Curves in the plane and their number of holes

Question

Suppose that the closed, piecewise $C^1$-curve $f(\mathbb T)$ has exactly $n$ points that are run through twice, all other points are run through once. Is it true that the compact set $f(\mathbb T)$ has exactly $n+1$ holes? For $n=1$ this is true and is based on the well-known fact that a crosscut in a Jordan domain determines two Jordan subdomains (Proposition 2.12 in Pommerenke's book). More generally, if for a finite decomposition $ \mathbb T=\cup_{j=1}^n I_j$ of $\mathbb T$ into closed arcs with pairwise disjoint interiors, $f$ is injective on the interior of each $I_j$, what is the maximum number of holes of the closed set $f(\mathbb T)$?

Answer 1

Here's a slightly more hands-on point of view, using just the Jordan curve theorem (although of course in the end it does come down to the same thing as Euler's formula somehow, as described by Alex.)

Take a closed curve $\gamma\colon [0,1]\to S^2$, and assume that there are only finitely many pairs of points in $[0,1]$ that have the same image under $\gamma$. (Here $S^2$ is the sphere, i.e. the one-point compactification of the plane.)

Now define a finite sequence $(x_n)_{n=0}^{N}$ in $[0,1]$ by letting $x_0 = 0$ and letting $x_{n+1}>x_n$ be minimal with $\gamma(x_{n+1})\in \gamma([0,x_{n+1}))$. By assumption, this yields a finite sequence with $x_N = 1$.

Now, I claim that $\gamma([0,x_n])$ separates the sphere $S^2$ (compactification of the plane) into $n+1$ connected components, which are all simply-connected (hence topologically equivalent to the plane). This is trivial for $n=0$. If true for $n<N$, then it follows for $n+1$. Indeed, $\gamma( (x_n,x_{n+1}) )$ lies in some complementary component $U$ of $\gamma([0,x_n])$. Either this curve is injective, in which case it tends to $\partial U$ in both directions. Under the homeomorphism of $U$ to the plane, it is thus a Jordan curve in the plane, and separates $U$ in exactly two components.

If the curve is not injective, then there is a point $x'_n \in (x_n,x_{n+1})$ with $\gamma(x'_n) = \gamma(x_{n+1})$. Then $\gamma( [x_n', x_{n+1}])$ is a Jordan curve in $U$, and $\gamma([x_n,x_n'])$ is an arc connecting a point of $\partial U$ to this Jordan curve. Again, we see that $\gamma(x_n,x_{n+1})$ separates $U$ into two components, and the total number of components increases by $1$.

So we have proved that $\gamma$ separates the plane into $N+1$ components. In your setting, we easily see that $N=n+1$ (there is one $x_n$ for each of your multiple points, plus $x_0 = 0$ and $x_{n+1}=1$). So, there are n+2 components, or n+1 "holes" in your terminology.

Nb. You will notice that this is essentially the same argument as in the proof of Euler's formula: adding a new edge with one existing and one new vertex does not create new faces, while adding a an edge between two existing vertices DOES create an extra face.

Answer 2

Your assumption, if I understood it correctly is that the curve has only double self-crossings.

Consider your curve on the sphere (plane with infinite point added). It defines a cell decomposition of the sphere into vertices, edges and faces. Their numbers satisfy the Euler formula: $$v-e+f=2$$ Your assumption implies that each vertex is of degree $4$, so we must have $e=4v/2=2v$. Substituting this to Euler's formula, we obtain $f=v+2$, and since your "holes" are bounded faces, and there is one unbounded face, the number of holes is $v+1$.