Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a topological manifold and $N$ is a subset of $M$. Let $f_t$ be an isotopy from $id $ to $f$ which is a homeomorphism on $M$. Suppose $f(N)=N$ and there is an isotopy $g_t$ on $N$ such that $g_0=id|_N$ and $g_1=f|_N$. Assuming $f_t$ and $g_t$ satisfy the following properties.

  1. For any $x\in N$, $\gamma_x: [0,1] \to M$ defined by $$\gamma_x(t):=\begin{cases} f_{2t}(x) & \text{if } t\in [0,1/2]\\ g_{2-2t}( x)&\text{if } t\in[1/2,1] \end{cases}$$

is a trivial loop in $\pi_1(M,x)$.

  1. For any $x,y\in N,\;x\neq y,$ $$\beta_x:=\{(\gamma_x(t),t)\in M\times S^1\,|\,t\in [0,1]\}$$ is a trivial loop in $\pi_1(M\times S^1-\beta_y,(x,0))$. ($\beta_y$ is defined similarly.)

Then I want to prove:

There is an extension of $g_t$ on $M$. Precisely, there is an isotopy $ \hat g_t $ from $id$ to $f$ such that when restricted in $N$, $\hat g_t|_N=g_t$.

I am not sure the conditions are sufficient to prove the conclusion. So proof or counter example are both welcome.

Thanks a lot.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.