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:


$F: X\times I \rightarrow X$

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

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...

  • $\begingroup$ @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. $\endgroup$ – Niloo May 17 at 14:09
  • $\begingroup$ I did not downvote this post, name. $\endgroup$ – 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 $A$. ]

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

| cite | improve this answer | |
  • $\begingroup$ Thanks for your rewriting, It organized the question better. $\endgroup$ – Niloo Jun 12 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.