I have the following situation: I have a graph $G$ embedded into $\mathbb{R}^2$, with $(0,0)$ a vertex, and I have a diffeomorphism $g$ of the plane. Let's call $G' = g(G)$ the new graph.
I suppose the for each edge $e$ of $G$, $e$ is embedded in $\mathbb{R}^2$ diffeomorphically.
Moreover, suppose that an edge of $G$ exiting $(0,0)$ is fixed by $g$.
Can I find a new diffeomorphism $g'$ of the plane, such that $g'(G) \subseteq G'$, and $g'_\ast (dx \wedge dy)|_G = dx\wedge dy$?
I tried many different strategies:
- change $g$ thorugh Moser's trick, but I cannot impose that $G$ is sent into $G'$
- manually "retract" the edges of $g(G)$, in order to change the volume form $g_\ast(dx \wedge dy)$. If I can extend this to "diffeomorphism mod the 1-scheletron", I can meet the conditions in the classical article of Munkres, to extend it to a diffeomorph of the plane. (Obstructions to the Smoothing of Piecewise-Differentiable Homeomorphism, Ann. Math)
- I can extend a continous isotopy of graphs to a family of omeomorphism of the plane. However, the obtained omeomorphism can be really bad, far from the condition of Munkres. This last approach was thought after reading the followin question: Extensions of non-smooth isotopies of not-submanifolds on surfaces
Can domebody give me a hint? It's also possible that the question has negative answer, or only a local one. The important part is that the edge fixed by $g$ is not touched.
