# How to prove this isotopy exist?

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.

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