How to define thickening and right/left neighbourhood of a curve? (reference request)

Question

For a (piecewise) smooth (nonself intersecting) closed curve $C$ on an orientable surface $Q$

I want to define (not uniquely) $r>0$ and disjoint subsets $L,R$ of $Q$ "on each side" of $C$ such that the set $ \{ B(v,r): \ \ v \in C \} $ (where $B(v,r)$ is the open ball around $v$ of radius $r$ in $Q$ ) is the union of $L$ and $R$. And for any (piecewise) smooth curve $f:[0,1] \rightarrow Q$ such that $f(x) \notin C$ for any $x \in [0,1)$, $f(1) \in C$ satisfies that for some $ \beta \in (0,1) $ either the image of the open interval $(\beta,1)$ $f( (\beta,1) ) $ lies in $L $, that is the curve "reaches" $C$ from the left $L$ or $f( (\beta,1) ) \in R $, that is the curve reaches $C$ from the right $R$.

What's the proper/rigorous way to define this? references appreciated.

Also want to state that I can continuously deform the curve $C$ so that it lies inside $L$ in a rigorous way i.e. shift the curve to the left a bit.

Also want that any (piecewise) smooth curve $\phi :[0,1] -> Q $ such that the image of the open interval $\phi((0,1)) $ is disjoint from $ C$, $\phi(1-t) $ reaches $C$ from the left and $\phi(t)$ reaches $C$ from the right, that is,

for some $0<\beta_1 < \beta_2 <1 $, the image of the open interval $(0, \beta_1)$, $\phi((0, \beta_1)) $ lies in $L$ and the image of the open interval $( \beta_2, 1)$, $\phi(( \beta_2, 1)) $ lies in $R$,

$\phi$ cannot be contained in a region of $Q$ homeomorphic to an open disk.

Answer 1

You have no chance to do this with arbitrary continuous maps, since they can be wild. The best setting are smooth embeddings (of an interval into a manifold). For those you can define tubular neighbourhoods. Smooth immersions are locally embeddings, so for those you can define your neighbourhood at least locally and glue them together with a partition of unity. Tubular neighbourhoods are exactly the geometric image that you have in your head. Plus, a tubular neighbourhood comes with a diffeomorphism from the tubular neighbourhood to a neighbourhood of the zero-section in the normal bundle, i.e. you have a nice way to describe points in the neighbourhood.

See ncatlab.org/nlab/show/tubular+neighborhood+theorem for the precise statement. It is more intuitive when you work with Riemannian manifolds. You can then define the normal bundle of S⊂M to be the orthogonal complement of TS in TM. Good references are Lee's book on smooth manifolds and Lee's book on Riemannian manifolds. I have heard somewhere that tubular neighbourhoods are essentially unique, but I do not know a good reference for this fact.

Answer 2

A further complication that you might wish to consider as a counter-example to your conjecture: note that a parametrized curve that travels "forward" and then "backward" along a line segment in the plane can be technically considered to be a piece-wise smooth parametrized curve.

Thus you may wish to impose a stronger hypothesis: that the closed curve is regular in the sense that it admits a continuously differentiable parametrization whose derivative (velocity vector) is always nonzero. The curve can then be parametrized by arc length. In this case, the associated velocity vector is a continuously varying unit-length tangent vector $T$. Then select an associated continuously-varying oriented normal $N$. That $N$ will provide the direction you seek for defining your one-sided region.

Perhaps the introductory chapters of Manfredo DoCarmo's classic textbook on differential geometry will have the details that you seek. [Differential Geometry of Curves and Surfaces: Revised and Updated Second Edition]