I am interested in the following variant of the usual Isotopy Extension Theorem:


$\textbf{Question}:$Let $K$ be a graph (1-complex) embedded on a surface $S$, and $i$ be an isotopy of $K$. Does $i$ extend to an isotopy of $S$ (generally called an ambient isotopy or a diffeotopy) ? If $i$ fixes a set of points $V\subseteq K$, does it extend to an ambient isotopy also fixing $V$ ?


$\textbf{Background}$: If $i$ is smooth and $K$ is a submanifold, it is the Isotopy Extension theorem one can find in the literature (for example in Hirsch), which is proved by integrating a vector field. I think that the same argument can be applied in the case of a 1-complex instead of a submanifold since one can still define a tubular neighborhood in this case, but it is a bit far from my field so I am not sure. 


I don't know how to get rid of the "smooth" hypothesis. On surfaces, classical results by Munkres (every homeomorphism is homotopic to a diffeomorphism) and Boldsen (or probably others before) (homotopic diffeomorphisms are smoothly isotopic) allow us to suppose everything about ambient isotopies is smooth. But here I start with a non ambient one and try to extend it, so I don't see how to apply them.


Any help would be appreciated.


P.s :This is a crosspost from math.se :<https://math.stackexchange.com/questions/187453/variants-of-isotopy-extensions>, where I did not get any answer.



Extensions of non-smooth isotopies of not-submanifolds on surfaces
The answer is yes, see the appendix of


<http://www.math.jussieu.fr/~lerouxf/RECHERCHE/TEXTES/0-LE%20ROUX-These-97.pdf>


This reference contains statement and proofs in the case of the 2-sphere, see Theorem A.3.1.
The proof can be adapted for other surfaces.


Frederic


