

For the benefit of other readers (and for the original poster), I'm going to rewrite this. I'm not going to provide an answer, just a rewrite of the question, because frankly, I think it might well be more useful. [BTW, I think that the claim made in the question is false, which is a another good reason not to provide an answer. Perhaps OP, knowing that the claim might be false, can find the explicit homotopy rel $A$. ]

$$ \newcommand{\new}[1]{\color{#1}} $$ Consider $$ X= \{ x\in R^{2} : 1\leq |x| \leq 2 \} \subset \Bbb R^2$$ with the subspace topology.

Let $A= \partial X$, so that

$$ A= S^{1} \cup 2S^{1}, $$ where $2S^1$ is shorthand for $\{x : |x| = 2 \}$.

Define $$ F: X\times [0, 1] \rightarrow X: (x, t) \mapsto e^{2\pi t i (|x|-1)} x $$ and for each $t \in I = [0,1]$ define $F_{t}(x)$ by $$ F_{t}: X\rightarrow X: x \mapsto F(x, t). $$

It is easy to show that $F$ is a homotopy (in $X$) between $F_{0}$ and $F_{1}$.

Furthermore, for $x \in S^1$, we have \begin{align} F_0(x) &= \exp(2\pi 0 (|x|-1)) x = x \\ F_1(x) &= \exp(2 \pi 1(|x|-1)) x \\ &= \exp(2 \pi 1(1-1)) x \\ &= \exp(0) x = x \end{align} and for $x \in 2S^2$ (i.e., the set of points $x$ with $|x| = 2$), we have \begin{align} F_0(x) &= \exp(2\pi 0 (|x|-1)) x = x \\ F_1(x) &= \exp(2 \pi 1(2-1)) x \\ &= \exp(2 \pi 1(2-1)) x \\ &= \exp(2\pi) x = x \end{align} as well. In other words, the homotopy $F$ fixes the boundary $A$ pointwise for $t = 0$ and $t = 1$.

I want to show that $F_0$ and $F_1$ are not homotopic rel $A$, i.e., that there is no continuous function $H: X \times I \to X$ such that

\begin{align} H(x, 0) &=F_{0}(x) & \text{for $x \in X$}\\ H(x, 1) &=F_{1}(x) & \text{for $x \in X$}\\ H(a, t)&=F_{0}(a)=F_{1}(a) & \text{for ${a \in A, 0 \le t \le 1}$} \end{align}

I tried to show that but I find no contradiction if those two function be homotopic.

(I've deleted the remainder, which merely observed that the restriction of $F$ to $S^1$ is a homotopy from $F_0$ to $F_1$ (restricted to $S^1$), and then that this statement doesn't extend to the rest of $A$ (or something).)