Let $M, N$ manifolds without bundary and $f: M\to N$ and $g: M\to N$ embeddings. We say that a differentiable map $h: [0,1]\times M\to N$ is an isotopy between $f$ and $g$ if each of the maps $h_t:M\to N$ with $h_t(x) = h(t,x)$ is an embedding and $h_0=f, h_1=g$.
By a diffeotopy of a manifold $N$ we mean a differentiable map $H:[0,1]\times N\to N$ such that $H_0 =\operatorname{Id}$ and each $H_t:N\times N$ is a diffeomorphism, and we say that two $h$-isotopic embeddings $f,g$ are diffeotopic in N if there exists a diffeotopy $H$ of $N$ such that for al $t$ we have that $h_t = H_t\circ h_0$.
My question is
Suppose that we have two diffeotopic embeddings $f$ and $g$ such that the complement of $f(M)$ is connected. Is it true that the complement of $g(M)$ is connected?
I think that the result is intuitively true, but I'm having trouble proving it. Any help would be appreciated.
Doesn't $x\mapsto H(1,x)$ define a homeomorphism between the complement of $f(M)$ and $g(M)$?