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

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$$

• @amWhy I wanted to merge them in one post but thy are different! In one I need to show those function are homotopic in another I need to show those function are not homotoic in another situation and if I know one answer it does not help me find the other problem answer and the proof technicality are separate too. If you convinced please take your minus vote back.Thank you. – Niloo May 17 at 14:09
• I did not downvote this post, name. – amWhy May 17 at 14:10

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 $$A$$. ]

$$\newcommand{\new}{\color{#1}}$$ 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).)

• Thanks for your rewriting, It organized the question better. – Niloo Jun 12 at 11:04