Let $X$ be the union of circles $C_n$ with center at $(n,0)$ and radius $1/2$ for $n \in \mathbb{N}$. I want to show that the fundamental group of $X$ is a free group with countable generators. I think that if $Y=X \cap \{(x,y) \in \mathbb{R}^2: y\geq 0\}$, then $X/Y$ is a wedge sum of circles and $\pi(X)$ is isomorphic to $\pi(X/Y)$. But I don't know how can I show that.
Can you help me, please?
$X$ (with the subspace topology in $\mathbb{R}^2$) can be given the structure of a CW complex. The $0$-cells are $(\mathbb{R}\times\{0\})\cap X$, and the $1$-cells are what is remaining.
The set $Y$ that you give is a CW subcomplex of $X$. Page 11 of Hatcher's Algebraic Topology says that then $X\to X/A$ is a homotopy equivalence since $A$ is contractible. As a CW complex, $X/A$ is a wedge sum of countably many circles. One can see this by realizing that the quotient has a single $0$-cell, and so all the $1$-cells must each contribute a circle wedge summand.
It is a general fact that the fundamental group of a graph (i.e., a CW $1$-complex) has generators in correspondence to the edges in the complement of a spanning tree, which is exactly what $Y$ happens to be. Section 1.A in Hatcher's book has more details.