Assume that a connected subset $\Gamma\subset \mathbb{R}^2$ can be written as the following $$ \Gamma=\bigcup \limits_{i=1}^m S_i=\bigcup \limits_{j=1}^k \tilde S_j $$ such that each $S_i$ (resp. $\tilde S_j$) is homeomorphic to $\mathbb{S}^1$ and $S_i\cap S_{i'}$ (resp. $\tilde S_j\cap \tilde S_{j'}$) is the empty set or a finite set for any $i\not =i'$ (resp. $j\not =j'$). $\Gamma$ can be seen as a planar graph in the obvious way, where the vertices are the points that appear as intersection of two different $S_i$'s (or $\tilde S_j$'s).
So, I have two questions:
1) What is the relation between $b_1(\Gamma)$ (the first Betti number of $\Gamma$) and $m$?
2) Is it true that $m\equiv k$ (mod 2)?
Edit 1: Please consider the above questions in the case when $\Gamma \subset \mathbb S^2$ and $a(\Gamma)=\Gamma$, where $a\colon \mathbb S^2\to \mathbb S^2$ is the antipodal map (i.e., $a(x)=-x$).
Consider three mutually externally tangent circles. (Take an equilateral triangle of side-length 2, and about each vertex draw a circle of radius 1; throw away the triangle.) In this case $m = 3$.
Using the tangent points as "dividers", each of these three circles is split into two halves -- an inner and an outer half. The union of the three outer halves looks like a circle; so does the union of the three inner halves. This gives a decomposition of the same connected subset as a union of $k = 2$ circles. So the answer to question 2 is "no".