Here is the original problem:
Exercise
Let $X ⊂ R^2$ be a connected graph that is the union of a finite number of straight line segments. Show that $π_1 (X)$ is free with a basis consisting of loops formed by the boundaries of the bounded complementary regions of $X$, joined to a base point by suitably chosen paths in $X$. [Assume the Jordan curve theorem for polygonal simple closed curves, which is equivalent to the case that $X$ is homeomorphic to $S_1$.]
I want to do this by using the following proposition:
Proposition :
For a connected graph $X$ with maximal tree $T$ , $π_1 (X)$ is a free
group with basis the classes $[f_α ]$ corresponding to the edges $e_α$ of $X − T$ .
By using this proposition and Jordan curve theorem I can deduce that $π_1 (X)$ is free with a basis consisting of loops formed by the boundaries of the bounded but may not complementary regions of $X$. Simply because when I choose different edge $e_\alpha$, and get closed curves respectively, I cannot promise that their interior are complementary.
So is it possible for me to finish my proof by
- Choosing a special maximal tree such that the regions are complementary.
- Or from the basis I have got to compute the basis I want (the boundaries of those complementary regions).
- Or there is another clear way to prove this or do this exercise.
Edit: To clarify why I got stuck, here is a simple example:
If I choose $\{BA,AC,CD\}$ as the maximal tree, then by adding $BC$ we get a region $ABC$, by adding $BD$ we get a region $ACDB$, the rectangle. Their boundaries represent the basis we get by using the proposition above. However, the corresponding two regions are not complementary.
Anyway, in this situation, it is possible to do some modifying.
For example, if the maximal tree we choose is $\{AB,BC,CD\}$, I'm very glad to see that $ABC$ and $BCD$ are what we want. However, I don't know how to apply this method to a general situation. In fact, I find the special maximal tree totally from intuition.